text:
\begin{document}

\begin{frontmatter}

\title{Fuzzy $\alpha$-cut and related structures}

\author[mymainaddress]{Purbita Jana\corref{mycorrespondingauthor}}
\cortext[mycorrespondingauthor]{Corresponding author}
\ead{purbita\_presi@yahoo.co.in}

\author[mysecondaryaddress]{Mihir .K. Chakraborty}
\ead{mihirc4@gmail.com}
\fntext[myfootnote]{The corresponding author acknowledge Department of Science $\&$ Technology, Government of India for financial support vide reference no. SR/WOS-A/PM-1010/2014 under Women Scientist Scheme to carry out this work.}

\address[mymainaddress]{Department of Pure Mathematics, University of Calcutta}
\address[mysecondaryaddress]{School of Cognitive Science, Jadavpur University}

\begin{abstract}
This paper deals with a new notion called fuzzy $\alpha$-cut and its properties. A notion called localic frame is also introduced. Algebraic structures arising out of the family of fuzzy $\alpha$-cuts have been investigated. It will be seen that this family forms a localic frame. Some significance and usefulness of fuzzy $\alpha$-cuts are discussed.
\end{abstract}

\begin{keyword}
$L$-fuzzy set, $\alpha$-cut, Fuzzy $\alpha$-cut, Frame, Graded frame, G$\ddot{o}$del arrow.
\end{keyword}

\end{frontmatter}

\section{Introduction}
Though well known now-a-days, we would like to start with a little bit of history. Fuzzy set was first introduced and studied by Lotfi Zadeh \cite{LZ} in 1965, which can be considered in a sense a generalisation of ordinary set. It is well known that in informal set theory a (crisp) set $A$ is considered as a subset of a universal set $U$ and is fully determined by a function from $U$ to $\{0,1\}$ called the characteristic function of $A$ (denoted by $\chi_A$). Whereas a fuzzy set is a function from $U$ to $[0,1]$, and in this case the function is known as membership function. In 1967, J. Goguen \cite{GJ} generalised this notion one step further by considering the function from $U$ to $L$ (a complete lattice) and called it $L$-fuzzy set. Subsequently, there had been many other generalisations of the original proposal of Zadeh \cite{AK1, BW, CE}. In this paper we will consider $L$ as a frame (c.f. Definition \ref{fm}), $1_L$ and $0_L$ being the top and the bottom elements respectively.

 In 1971, Zadeh proposed a representation theorem of  fuzzy sets using the notion of $\alpha$-cuts (c.f. Definition \ref{1'} considering $L$ as [0,1]), known as first decomposition theorem \cite{KY} in the literature. $\alpha$-cuts of a fuzzy set are crisp sets. In this paper we delve into the notion of fuzzy $\alpha$-cut (c.f. Definition \ref{2'}), which was introduced by the present authors in \cite{MP2}. Recently we have noticed that the notion of fuzzy $\alpha$-cut exists in the literature as `level fuzzy sets' introduced in \cite{TR}. In this regard the authors are grateful to the editor of this journal for his valuable advise on an earlier version of this paper.  A fuzzy $\alpha$-cut of a fuzzy set gives a fuzzy subset of the given fuzzy set. We will see that fuzzy $\alpha$-cut of a fuzzy mathematical structure is a fuzzy mathematical substructure.

In this paper we have proved that a family of fuzzy $\alpha$-cuts over a frame  forms a localic frame. Consequently they generate a model of so called fuzzy geometric logic with graded consequence \cite{MP3}. The notion of graded frame and fuzzy geometric logic with graded consequence was introduced in \cite{MP3}. Such an algebraic structure and a logic were proposed to serve the purpose of giving an answer to the question-``From which logic fuzzy topology can be generated?" This question came up parallel to a similar idea provided in Vickers's book \cite{SV} ``Topology via logic", viz. from which logic classical topology can be generated?. It is to be noted that as an answer, fuzzy geometric logic was invented and as a further generalisation fuzzy geometric logic with graded consequence was introduced. Notion of graded frame came into the picture as Lindenbaum type algebra of fuzzy geometric logic with graded consequence \cite{MP3}. The notion of localic frame (c.f. Definition \ref{gfl}) is introduced here which is a further generalisation of graded frame by taking a frame-valued binary relation instead of $[0,1]$ -valued binary relation as $[0,1]$ is a particular frame.

As for usefulness and significance of the notion of fuzzy $\alpha$-cuts, we claim that they give natural substructures of fuzzy topological spaces and fuzzy algebraic structures. Classical fuzzy topological spaces are defined by taking a crisp set and fuzzy open sets. There are two major streams of research, one following Chang's definition \cite{CL} and the other following Lowen's definition \cite{RL}. In both cases, subspaces are defined on crisp subsets of the base set. On the other hand in case of fuzzy algebraic structures (e.g. Rosenfeld \cite{AR}), one starts with a classical algebraic structure and defines fuzzy substructures. In neither construction topological or algebraic, the starting base set is taken to be fuzzy. While in the topological case, the base set as well as all the subspaces are to be taken crisp, in the algebraic case though the sub algebras are fuzzy, one has to begin with a classical crisp algebraic structure.

In \cite{MA} and \cite{MB} there had been proposals to develop both the kinds of structures on fuzzy sets. Besides, in \cite{MB} the algebraic compositions also have been fuzzy right from the start. Fuzzy $\alpha$-cuts being fuzzy subsets of a fuzzy set, using the above proposals it would be possible to define topological and algebraic substructures on them. These will be quite natural substructures. We shall present these constructions in section 4.

Secondly, fuzzy $\alpha$-cuts will provide fuzzy sets as lower and upper approximations in the probabilistic rough set framework \cite{YY}. We expects these kinds of approximation will be useful in the domain of application of rough set theory \cite{PW}.

This paper is organised as follows. Section 2 emphasises upon the properties of fuzzy $\alpha$-cut. In section 3, various algebraic structures and a notion of G$\ddot{o}$del-like arrow along with its properties are discussed. Algebraic structures formed by the family of fuzzy $\alpha$-cuts are also studied in this section. The significance of fuzzy $\alpha$-cuts is provided in section 4. Section 5 presents some concluding remarks.

We give below some preliminary definitions that would be required in the sequel.

\begin{definition}[Frame]\label{fm}
A \textbf{frame} \index{frame} is a complete lattice such that, $$x\wedge\bigvee Y=\bigvee\{x\wedge y :y\in Y\}.$$
i.e., the binary meet distributes over arbitrary join.
\end{definition}
\begin{definition}[$\alpha$-cut of a fuzzy set]\label{1'}
Let $(X,\tilde{A})$ be an $L$-fuzzy set, where $X$ is the base set and $L$ is a frame. Then for $\alpha \in L$, the \textbf{$\alpha$-cut} of
$(X,\tilde{A})$ is the ordinary set $\{ x\in X\mid \tilde{A}(x)\geq \alpha \}$ and is denoted by $\alpha_{\tilde{A}}$.
\end{definition}

\begin{definition}[G$\ddot{o}$del arrow]\cite{KY}
G$\ddot{o}$del arrow is defined as follows:
\begin{center}
$a\rightarrow b$ $=\begin{cases}
        1 & \emph{if $a\leq b$} \\
        b & \emph{if $a>b$}.
    \end{cases}$
    \end{center}

for all $a,b\in[0,1]$.
\end{definition}
G$\ddot{o}$del arrow can be generalised to the following.
\begin{definition}[G$\ddot{o}$del-like arrow]\label{gga}
Let $L$ be any frame. Then the G$\ddot{o}$del-like arrow is defined as follows:\\
\begin{center}
$a\rightarrow b$ $=\begin{cases}
        1_L & \emph{if $a\leq b$} \\
        b & \emph{otherwise}.
    \end{cases}$
\end{center}
for all $a,b\in L$.

\end{definition}
There is another kind of implication in $L$ called residuated implication defined as below.
\begin{definition}[Residuated implication]\label{gha}
Let $L$ be any any frame. Then the residuated implication is defined by $a\rightarrow b=sup\{c\in L\mid c\wedge a\leq b\}$ for all $a,b\in L$.
\end{definition}
The relationship between these two types of implications is discussed in subsection \ref{pga}.

\section{Fuzzy $\alpha$-cut and its properties}
In this section we will define the notion of fuzzy $\alpha$-cut and provide some of the algebraic properties of fuzzy $\alpha$-cut. For the corresponding classical notion we refer to \cite{KY}.

\begin{definition}[Fuzzy $\alpha$-cut of a fuzzy set]\label{2'}\cite{MP2}
Let $(X,\tilde{A})$ be an $L$-fuzzy set. Then for $\alpha \in L$, the \textbf{fuzzy $\alpha$-cut} of
$(X,\tilde{A})$ is the fuzzy subset $(X,\leftidx{^{\tilde{A}}}{\alpha}{})$ where $\leftidx{^{\tilde{A}}}{\alpha}{}$ is defined as follows:
\begin{center}
$\leftidx{^{\tilde{A}}}{\alpha}{}(x)$ $=\begin{cases}
        \tilde{A}(x) & \emph{if $\tilde{A}(x)\geq \alpha $} \\
        0_L & \emph{otherwise}.\end{cases}$
\end{center}
We will denote fuzzy $\alpha$-cut of an $L$-fuzzy set $(X,\tilde{A})$ simply by $\leftidx{^{\tilde{A}}}{\alpha}{}$ if the base set $X$ is understood.
\end{definition}
It is to be noted that fuzzy $\alpha$-cut of a fuzzy set is also known as level fuzzy set \cite{TR}. As mentioned in the introduction, present authors were not aware of this paper published long back in 1977 and not used frequently in subsequent literature. In the paper \cite{TR} the author defined the algebraic operations viz. intersection, union, complementation of level sets as is done in fuzzy set theory by min, max and $1-(\cdot)$, in the value set $[0,1]$ and established certain elementary properties. In our paper, however, these operations are presumed since these are none else than the corresponding operations of fuzzy subsets. We have rather proved some non-trivial results in this section where the value set $L$ is taken to be a frame.
\begin{example}
Consider the fuzzy set $\tilde {A}$ defined on the interval $X=[0,10]$ of real numbers by the membership function $\tilde{A}(x)=\frac{x}{x+2}$. Then
\begin{center}
$\leftidx{^{\tilde{A}}}{0.2}{}(x)$ $=\begin{cases}
        \frac{x}{x+2} & \emph{for $x\in [0.5,10] $} \\
        0 & \emph{otherwise}.\end{cases}$
\end{center}
\end{example}
Let $(X,\tilde{A})$, $(X,\tilde{B})$ be two $L$-fuzzy sets. Then $\tilde{A}\subseteq \tilde{B}$ if and only if $\tilde{A}(x)\leq \tilde{B}(x)$, for any $x\in X$.

 As we are dealing with $L$-fuzzy sets, we are considering the generalised version of the definition of $\alpha$-cut and fuzzy $\alpha$-cut by generalising the value set $[0,1]$ to a frame $L$.

\begin{theorem} Let $(X,\tilde{A})$, $(X,\tilde{B})$ be two fuzzy sets. Then for any $\alpha$, $\alpha_1$, $\alpha_2\in L$ the following properties hold:
1. $\leftidx{^{\tilde{A}}}{\alpha}{}\subseteq \chi_{\alpha_{\tilde{A}}}$; 2. $\alpha_1\leq \alpha_2$ implies $\leftidx{^{\tilde{A}}}{\alpha_1}{}\supseteq \leftidx{^{\tilde{A}}}{\alpha_2}{}$; \ \ 3.  $\leftidx{^{(\tilde{A}\cap\tilde{B})}}{\alpha}{}=\leftidx{^{\tilde{A}}}{\alpha}{}\cap \leftidx{^{\tilde{B}}}{\alpha}{}$ and $\leftidx{^{(\tilde{A}\cup\tilde{B})}}{\alpha}{}=\leftidx{^{\tilde{A}}}{\alpha}{}\cup \leftidx{^{\tilde{B}}}{\alpha}{}$.

\end{theorem}
\begin{proof} We demonstrate the proof of 3 only. For the first part of 3 we proceed as follows:

\begin{align*}
\leftidx{^{(\tilde{A}\cap\tilde{B})}}{\alpha}{}(x) & =\begin{cases}
        (\tilde{A}\cap\tilde{B})(x) & \emph{if $(\tilde{A}\cap\tilde{B})(x)\geq \alpha $} \\
        0_L & \emph{otherwise}.\end{cases}\\
        & =\begin{cases}
        \tilde{A}(x)\wedge\tilde{B}(x) & \emph{if $\tilde{A}(x)\wedge\tilde{B}(x)\geq \alpha $} \\
        0_L & \emph{otherwise}.\end{cases}\\
        & =\begin{cases}
        \tilde{A}(x)\wedge\tilde{B}(x) & \emph{if $\tilde{A}(x)\geq \alpha$ and $\tilde{B}(x)\geq \alpha $} \\
        0_L & \emph{otherwise}.\end{cases}\\
        & =\begin{cases}
        \tilde{A}(x)\wedge\tilde{B}(x) & \emph{if $\tilde{A}(x)\geq \alpha$ and $\tilde{B}(x)\geq \alpha $} \\
        0_L & \emph{if $\tilde{A}(x)\geq \alpha$ and $\tilde{B}(x)< \alpha $} \\
        0_L & \emph{if $\tilde{A}(x)< \alpha$ and $\tilde{B}(x)\geq \alpha $} \\
        0_L & \emph{if $\tilde{A}(x)< \alpha$ and $\tilde{B}(x)< \alpha $}.\end{cases}\\
        & =\begin{cases}
        \tilde{A}(x)\wedge\tilde{B}(x) & \emph{if $\tilde{A}(x)\geq \alpha$ and $\tilde{B}(x)\geq \alpha $} \\
        \tilde{A}(x)\wedge 0_L & \emph{if $\tilde{A}(x)\geq \alpha$ and $\tilde{B}(x)< \alpha $} \\
        0_L\wedge\tilde{B}(x) & \emph{if $\tilde{A}(x)< \alpha$ and $\tilde{B}(x)\geq \alpha $} \\
        0_L & \emph{if $\tilde{A}(x)< \alpha$ and $\tilde{B}(x)< \alpha $}.\end{cases}\\
 & = \leftidx{^{\tilde{A}}}{\alpha}{}(x)\wedge\leftidx{^{\tilde{B}}}{\alpha}{}(x)
       =(\leftidx{^{\tilde{A}}}{\alpha}{}\cap\leftidx{^{\tilde{B}}}{\alpha}{})(x).
\end{align*}

Similarly the second equality holds.
\end{proof}
For any mapping $f:X\longrightarrow Y$, the image of the fuzzy subset $(X,\tilde{A})$ of $X$ is the fuzzy subset $(Y,f(\tilde{A}))$ and defined by \cite{KY}
$$f(\tilde{A})(y)=\bigvee_{x\in X}\{\tilde{A}(x)\mid y=f(x)\}.$$
Thus $f(\leftidx{^{\tilde{A}}}{\alpha}{})$ gives the fuzzy subset $(Y,\leftidx{^{\tilde{A}}}{\alpha}{})$ of $Y$. We now have the following theorem.
 \begin{theorem}
 Let $f:X\longrightarrow Y$ be a mapping. Then for any $L$-fuzzy set $(X,\tilde{A})$ and $\alpha\in L$, $f(\leftidx{^{\tilde{A}}}{\alpha}{})= \leftidx{^{(f(\tilde{A}))}}{\alpha}{}$.
 \end{theorem}
 \begin{proof}
 For any $y\in Y$, we have the following:
 \begin{align*}
 f(\leftidx{^{\tilde{A}}}{\alpha}{})(y)  & =\bigvee_x\{\leftidx{^{\tilde{A}}}{\alpha}{}(x)\mid y=f(x)\} \\
& =\begin{cases}
        0_L & \emph{if $\tilde{A}(x)< \alpha $ for all $x\in X$} \\
        \bigvee_x\{\tilde{A}(x)\mid y=f(x)\} & \emph{otherwise}.\end{cases}\\
        & =\begin{cases}
        \bigvee_x\{\tilde{A}(x)\mid y=f(x)\} & \emph{if $\bigvee_x\{\tilde{A}(x)\mid y=f(x)\}\geq \alpha $} \\
        0_L & \emph{otherwise}.\end{cases}\\
         & =\begin{cases}
       (f(\tilde{A}))(y) & \emph{if $(f(\tilde{A}))(y)\geq \alpha $} \\
        0_L & \emph{otherwise}.\end{cases}\\
        & =\leftidx{^{(f(\tilde{A}))}}{\alpha}{}(y).
 \end{align*}
 Hence $f(\leftidx{^{\tilde{A}}}{\alpha}{})= \leftidx{^{(f(\tilde{A}))}}{\alpha}{}$.
 \end{proof}
 \begin{proposition}
 Let $f:X\longrightarrow Y$ be a mapping. Then for any $L$-fuzzy set $(X,\tilde{A})$ and $\alpha, \beta\in L$, $\alpha\leq \beta\Rightarrow f(\leftidx{^{\tilde{A}}}{\alpha}{})\subseteq f(\leftidx{^{\tilde{A}}}{\beta}{})$.
 \end{proposition}
 \begin{proposition}
 Let $f:X\longrightarrow Y$ and $g:Y\longrightarrow Z$. Then for any $L$-fuzzy set $(X,\tilde{A})$ and $\alpha\in L$, $(g\circ f)(\leftidx{^{\tilde{A}}}{\alpha}{})=\leftidx{^{(g\circ f(\tilde{A}))}}{\alpha}{}$.
 \end{proposition}
 \begin{proof}
 $(g\circ f)(\leftidx{^{\tilde{A}}}{\alpha}{})=g(f(\leftidx{^{\tilde{A}}}{\alpha}{}))=g(\leftidx{^{(f(\tilde{A}))}}{\alpha}{})=\leftidx{^{(g(f(\tilde{A})))}}{\alpha}{}=\leftidx{^{(g\circ f(\tilde{A}))}}{\alpha}{}$.
\end{proof}
$(Z,(g\circ f)\leftidx{^{\tilde{A}}}{\alpha}{})$ is a fuzzy subset of $Z$.
\begin{proposition}
 Let $f:X\longrightarrow Y$, $g:Y\longrightarrow Z$ and $h:Z\longrightarrow W$. Then for any $L$-fuzzy set $(X,\tilde{A})$ and $\alpha\in L$, $(h\circ(g\circ f))(\leftidx{^{\tilde{A}}}{\alpha}{})=((h\circ g)\circ f)(\leftidx{^{\tilde{A}}}{\alpha}{})$.
 \end{proposition}
 \begin{proposition}
 Let $f:X\longrightarrow Y$ be a mapping and $id_X:X\longrightarrow X$ be the identity mapping. Then for any $L$-fuzzy set $(X,\tilde{A})$ and $\alpha\in L$, $(f\circ id_X)(\leftidx{^{\tilde{A}}}{\alpha}{})=f(\leftidx{^{\tilde{A}}}{\alpha}{})$.
 \end{proposition}
 \begin{proposition}
 Let $f:X\longrightarrow Y$ be a mapping and $id_Y:Y\longrightarrow Y$ be the identity mapping. Then for any $L$-fuzzy set $(X,\tilde{A})$ and $\alpha\in L$, $(id_Y\circ f)(\leftidx{^{\tilde{A}}}{\alpha}{})=f(\leftidx{^{\tilde{A}}}{\alpha}{})$.
 \end{proposition}
 \begin{proof}
 $\alpha\leq\beta\Rightarrow \leftidx{^{\tilde{A}}}{\alpha}{}\subseteq \leftidx{^{\tilde{A}}}{\beta}{}\Rightarrow f(\leftidx{^{\tilde{A}}}{\alpha}{})\subseteq f(\leftidx{^{\tilde{A}}}{\beta}{})$.
 \end{proof}
\section{Algebraic structure of the family of fuzzy $\alpha$-cuts}
We shall establish that the family of fuzzy $\alpha$-cuts forms a localic frame [c.f. Definition \ref{gfl}] with respect to an $L$-fuzzy relation $R$ defined in terms of the G$\ddot{o}$del-like arrow [c.f. Definition \ref{gga}] in $L$.

\begin{definition}[Localic Frame]\label{gfl}\index{graded frame}
A \textbf{localic frame} is a 5-tuple $(A,\top,\wedge,\bigvee,R_L)$, where $A$ is a non-empty set, $\top\in A$, $\wedge$ is a binary operation, $\bigvee$ is an operation on arbitrary subset of $A$, $R_L$ is an $L$-valued fuzzy binary relation on $A$ satisfying the following conditions:
\begin{enumerate}
\item $R_L(a,a)=1_L$ (fuzzy reflexivity);
\item $R_L(a,b)=1_L=R_L(b,a)\Rightarrow a=b$ (fuzzy antisymmetry);
\item $R_L(a,b)\wedge R_L(b,c)\leq R_L(a,c)$ (fuzzy transitivity);
\item $R_L(a\wedge b,a)=1_L=R_L(a\wedge b,b)$;
\item $R_L(a,\top)=1_L$;
\item $R_L(a,b)\wedge R_L(a,c)=R_L(a,b\wedge c)$;
\item $R_L(a,\bigvee S)=1_L$ if $a\in S$;
\item $inf\{R_L(a,b)\mid a\in S\}=R_L(\bigvee S,b)$;
\item $R_L(a\wedge\bigvee S,\bigvee\{a\wedge b\mid b\in S\})=1_L$;
\end{enumerate}
for any $a,b,c\in A$ and $S\subseteq A$. We will denote a localic frame by $(A,R_L)$.
\end{definition}
In particular, $(A,\top,\wedge,\bigvee,R_{[0,1]})$ is a graded frame \cite{MP3}. Thus localic frame is a generalisation of graded frame. It is to be noted that an algebraic structure satisfying the first three conditions of Definition \ref{gfl} is known as localic poset \cite{JA}. That is, a localic poset is a set endowed with fuzzy partial order relation. However, this is not the only definition of fuzzy partial order. For more general definitions see \cite{MS}.

\begin{theorem}\label{theorem4}
$(\{\leftidx{^{\tilde{A}}}{\alpha}{}\mid \alpha \in L\},\subseteq, \cap, \bigcup)$ is a frame.
\end{theorem}
\begin{proof}
Here we only show the distributive property i.e., $$\leftidx{^{\tilde{A}}}{\alpha}{}\cap\bigcup_i\leftidx{^{\tilde{A}}}{\alpha_i}{}=\bigcup_i(\leftidx{^{\tilde{A}}}{\alpha}{}\cap\leftidx{^{\tilde{A}}}{\alpha_i}{}).$$
\begin{align*}
(\leftidx{^{\tilde{A}}}{\alpha}{}\cap\bigcup_i\leftidx{^{\tilde{A}}}{\alpha_i}{})(x) & = \leftidx{^{\tilde{A}}}{\alpha}{}(x)\wedge((\bigcup_i\leftidx{^{\tilde{A}}}{\alpha_i}{})(x))\\
& = \leftidx{^{\tilde{A}}}{\alpha}{}(x)\wedge(\bigvee_i\leftidx{^{\tilde{A}}}{\alpha_i}{}(x))\\
& = \bigvee_i(\leftidx{^{\tilde{A}}}{\alpha}{}(x)\wedge\leftidx{^{\tilde{A}}}{\alpha_i}{}(x))\ \text{[as $L$ is a frame]} \\
& = \bigvee_i(\leftidx{^{\tilde{A}}}{\alpha}{}\cap\leftidx{^{\tilde{A}}}{\alpha_i}{})(x))\\
& = (\bigcup_i(\leftidx{^{\tilde{A}}}{\alpha}{}\cap\leftidx{^{\tilde{A}}}{\alpha_i}{}))(x).
\end{align*}
This completes the proof.
\end{proof}

\subsection{Prelinear and Semilinear Frame}\label{PSF}
In this subsection we will consider prelinear frame and semilinear frame. This subsection includes detailed study of the above mentioned notions with examples.
\begin{definition}[Prelinear Frame]\label{pfm}\cite{RB}
A frame $L$ together with a binary operation $\rightarrow$ is said to be a prelinear if for each $l_1,l_2\in L$, $(l_1\rightarrow l_2)\vee (l_2\rightarrow l_1)=\top$, where $\top$ is the top element of $L$.
\end{definition}
For our purpose we will take $\rightarrow$ as G$\ddot{o}$del-like arrow. In this section henceforth all the arrows are G$\ddot{o}$del-like arrow.

\underline{\textbf{Note:}} For the notion of prelinearity in more general set up we refer
to \cite{PH}. Whenever there is an $\rightarrow$ satisfying the property $l_1\leq l_2$
implies $l_1\rightarrow l_2=\top$, linearity of the order implies prelinearity. Our purpose here will be served under the assumption of a notion more general than prelinearity viz. semilinearity (c.f. Definition \ref{sfm}). The purpose is to generalize the notion of graded frame \cite{MP3}  where the value set is taken as $[0,1]$.

\begin{definition}[Semilinear Frame]\label{sfm}
A semilinear frame $L=(L,\wedge,\bigvee,\rightarrow)$ is a frame $(L,\wedge,\bigvee)$ together with a binary operation $\rightarrow$ such that for all $l_1,l_2,l_3\in L$, $(l_1\rightarrow l_2)\wedge (l_1\rightarrow l_3)= (l_1\rightarrow l_2\wedge l_3)$.
\end{definition}
It can be verified by considering all possible cases that any frame with up to 4-elements is always preilinear.
We give below an example of a 5-element lattice which is the smallest semilinear but not prelinear frame.
\begin{example}\label{snp}
The following frame is not prelinear but semilinear.
\begin{center}
\begin{tikzpicture}
  \matrix (galois)
     [matrix of nodes,
      nodes in empty cells,
      nodes={outer sep=0pt,circle,minimum size=4pt},
      column sep={1cm,between origins},
      row sep={1cm,between origins}]
   {
    && $\top$ &&\\
    &  & a & \\
    & b &  & c\\

    && $\bot$ &&\\
   };

      \draw (galois-1-3) -- (galois-2-3);

      \draw (galois-2-3) -- (galois-3-2);
      \draw (galois-2-3) -- (galois-3-4);

       \draw (galois-3-2) -- (galois-4-3);

       \draw (galois-3-4) -- (galois-4-3);

\end{tikzpicture}
\end{center}
For this frame $(b\rightarrow c)\vee (c\rightarrow b)=c\vee b=a\neq\top$. Hence it is not prelinear.
\end{example}

The following is an example of a frame with six elements which is not semilinear.
It is to be noted that the following frame is the smallest non semilinear frame. In other words a non-semilinear distributive lattice contains at least six elements.

\begin{example}
The following frame is not semilinear.
\begin{center}
\begin{tikzpicture}
  \matrix (galois)
     [matrix of nodes,
      nodes in empty cells,
      nodes={outer sep=0pt,circle,minimum size=4pt},
      column sep={1cm,between origins},
      row sep={1cm,between origins}]
   {
    && $\top$ &&\\
    & a &  & d\\
     &  & b & c\\

    && $\bot$ &&\\
   };

     \draw (galois-1-3) -- (galois-2-2);

     \draw (galois-1-3) -- (galois-2-4);

      \draw (galois-2-2) -- (galois-3-3);

       \draw (galois-2-4) -- (galois-3-3);
      \draw (galois-2-4) -- (galois-3-4);
       \draw (galois-2-2) -- (galois-4-3);
      \draw (galois-3-3) -- (galois-4-3);
       \draw (galois-3-4) -- (galois-4-3);

\end{tikzpicture}
\end{center}
For this frame $(b\rightarrow a)\wedge (b\rightarrow c)=\top\wedge c=c$, whereas $b\rightarrow (a\wedge c)=b\rightarrow \bot=\bot$. So, Property \ref{prop5} fails.
\end{example}
\begin{example}
The following frame is the smallest Boolean algebra which is not semilinear.
\begin{center}
\begin{tikzpicture}
  \matrix (galois)
     [matrix of nodes,
      nodes in empty cells,
      nodes={outer sep=0pt,circle,minimum size=4pt},
      column sep={1cm,between origins},
      row sep={1cm,between origins}]
   {
    && $\top$ &&\\
    & d & e & f\\
     & a & b & c\\

    && $\bot$ &&\\
   };

     \draw (galois-1-3) -- (galois-2-2);
      \draw (galois-1-3) -- (galois-2-3);
     \draw (galois-1-3) -- (galois-2-4);

     \draw (galois-2-2) -- (galois-3-2);
      \draw (galois-2-2) -- (galois-3-3);
      \draw (galois-2-3) -- (galois-3-2);
      \draw (galois-2-3) -- (galois-3-4);
       \draw (galois-2-4) -- (galois-3-3);
      \draw (galois-2-4) -- (galois-3-4);
       \draw (galois-3-2) -- (galois-4-3);
      \draw (galois-3-3) -- (galois-4-3);
       \draw (galois-3-4) -- (galois-4-3);

\end{tikzpicture}
\end{center}
For this frame $(a\rightarrow c)\wedge (a\rightarrow d)=c\wedge \top=c$, whereas $a\rightarrow (c\wedge d)=a\rightarrow \bot=\bot$. So, Property \ref{prop5} fails.
\end{example}

One can see that the concepts of prelinearity and semilinearity are based on the underlying lattice of the frame which is distributive. While prelinearity is an well known concept, semilinearity is not so and which is a more general concept [c.f. Property \ref{prop5'}]

Before proceeding to the next theorem let us enlist below some properties of G$\ddot{o}$del-like arrow \cite{KY} that would be used in the sequel.
\subsection{Properties of G$\ddot{o}$del-like arrow}\label{pga}
In this subsection some required properties of G$\ddot{o}$del-like arrow are listed along with verification of some of them.
\begin{property}\label{prop1}
$a\rightarrow a=1_L$, for any $a\in L$.
\end{property}

\begin{property}\label{prop2}
$(a\rightarrow b)\wedge (b\rightarrow c)\leq (a\rightarrow c)$, for any $a,b,c\in L$.
\end{property}
\begin{proof}
It may be observed that the values of $a\rightarrow b$ is either $1_L$ or $b$. Similarly for $b\rightarrow c$ the values are either $1_L$ or $c$ and for $a\rightarrow c$ values are either $1_L$ or $c$. Now the following cases may arise:\\
Case 1: $a\rightarrow b=1_L$ and $b\rightarrow c=1_L$.\\
Here $a\leq b$ and $b\leq c$ and hence as $L$ is transitive, $a\leq c$. Consequently $a\rightarrow c=1_L$. Therefore $(a\rightarrow b)\wedge (b\rightarrow c)=1_L\wedge 1_L=1_L=(a\rightarrow c)$.\\
Case 2: $a\rightarrow b=1_L$ and $b\rightarrow c=c$.\\
We have $(a\rightarrow b)\wedge (b\rightarrow c)=1_L\wedge c=c\leq c\ (or\ 1_L)=(a\rightarrow c)$.\\
Case 3: $a\rightarrow b=b$ and $b\rightarrow c=1_L$.\\
As $b\leq c$,  $(a\rightarrow b)\wedge (b\rightarrow c)=b\wedge 1_L=b\leq c\ (or\ 1_L)=(a\rightarrow c)$.\\
Case 4: $a\rightarrow b=b$ and $b\rightarrow c=c$.\\
In this case $(a\rightarrow b)\wedge (b\rightarrow c)=b\wedge c=c\leq c\ (or\ 1_L)=(a\rightarrow c)$.
\end{proof}
\begin{property}\label{prop3}
 $a\leq b$ implies $(a\rightarrow x)\geq (b\rightarrow x)$, for any $a,b,x\in L$.
 \end{property}

\begin{property}\label{prop4}
$a\leq b$ implies $(x\rightarrow a)\leq (x\rightarrow b)$, for any $a,b,x\in L$.
\end{property}

\begin{property}\label{prop5'}
If $L$ is prelinear then it is semilinear.
\end{property}
\begin{proof}
If possible let  $L$ is prelinear i.e., $(a\rightarrow b)\vee (b\rightarrow a)=\top$, for any $a,b\in L$ and for some $a,b,c\in L$, $(a\rightarrow b)\wedge (a\rightarrow c)\neq a\rightarrow (b\wedge c)$. Then two cases may arise. Case 1: $a< b$, $(a,c)$ and $(b,c)$ are incomparable [where $(a,b)$ represents the pair of points from $L$]. Case 2: $a< c$, $(a,b)$ and $(b,c)$ are incomparable.

Case 1: In this case notice that $a\vee c=\top$ as $(c\rightarrow a)\vee (a\rightarrow c)=\top$ and $a,c$ are incomparable. Hence $b\wedge(c\vee a)=b\wedge \top=b$.
Now $(b\wedge c)\vee (b\wedge a)=(b\wedge c)\vee a$. The following three cases may arise under this situation. Either $b\wedge c\leq a$ or $b\wedge c>a$ or the pair $(b\wedge c,a)$ is incomparable. If $b\wedge c\leq a$, then $(b\wedge c)\vee a=a\neq b$. When $b\wedge c>a$ then $(b\wedge c)\vee a=b\wedge c\neq b$ as if $b\wedge c=b$ then $b\leq c$, a contradiction. As $L$ is prelinear $(a\rightarrow(b\wedge c))\vee ((b\wedge c)\rightarrow a)=\top$. When the pair $(b\wedge c,a)$ is incomparable then $(a\rightarrow(b\wedge c))\vee ((b\wedge c)\rightarrow a)=(b\wedge c)\vee a=\top\neq b$ (as $b$ and $c$ are incomparable).

Hence for either cases $b\wedge(c\vee a)=b\neq (b\wedge c)\vee (b\wedge a)$, but $L$ is distributive.

Similarly for Case 2 also we get a contradiction.
\end{proof}
\begin{corollary}\label{prop5}
 If $L$ is totally ordered frame then $(a\rightarrow b)\wedge (a\rightarrow c)=a\rightarrow (b\wedge c)$, for any $a,b,c\in L$.
 \end{corollary}

\begin{property}\label{prop6}
$inf_i\{(a_i\rightarrow b)\}=sup_i\{a_{i}\}\rightarrow b$, for any $a_i,b\in L$.
\end{property}
\begin{proof}
$sup_i\{a_i\}\rightarrow b=\begin{cases}
        1 & \text{if}\ \ sup_i\{a_i\}\leq b \\
        b & \text{otherwise}.
    \end{cases}$\\
Now for $sup_i\{a_i\}\leq b$ we have $a_i\leq sup_i\{a_i\}\leq b$.\\
Hence for this case $(a_i\rightarrow b)=1$, for each $i$ and consequently $inf_i\{a_i\rightarrow b\}=1$.\\
If $sup_i\{a_i\}>b$ then there exist atleast one $a_i$ such that $a_i>b$ and rest will be either bellow $b$ or equal to $b$. Now for the case $a_i>b$, $a_i\rightarrow b=b$ and for all other cases $a_i\rightarrow b=1$. As $b\leq 1$, $inf_i\{a_i\rightarrow b\}=b$. If $sup\{a_i\}$ and $b$ are incomparable then atleast one of the $a_i$'s, say $a_j$ is incomparable to $b$ and consequently $a_j\rightarrow b=b$. Hence $inf_i\{a_i\rightarrow b\}=b$.
\end{proof}
\begin{property}\label{prop7}
$a\leq b$ iff $a\rightarrow b=1_L$.
\end{property}

\begin{property}\label{prop8}
 $a\wedge (a\rightarrow b)\leq b$.
 \end{property}
 It is to be noted that all the above properties are true for generalised G$\ddot{o}$del arrow as these properties are satisfied by any  arrow with residuation property. That means residuated arrow satisfies semilinear property but there are semilinear arrows which are not residuated arrows.

We now proceed to the main theorem which constitutes subsection \ref{TH}
\subsection{Main Theorem}\label{TH}
\begin{theorem}
Let $L$ be semilinear frame. Then
$(\{\leftidx{^{\tilde{A}}}{\alpha}{}\mid \alpha \in L\},\leftidx{^{\tilde{A}}}{0_L}{}, \cap, \bigcup,R_L)$ is a localic frame, where $R_L(\leftidx{^{\tilde{A}}}{\alpha_1}{}, \leftidx{^{\tilde{A}}}{\alpha_2}{})=inf_x\{\leftidx{^{\tilde{A}}}{\alpha_1}{}(x)\rightarrow \leftidx{^{\tilde{A}}}{\alpha_2}{}(x)\}$ for $\alpha_1$, $\alpha_2\in L$ and `$\rightarrow$' is the G$\ddot{o}$del-like arrow.
\end{theorem}

\begin{proof}
Let us check the properties for $(\{\leftidx{^{\tilde{A}}}{\alpha}{}\mid \alpha \in L\},\leftidx{^{\tilde{A}}}{0_L}{}, \cap, \bigcup,R_L)$ to be a localic frame.
\begin{enumerate}
\item $R_L(\leftidx{^{\tilde{A}}}{\alpha}{},\leftidx{^{\tilde{A}}}{\alpha}{})=inf_x\{\leftidx{^{\tilde{A}}}{\alpha}{}(x)\rightarrow \leftidx{^{\tilde{A}}}{\alpha}{}(x)\}=1_L$ [from Property \ref{prop1} $\leftidx{^{\tilde{A}}}{\alpha}{}(x)\rightarrow\leftidx{^{\tilde{A}}}{\alpha}{}(x)=1_L$, for all $x$].

\item Let $R_L(\leftidx{^{\tilde{A}}}{\alpha_1}{},\leftidx{^{\tilde{A}}}{\alpha_2}{})=1_L=R_L(\leftidx{^{\tilde{A}}}{\alpha_2}{},\leftidx{^{\tilde{A}}}{\alpha_1}{})$. So, $inf_x\{\leftidx{^{\tilde{A}}}{\alpha_1}{}(x)\rightarrow \leftidx{^{\tilde{A}}}{\alpha_2}{}(x)\}=1_L=inf_x\{\leftidx{^{\tilde{A}}}{\alpha_2}{}(x)\rightarrow \leftidx{^{\tilde{A}}}{\alpha_1}{}(x)\}$. Therefore $\leftidx{^{\tilde{A}}}{\alpha_1}{}(x)\leq\leftidx{^{\tilde{A}}}{\alpha_2}{}(x)$ and $\leftidx{^{\tilde{A}}}{\alpha_2}{}(x)\leq\leftidx{^{\tilde{A}}}{\alpha_1}{}(x)$, for all $x$. So, $\leftidx{^{\tilde{A}}}{\alpha_1}{}(x)=\leftidx{^{\tilde{A}}}{\alpha_2}{}(x)$, for any $x$. Hence $\leftidx{^{\tilde{A}}}{\alpha_1}{}=\leftidx{^{\tilde{A}}}{\alpha_2}{}$.

\item From Property \ref{prop2}, we have $(\leftidx{^{\tilde{A}}}{\alpha_1}{}(x)\rightarrow \leftidx{^{\tilde{A}}}{\alpha_2}{}(x))\wedge (\leftidx{^{\tilde{A}}}{\alpha_2}{}(x)\rightarrow \leftidx{^{\tilde{A}}}{\alpha_3}{}(x))\leq (\leftidx{^{\tilde{A}}}{\alpha_1}{}(x)\rightarrow\leftidx{^{\tilde{A}}}{\alpha_3}{}(x))$, for all $x$. Hence $inf_x\{(\leftidx{^{\tilde{A}}}{\alpha_1}{}(x)\rightarrow\leftidx{^{\tilde{A}}}{\alpha_2}{}(x))\wedge(\leftidx{^{\tilde{A}}}{\alpha_2}{}(x)\rightarrow\leftidx{^{\tilde{A}}}{\alpha_3}{}(x))\}\leq (\leftidx{^{\tilde{A}}}{\alpha_1}{}(x)\rightarrow\leftidx{^{\tilde{A}}}{\alpha_3}{}(x))$, for any $x$ and consequently $inf_x\{(\leftidx{^{\tilde{A}}}{\alpha_1}{}(x)\rightarrow\leftidx{^{\tilde{A}}}{\alpha_2}{}(x))\wedge(\leftidx{^{\tilde{A}}}{\alpha_2}{}(x)\rightarrow\leftidx{^{\tilde{A}}}{\alpha_3}{}(x))\}\leq inf_x\{\leftidx{^{\tilde{A}}}{\alpha_1}{}(x)\rightarrow\leftidx{^{\tilde{A}}}{\alpha_3}{}(x)\}$. Therefore,
\begin{align*}
& R_L(\leftidx{^{\tilde{A}}}{\alpha_1}{},\leftidx{^{\tilde{A}}}{\alpha_2}{})\wedge R_L(\leftidx{^{\tilde{A}}}{\alpha_2}{},\leftidx{^{\tilde{A}}}{\alpha_3}{})\\ & = inf_x\{\leftidx{^{\tilde{A}}}{\alpha_1}{}(x)\rightarrow \leftidx{^{\tilde{A}}}{\alpha_2}{}(x)\}\wedge inf_x\{\leftidx{^{\tilde{A}}}{\alpha_2}{}(x)\rightarrow \leftidx{^{\tilde{A}}}{\alpha_3}{}(x)\}\\
 & \leq inf_x\{(\leftidx{^{\tilde{A}}}{\alpha_1}{}(x)\rightarrow \leftidx{^{\tilde{A}}}{\alpha_2}{}(x))\wedge (\leftidx{^{\tilde{A}}}{\alpha_2}{}(x)\rightarrow \leftidx{^{\tilde{A}}}{\alpha_3}{}(x))\}\\
 & \leq inf_x\{\leftidx{^{\tilde{A}}}{\alpha_1}{}(x)\rightarrow\leftidx{^{\tilde{A}}}{\alpha_3}{}(x)\}\ \ \ \ \ [\text{using Property \ref{prop2}}]\\
 & = R_L(\leftidx{^{\tilde{A}}}{\alpha_1}{},\leftidx{^{\tilde{A}}}{\alpha_3}{}).
 \end{align*}

 \item $R_L(\leftidx{^{\tilde{A}}}{\alpha_1}{}\wedge\leftidx{^{\tilde{A}}}{\alpha_2}{},\leftidx{^{\tilde{A}}}{\alpha_1}{})=inf_x\{(\leftidx{^{\tilde{A}}}{\alpha_1}{}\wedge \leftidx{^{\tilde{A}}}{\alpha_2}{})(x)\rightarrow \leftidx{^{\tilde{A}}}{\alpha_1}{}(x)\}=1_L$, as $\leftidx{^{\tilde{A}}}{\alpha_1}{}\cap\leftidx{^{\tilde{A}}}{\alpha_2}{}\subseteq \leftidx{^{\tilde{A}}}{\alpha_1}{}$. Similarly $R_L(\leftidx{^{\tilde{A}}}{\alpha_1}{}\wedge\leftidx{^{\tilde{A}}}{\alpha_2}{},\leftidx{^{\tilde{A}}}{\alpha_2}{})=1_L$.

 \item $R_L(\leftidx{^{\tilde{A}}}{\alpha_1}{},\leftidx{^{\tilde{A}}}{0_L}{})=inf_x\{\leftidx{^{\tilde{A}}}{\alpha_1}{}(x)\rightarrow \leftidx{^{\tilde{A}}}{0_L}{}(x)\}=1_L$, as $\leftidx{^{\tilde{A}}}{0_L}{}(x)=\tilde{A}(x)$ and $\leftidx{^{\tilde{A}}}{\alpha_1}{}(x)\leq\tilde{A}(x)$, for any $x$.

 \item We have,
 \begin{align*}
 & R_L(\leftidx{^{\tilde{A}}}{\alpha_1}{},\leftidx{^{\tilde{A}}}{\alpha_2}{})\wedge R_L(\leftidx{^{\tilde{A}}}{\alpha_1}{},\leftidx{^{\tilde{A}}}{\alpha_3}{})\\
 & = inf_x\{\leftidx{^{\tilde{A}}}{\alpha_1}{}(x)\rightarrow \leftidx{^{\tilde{A}}}{\alpha_2}{}(x)\}\wedge inf_x\{\leftidx{^{\tilde{A}}}{\alpha_1}{}(x)\rightarrow\leftidx{^{\tilde{A}}}{\alpha_3}{}(x)\}\\
 & = inf_x\{(\leftidx{^{\tilde{A}}}{\alpha_1}{}(x)\rightarrow\leftidx{^{\tilde{A}}}{\alpha_2}{}(x))\wedge (\leftidx{^{\tilde{A}}}{\alpha_1}{}(x)\rightarrow \leftidx{^{\tilde{A}}}{\alpha_3}{}(x))\}\\
 & = inf_x\{\leftidx{^{\tilde{A}}}{\alpha_1}{}(x)\rightarrow (\leftidx{^{\tilde{A}}}{\alpha_2}{}(x)\wedge\leftidx{^{\tilde{A}}}{\alpha_3}{}(x))\}\ \ \ \ \ [\text{using Property \ref{prop5}}]\\
 & = inf_x\{\leftidx{^{\tilde{A}}}{\alpha_1}{}(x)\rightarrow (\leftidx{^{\tilde{A}}}{\alpha_2}{}\cap\leftidx{^{\tilde{A}}}{\alpha_3}{})(x)\}\\
 & = R_L(\leftidx{^{\tilde{A}}}{\alpha_1}{},\leftidx{^{\tilde{A}}}{\alpha_2}{}\cap\leftidx{^{\tilde{A}}}{\alpha_3}{}).
 \end{align*}

 \item Let $\leftidx{^{\tilde{A}}}{\alpha}{}\in \{\leftidx{^{\tilde{A}}}{\alpha_i}{}\}_i$, then
 \begin{align*}
 R_L(\leftidx{^{\tilde{A}}}{\alpha}{},\bigcup_i\leftidx{^{\tilde{A}}}{\alpha_i}{}) & = inf_x\{\leftidx{^{\tilde{A}}}{\alpha}{}(x)\rightarrow (\bigcup_i\leftidx{^{\tilde{A}}}{\alpha_i}{})(x)\}\\
 & = inf_x\{\leftidx{^{\tilde{A}}}{\alpha}{}(x)\rightarrow \bigvee_i\leftidx{^{\tilde{A}}}{\alpha_i}{}(x)\}\\
 & = 1_L\ \ \ \ \ [\text{using Property \ref{prop7}}].
 \end{align*}

 \item Here we have,
 \begin{align*}
 inf_i\{R_L(\leftidx{^{\tilde{A}}}{\alpha_i}{},\leftidx{^{\tilde{A}}}{\alpha}{})\}
 & = inf_i\{inf_x\{\leftidx{^{\tilde{A}}}{\alpha_i}{}(x)\rightarrow \leftidx{^{\tilde{A}}}{\alpha}{}(x)\}\}\\
 & =  inf_x\{inf_i\{\leftidx{^{\tilde{A}}}{\alpha_i}{}(x)\rightarrow \leftidx{^{\tilde{A}}}{\alpha}{}(x)\}\}\\
 & = inf_x\{\bigvee_i(\leftidx{^{\tilde{A}}}{\alpha_i}{}(x))\rightarrow \leftidx{^{\tilde{A}}}{\alpha}{}(x)\} \ \ \ \ \ [using Property \ref{prop6}]\\
 & = inf_x\{(\bigcup_i\leftidx{^{\tilde{A}}}{\alpha_i}{})(x)\rightarrow \leftidx{^{\tilde{A}}}{\alpha}{}(x)\}\\
 & = R_L(\bigcup_i\leftidx{^{\tilde{A}}}{\alpha_i}{},\leftidx{^{\tilde{A}}}{\alpha}{}).
 \end{align*}

 \item From Theorem \ref{theorem4} we have $\leftidx{^{\tilde{A}}}{\alpha}{}\cap\bigcup_i\leftidx{^{\tilde{A}}}{\alpha_i}{}=\bigcup_i(\leftidx{^{\tilde{A}}}{\alpha}{}\cap\leftidx{^{\tilde{A}}}{\alpha_i}{})$. So,
 $R_L(\leftidx{^{\tilde{A}}}{\alpha}{}\cap\bigcup_i\leftidx{^{\tilde{A}}}{\alpha_i}{},\bigcup_i(\leftidx{^{\tilde{A}}}{\alpha}{}\cap\leftidx{^{\tilde{A}}}{\alpha_i}{}))=inf_x\{(\leftidx{^{\tilde{A}}}{\alpha}{}\cap\bigcup_i\leftidx{^{\tilde{A}}}{\alpha_i}{})(x)\rightarrow(\bigcup_i(\leftidx{^{\tilde{A}}}{\alpha}{}\cap\leftidx{^{\tilde{A}}}{\alpha_i}{}))(x)=1_L$.
 \end{enumerate}
 Hence $(\{\leftidx{^{\tilde{A}}}{\alpha}{}\mid \alpha \in L\},\leftidx{^{\tilde{A}}}{0_L}{}, \cap, \bigcup,R_L)$ is a localic frame.
\end{proof}
\begin{corollary}
$(\{\leftidx{^{\tilde{A}}}{\alpha}{}\mid \alpha \in [0,1]\},\leftidx{^{\tilde{A}}}{0}{}, \cap, \bigcup,R_{[0,1]})$ is a graded frame, where `$\rightarrow$' is the G$\ddot{o}$del arrow and $R_{[0,1]}(\leftidx{^{\tilde{A}}}{\alpha_1}{}, \leftidx{^{\tilde{A}}}{\alpha_2}{})=inf_x\{\leftidx{^{\tilde{A}}}{\alpha_1}{}(x)\rightarrow \leftidx{^{\tilde{A}}}{\alpha_2}{}(x)\}$ for $\alpha_1$, $\alpha_2\in [0,1]$.
\end{corollary}

It may be noted that the 5-tuple $(\{\leftidx{^{\tilde{A}}}{\alpha}{}\mid \alpha \in L\},\leftidx{^{\tilde{A}}}{0_L}{}, \cap, \bigcup,R_L)$ is a localic preordered set, where $L$ is a  frame, $R_L(\leftidx{^{\tilde{A}}}{\alpha_1}{}, \leftidx{^{\tilde{A}}}{\alpha_2}{})=inf_x\{\leftidx{^{\tilde{A}}}{\alpha_1}{}(x)\rightarrow \leftidx{^{\tilde{A}}}{\alpha_2}{}(x)\}$ for $\alpha_1$, $\alpha_2\in L$ and `$\rightarrow$' is the G$\ddot{o}$del-like arrow, as it satisfies all the properties to be a localic frame except the property namely $$R_L(\leftidx{^{\tilde{A}}}{\alpha_1}{},\leftidx{^{\tilde{A}}}{\alpha_2}{})\wedge R_L(\leftidx{^{\tilde{A}}}{\alpha_1}{},\leftidx{^{\tilde{A}}}{\alpha_3}{})=R_L(\leftidx{^{\tilde{A}}}{\alpha_1}{},\leftidx{^{\tilde{A}}}{\alpha_2}{}\cap \leftidx{^{\tilde{A}}}{\alpha_3}{}).$$
 \section{Some applications of fuzzy $\alpha$-cut}
 In this section we shall show some usages of the notion of fuzzy $\alpha$-cuts.
\subsection{Topological Structure}
Usually a fuzzy topological space is defined as a crisp set having fuzzy open sets \cite{CL, RL}. In 1992, Chakraborty and Ahsanullah proposed a notion of fuzzy topology on fuzzy sets \cite{MA}. This generalisation allows for defining topological subspaces on fuzzy subsets of the original fuzzy topological space. With respect to the classical definition subspaces have to be defined on crisp subsets of the original set. In our recent work on fuzzy topological systems \cite{MP2} we needed to use $\mathscr{L}$-topological spaces where the value set $L$ is a frame. This is one further step towards generalisation of \cite{MA}. To make this paper self contained we give the definition below.
\begin{definition}[$\mathscr{L}$-Topological Space]\cite{MA}\label{ltop}
 Let $(X,\tilde{A})$ be an $L$-fuzzy set and $\tau$ a collection of fuzzy subsets of $(X, \tilde{A})$ such that
 \begin{enumerate}
\item $(X,\tilde{\emptyset})$ and $(X,\tilde{A})$ are in $\tau$, where $\tilde{\emptyset} : X \longrightarrow L$ is such that $\tilde{\emptyset}(x) = 0_L$, for all $x \in X$, where $0_L$ is the least element of the frame $L$;
\item $(X,\tilde{A_1})$, $(X,\tilde{A_2})$ are in $\tau$ implies $(X,\tilde{A_1} \cap\tilde{A_2})$ is in $\tau$, where $(\tilde{A_1} \cap\tilde{A_2})(x)$ = $\tilde{A_1}(x) \wedge \tilde{A_2}(x)$, for all $x \in X$;
\item $(X,\tilde{A_i})\in\tau$ implies $(X,\bigvee_{i\in I}\tilde{A_i}) \in \tau$, where $\bigvee_{i\in I}\tilde{A_i} : X \longrightarrow L$ is such that $(\bigcup_{i\in I} \tilde{A_i})(x) = \bigvee_{i\in I} \tilde{A_i}(x)$, for all $x \in X$.
 \end{enumerate}
Then $(X, \tilde{A}, \tau)$ is an $\mathscr{L}$-topological space.
 \end{definition}
 One can easily see that the fuzzy $\alpha$-cuts of a fuzzy topological space are fuzzy topological subspaces. More specifically, we have.
 \begin{theorem}\cite{MP2}
 Let $(X,\tilde{A},\tau)$ be an $\mathscr{L}$-topological space and a fuzzy $\alpha$-cut of $(X,\tilde{A})$ i.e., $(X,\leftidx{^{\tilde{A}}}{\alpha}{})$ be taken. Let $\tau'$ be defined by $\tau' = \{(X,\tilde{T'}) \mid \tilde{T'} =\leftidx{^{\tilde{A}}}{\alpha}{} \cap\tilde{T},\tilde{T}\in\tau\}$. Then $(X,\leftidx{^{\tilde{A}}}{\alpha}{},\tau')$ also forms an $\mathscr{L}$-topological space and is an $\mathscr{L}$-topological subspace.
\end{theorem}
Hence fuzzy $\alpha$-cuts provide us with a natural class of fuzzy substructure of $\mathscr{L}$-topological space.

It should be noted that in subsequent years there has been a lot of serious work on fuzzy topological spaces from the angle of category theory \cite{MA, MB1, UH, RL}. But to our knowledge, the notion of fuzzy $\alpha$-cuts as substructures has not been discussed. It would be interesting to investigate what kind of sub objects these fuzzy $\alpha$-cuts give rise to.
\subsection{Algebraic Structure}
A similar approach was initially adopted in developing fuzzy algebraic structures. Rosenfeld's pioneering work in fuzzy groups starts with an ordinary group and proceeds to define fuzzy subgroups of that group. On the other hand in \cite{MB}, Chakraborty and Banerjee defined fuzzy operations on fuzzy sets thus obtaining a generalisation that was intended. They, however, placed their work in categorical framework. We shall adopt their idea basically but avoiding categorical language and then show the role of fuzzy $\alpha$-cuts in this context. It is to be noted that a binary operation on a crisp set $A$ (e.g. the group operation) is a mapping from $A\times A$ to $A$. We shall define a fuzzy binary operation on a fuzzy set $(X,\tilde{A})$ using fuzzy equality. It is also to be noted that the Cartesian product of two $L$-fuzzy sets $(X,\tilde{A})$ and $(Y,\tilde{B})$ is the $L$-fuzzy set $(X\times Y,\tilde{A}\times \tilde{B})$ where $(\tilde{A}\times\tilde{B})(x,y)=\tilde{A}(x)\wedge\tilde{B}(y)$. So a fuzzy binary composition on $(X,\tilde{A})$ has to be a kind of mapping from $(X\times X,\tilde{A}\times\tilde{A})$ to $(X,\tilde{A})$ where the pre-image is mapped to the image to some grade belonging to the frame $L$. For any $L$-fuzzy set $(X,\tilde{A})$ by $\mid\tilde{A}\mid$ is meant the support viz. $\{x\in X\mid\tilde{A}(x)>0_L\}$. Instead of using pre-fix notation for the operator $\oplus$ we shall use infix notation, i.e. we write
$$x_1\oplus x_2=x_3,\ \text{for}\ \oplus(x_1,x_2)=x_3$$
and additionally equality relation $(=)$ is graded. That is, for any $x\in X$, $y\in Y$, the expression $x=y$ gets a degree from $L$.
We shall write $gr(x\simeq y)$ to make a distinction between fuzzy equality and ordinary equality. Formally, we have Definition \ref{fbo}.

\begin{definition}[$L$-fuzzy binary operation]\label{fbo}
An $L$-fuzzy binary operation $\oplus:(X\times X,\tilde{A}\times\tilde{A})\longrightarrow (X,\tilde{A})$ is a map such that
\begin{enumerate}
\item $gr(x_1\oplus x_2\simeq x_3)\leq \tilde{A}\times \tilde{A}(x_1,x_2)\wedge\tilde{A}(x_3)$ for any $x_1,x_2,x_3\in X$, where $gr:(X\times X)\times X\longrightarrow L$ and $\tilde{A}\times\tilde{A}(x_1,x_2)=\tilde{A}(x_1)\wedge\tilde{A}(x_2)$;
\item for any $(a_1,a_2)\in\mid\tilde{A}\times\tilde{A}\mid$ there exist a unique $a\in\mid\tilde{A}\mid$ with $gr(a_1\oplus a_2\simeq a)=\tilde{A}(a_1)\wedge\tilde{A}(a_2)$ and $gr(a_1\oplus a_2\simeq a')=0_L$, if $a'(\neq a)\in \mid\tilde{A}\mid$.
\end{enumerate}
\end{definition}
Note that condition 1 is the fuzzy counterpart of the closure property of $\tilde{A}$ relative to the operation $\oplus$. Also $gr(x_1\oplus x_2\simeq x_3)$ represents the fuzzy equality that is, the grade in which the pair $(x_1,x_2)$ equals to $x_3$ by the fuzzy composition $\oplus$. That is, here we will talk about the degree of equality between $x_1\oplus x_2$ and $x_3$ of $X$.
\begin{definition}[$L$-fuzzy group]
An $L$-fuzzy group is a triple $(X,\tilde{A},\oplus)$ consisting of an $L$-fuzzy set $(X,\tilde{A})$ with $\mid\tilde{A}\mid\neq\emptyset$ and an $L$-fuzzy binary operation $\oplus$ such that
\begin{enumerate}
\item if for any $a,b,a_1,a_2,a_3,b_1,b_2\in \mid\tilde{A}\mid$,
\begin{align*}
gr(a_1\oplus a_2\simeq b_1) & = \tilde{A}(a_1)\wedge \tilde{A}(a_2),\\
gr(b_1\oplus a_3\simeq a) & = \tilde{A}(b_1)\wedge \tilde{A}(a_3),\\
gr(a_2\oplus a_3\simeq b_2) & = \tilde{A}(a_2)\wedge \tilde{A}(a_3) \ and\\
gr(a_1\oplus b_2\simeq b) & = \tilde{A}(a_1)\wedge \tilde{A}(b_2)\\
\end{align*}
then $a=b$;
\item there exist $e\in\mid\tilde{A}\mid$ such that $gr(a\oplus e\simeq a)=\tilde{A}(a)\wedge\tilde{A}(e)=gr(e\oplus a\simeq a)$ for any $a\in \mid\tilde{A}\mid$ and
\item for any $a\in \mid\tilde{A}\mid$, there exist $a^{-1}\in \mid\tilde{A}\mid$, such that $gr(a\oplus a^{-1}\simeq e)=\tilde{A}(a)\wedge\tilde{A}(a^{-1})=gr(a^{-1}\oplus a\simeq e)$ and $\tilde{A}(a)=\tilde{A}(a^{-1})$.
\end{enumerate}
$e\in\mid\tilde{A}\mid$ described in 2 is known as the identity whereas $a^{-1}\in \mid\tilde{A}\mid$ for each $a\in \mid\tilde{A}\mid$ illustrated in condition 3 are known as inverse of $a$ in the $L$-fuzzy group $(X,\tilde{A},\oplus)$. It is possible to show that identity and inverse of an element in the $L$-fuzzy group are unique.

It is to be noted that property 1 of being an $L$-fuzzy group represents the fuzzy version of associativity.
\end{definition}
Equipped with this definition of an $L$-fuzzy group, it will be observed how does fuzzy $\alpha$-cuts play a role.
\begin{definition}[$L$-fuzzy subgroup]
Let $(X,\tilde{A},\oplus)$ be an $L$-fuzzy group and $(X,\tilde{B})$ be a $L$-fuzzy subset of the $L$-fuzzy set $(X,\tilde{A})$. Then the fuzzy substructure $(X,\tilde{B},\oplus')$ where $\oplus'$ is an $L$-fuzzy binary operation on $(X,\tilde{B})$ defined by, $$gr(x_1\oplus'x_2\simeq x_3)=gr(x_1\oplus x_2\simeq x_3)\wedge\tilde{B}(x_1)\wedge\tilde{B}(x_2)\wedge\tilde{B}(x_3),$$ for any $x_1,x_2,x_3\in X$
is called a fuzzy subgroup of $(X,\tilde{A},\oplus)$ if $(X,\tilde{B},\oplus')$ is itself an $L$-fuzzy group, [$\oplus'$ is the restriction of $\oplus$ on the fuzzy subset $(X,\tilde{B})$ of $(X,\tilde{A})$].
\end{definition}

\begin{proposition}\label{iden}
Let $(X,\tilde{A},\oplus)$ be an $L$-fuzzy group. Then for any $a\in \mid\tilde{A}\mid$, $\tilde{A}(a)\leq \tilde{A}(e)$.
\end{proposition}
\begin{proof}
For any $a\in\mid\tilde{A}\mid$, there exist $a^{-1}\in\mid\tilde{A}\mid$ such that $gr(a\oplus a^{-1}\simeq e)=\tilde{A}(a)\wedge \tilde{A}(a^{-1})=\tilde{A}(a)$, as $\tilde{A}(a)=\tilde{A}(a^{-1})$. Also we know $gr(a\oplus a^{-1}\simeq e)\leq \tilde{A}(e)$ and consequently $\tilde{A}(a)\leq\tilde{A}(e)$, for any $a\in\mid\tilde{A}\mid$.
\end{proof}

\begin{theorem}
Let $(X,\tilde{A},\oplus)$ be an $L$-fuzzy group. Then  $(X,\leftidx{^{\tilde{A}}}{\alpha}{},\oplus')$ is an $L$-fuzzy sub group for any $\alpha\in L$ such that $\mid \leftidx{^{\tilde{A}}}{\alpha}{}\mid\neq\emptyset$.
\end{theorem}
\begin{proof}
Given that $(X,\tilde{A},\oplus)$ is an $L$-fuzzy group. Let us restrict the function $\oplus$ on $\leftidx{^{\tilde{A}}}{\alpha}{}$. Then for $x,y,z\in X$, if $gr(x\oplus y\simeq z)=\tilde{A}(x)\wedge \tilde{A}(y)$ then
$$gr(x\oplus' y\simeq z)=\tilde{A}(x)\wedge\tilde{A}(y)\wedge \leftidx{^{\tilde{A}}}{\alpha}{}(x)\wedge \leftidx{^{\tilde{A}}}{\alpha}{}(y)\wedge \leftidx{^{\tilde{A}}}{\alpha}{}(z).$$
Now as $\tilde{A}(x)\wedge\tilde{A}(y)\leq\tilde{A}(z)$, we have $\leftidx{^{\tilde{A}}}{\alpha}{}(x)\wedge\leftidx{^{\tilde{A}}}{\alpha}{}(y)\leq\leftidx{^{\tilde{A}}}{\alpha}{}(z).$ Therefore $\leftidx{^{\tilde{A}}}{\alpha}{}(x)\wedge\leftidx{^{\tilde{A}}}{\alpha}{}(y)\wedge\leftidx{^{\tilde{A}}}{\alpha}{}(z)=\leftidx{^{\tilde{A}}}{\alpha}{}(x)\wedge\leftidx{^{\tilde{A}}}{\alpha}{}(y)$. Hence for any $x,y,z\in X$ if $gr(x\oplus y\simeq z)=\tilde{A}(x)\wedge\tilde{A}(y)$ then $gr(x\oplus' y\simeq z)=\leftidx{^{\tilde{A}}}{\alpha}{}(x)\wedge\leftidx{^{\tilde{A}}}{\alpha}{}(y)$.

To show that $(X,\leftidx{^{\tilde{A}}}{\alpha}{},\oplus')$ is an $L$-fuzzy subgroup for $\alpha$ such that $\mid\leftidx{^{\tilde{A}}}{\alpha}{}\mid\neq\emptyset$ first of all notice that $\oplus'$ is indeed an $L$-fuzzy binary operation as the following holds.

(i) $gr(x\oplus'y\simeq z)=gr(x\oplus y\simeq z)\wedge\leftidx{^{\tilde{A}}}{\alpha}{}(x)\wedge\leftidx{^{\tilde{A}}}{\alpha}{}(y)\wedge\leftidx{^{\tilde{A}}}{\alpha}{}(z)\leq(\leftidx{^{\tilde{A}}}{\alpha}{}(x)\wedge\leftidx{^{\tilde{A}}}{\alpha}{}(y))\wedge\leftidx{^{\tilde{A}}}{\alpha}{}(z)$.

(ii) If $a,b\in\mid\leftidx{^{\tilde{A}}}{\alpha}{}\mid$ then $\leftidx{^{\tilde{A}}}{\alpha}{}(a)>0_L$, $\leftidx{^{\tilde{A}}}{\alpha}{}(b)>0_L$. Hence $\tilde{A}(a)\geq \alpha$ and $\tilde{A}(b)\geq\alpha$. As $\oplus$ is an $L$-fuzzy binary operation so for any $a,b\in\mid\leftidx{^{\tilde{A}}}{\alpha}{}\mid\subseteq\mid\tilde{A}\mid$ there exist unique $c\in\mid\tilde{A}\mid$ with $gr(a\oplus b\simeq c)=\tilde{A}(a)\wedge\tilde{A}(b)\leq\tilde{A}(c)$ and $gr(a\oplus b\simeq c')=0_L$, if $c'(\neq c)\in\mid\tilde{A}\mid$. Therefore $\tilde{A}(c)\geq \alpha$ and hence $\leftidx{^{\tilde{A}}}{\alpha}{}(c)=\tilde{A}(c)$. $c\in \mid\leftidx{^{\tilde{A}}}{\alpha}{}\mid$. Consequently for $(a,b)\in\mid\leftidx{^{\tilde{A}}}{\alpha}{}\times\leftidx{^{\tilde{A}}}{\alpha}{}\mid$, there exist unique $c\in\mid\leftidx{^{\tilde{A}}}{\alpha}{}\mid$ with $gr(a\oplus' b\simeq c)=\leftidx{^{\tilde{A}}}{\alpha}{}(a)\wedge\leftidx{^{\tilde{A}}}{\alpha}{}(b)$ and $gr(a\oplus' b\simeq c')=0_L$ if $c'(\neq c)\in\mid\leftidx{^{\tilde{A}}}{\alpha}{}\mid$.

As for any $x,y,z\in X$ if $gr(x\oplus y\simeq z)=\tilde{A}(x)\wedge\tilde{A}(y)$ then $gr(x\oplus' y\simeq z)=\leftidx{^{\tilde{A}}}{\alpha}{}(x)\wedge\leftidx{^{\tilde{A}}}{\alpha}{}(y)$, associativity holds good.

Now $(X,\tilde{A},\oplus)$ is an $L$-group and so there exist $e\in\mid\tilde{A}\mid$ such that for any $a\in \mid\tilde{A}\mid$, $gr(a\oplus e\simeq a)=\tilde{A}(a)\wedge\tilde{A}(e)$.
Hence for any $a\in\mid\leftidx{^{\tilde{A}}}{\alpha}{}\mid\subseteq\mid\tilde{A}\mid$, $gr(a\oplus' e\simeq a)=\leftidx{^{\tilde{A}}}{\alpha}{}(a)\wedge\leftidx{^{\tilde{A}}}{\alpha}{}(e)$. Now $a\in\mid \leftidx{^{\tilde{A}}}{\alpha}{}\mid$ implies $\leftidx{^{\tilde{A}}}{\alpha}{}(a)>0_L$ and hence $\tilde{A}(a)\geq \alpha$. Using Proposition \ref{iden}, we have $\alpha\leq\tilde{A}(a)\leq\tilde{A}(e)$. Hence $\leftidx{^{\tilde{A}}}{\alpha}{}(e)=\tilde{A}(e)$. So, $e\in\mid\leftidx{^{\tilde{A}}}{\alpha}{}\mid$.

Similarly it can be shown that for any $a\in\mid\leftidx{^{\tilde{A}}}{\alpha}{}\mid$, there exist $a^{-1}\in\mid\leftidx{^{\tilde{A}}}{\alpha}{}\mid$ such that $gr(a\oplus' a^{-1}\simeq e)=\leftidx{^{\tilde{A}}}{\alpha}{}(a)\wedge \leftidx{^{\tilde{A}}}{\alpha}{}(a^{-1})$, as $\tilde{A}(a^{-1})=\tilde{A}(a)\geq \alpha$.
\end{proof}
Thus fuzzy $\alpha$-cuts form natural fuzzy subgroups of the fuzzy group.

This method of defining fuzzy algebraic structures and their sub structures may be adopted for any kind of algebraic structure not necessarily fuzzy groups only.
\begin{example}
Let $X=\{x_1,x_2,x_3,x_4,x_5\}$, $\tilde{A}:X\longrightarrow L$ where
\begin{center}
\begin{tikzpicture}
  \matrix (galois)
     [matrix of nodes,
      nodes in empty cells,
      nodes={outer sep=0pt,circle,minimum size=4pt},
      column sep={1cm,between origins},
      row sep={1cm,between origins}]
   {
    && $1_L$ &&\\
    &  & $l_4$ & \\
   & $l_1$ &  & $l_2$\\

    && $l_3$ &&\\
    && $0_L$ &&\\
   };

      \draw (galois-1-3) -- (galois-2-3);

      \draw (galois-2-3) -- (galois-3-2);
      \draw (galois-2-3) -- (galois-3-4);

       \draw (galois-3-2) -- (galois-4-3);

       \draw (galois-3-4) -- (galois-4-3);
        \draw (galois-4-3) -- (galois-5-3);

\end{tikzpicture}
\end{center}
such that $\tilde{A}(x_i)=l_i$, for $i=\{1,2,3,4\}$ and $\tilde{A}(x_5)=0_L$. Here $\mid\tilde{A}\mid=\{x_1,x_2,x_3,x_4\}$. Let us define $\oplus$ as follows:
\begin{align*}
gr(x_1\oplus x_1\simeq x_4)=l_1 && gr(x_1\oplus x_2\simeq x_3)=l_3\\
gr(x_1\oplus x_3\simeq x_2)=l_3 && gr(x_1\oplus x_4\simeq x_1)=l_1 \\
gr(x_2\oplus x_1\simeq x_3)=l_3 && gr(x_2\oplus x_2\simeq x_4)=l_2 \\
gr(x_2\oplus x_3\simeq x_1)=l_3 && gr(x_2\oplus x_4\simeq x_2)=l_2 \\
gr(x_3\oplus x_1\simeq x_2)=l_3 && gr(x_3\oplus x_2\simeq x_1)=l_3 \\
gr(x_3\oplus x_3\simeq x_4)=l_3 && gr(x_3\oplus x_4\simeq x_3)=l_3 \\
gr(x_4\oplus x_1\simeq x_1)=l_1 && gr(x_4\oplus x_2\simeq x_2)=l_2 \\
gr(x_4\oplus x_3\simeq x_3)=l_3 && gr(x_4\oplus x_4\simeq x_4)=l_4 \\
\end{align*}
Then clearly $x_4=e$, $x_1^{-1}=x_1$, $x_2^{-1}=x_2$, $x_3^{-1}=x_3$ and $x_4^{-1}=x_4$. Hence $(X,\tilde{A},\oplus)$ is an $L$-fuzzy group.

Let $\alpha=l_1$, then $\mid\leftidx{^{\tilde{A}}}{l_1}{}\mid=\{x_1,x_4\}\neq\emptyset$ and $(X,\leftidx{^{\tilde{A}}}{l_1}{},\oplus')$ forms an $L$-fuzzy group and consequently becomes an $L$-fuzzy subgroup of $(X,\tilde{A},\oplus)$. Similarly for other $\alpha\in L$, where $\mid\leftidx{^{\tilde{A}}}{\alpha}{}\mid\neq\emptyset$, it can be shown that $(X,\leftidx{^{\tilde{A}}}{\alpha}{},\oplus')$ forms $L$-fuzzy subgroups of $(X,\tilde{A},\oplus)$.
\end{example}
\subsection{Probabilistic Rough Set Theory}
We will now observe another kind of usefulness of the notion of fuzzy $\alpha$-cuts in the context of rough set theory \cite{PW, YY}. An approximation space is a tuple $(X,R)$, consisting of a set of objects $X$ and an equivalence relation $R$, known as indiscernibility relation on $X$. For any $A\subseteq X$, the lower and upper approximations of $A$ in the approximation space $(X,R)$ are denoted by $\underline{A}$ and $\overline{A}$ respectively and defined as follows.
$$\underline{A}=\bigcup\{[x]\mid [x]\subseteq A\};$$
$$\overline{A}=\bigcup\{[x]\mid A\cap [x]\neq\emptyset\}.$$
A rough membership function of $A$, denoted by $\mu_A$, is a function from $X$ to $[0,1]$ such that $\mu_A(x)=\frac{\mid [x]\cap A\mid}{\mid [x]\mid}\leq 1$, where $\mid S\mid$ stands for the cardinality of the set $S$ and $[x]$ stands for the equivalence class of $x\in X$. In this definition $X$ is taken to be a finite set.

In \cite{YY}, we notice that for generalised probabilistic approximations, they considered a pair of parameters $\alpha,\beta\in [0,1]$ with $\alpha\geq\beta$ to ensure that the lower approximation is smaller than the upper approximation in order to be consistent with existing approximation operators.

In the theory of probabilistic rough sets a weight or grade from the set $[0,1]$ is attached with each granule. The grades of granules are obtained with the help of some rough membership function. In particular the grade of granule may be determined with the help of above described rough membership function. Notice that in \cite{YY}, while defining lower and upper approximations of a set $A$, $\alpha$-cuts and strict $\beta$-cuts are used with $0\leq\beta<\alpha\leq 1$ in the following way.
$$\underline{A_\alpha}=\{x\in X\mid\mu_A(x)\geq \alpha\};$$
$$\overline{A_\beta}=\{x\in X\mid\mu_A(x)> \beta\}.$$
These are crisp sets. Hence the grade disappears in the final approximations.

But while defining lower and upper approximations of a set $A$, if we use the concept of fuzzy $\alpha$-cuts and fuzzy $\beta$-cuts instead of $\alpha$-cuts and strict $\beta$-cuts then we will able to end up with the final approximations having grades. That is, the lower and upper approximations becomes fuzzy sets and defined as follows.
$$\underline{\leftidx{^{A}}{\alpha}{}}:X\longrightarrow [0,1]\ \text{s.t.}\ \underline{\leftidx{^{A}}{\alpha}{}}(x)=
\begin{cases}
\mu_A(x) & \text{if}\ \mu_A(x)\geq \alpha\\
0 & \text{otherwise}.
\end{cases}$$
$$\overline{\leftidx{^{A}}{\beta}{}}:X\longrightarrow [0,1]\ \text{s.t.}\ \overline{\leftidx{^{A}}{\beta}{}}(x)=
\begin{cases}
\mu_A(x) & \text{if}\ \mu_A(x)\geq \beta\\
0 & \text{otherwise}.
\end{cases}$$
It is quite expected that the above described notion of lower and upper approximations will play a significant role in probabilistic rough set theory. Here instead of two different types of cuts viz. $\alpha$-cuts and strict $\beta$-cuts one type of cut has been used uniformly in determining lower and upper approximations. In this paper we will not delve into this topic, but it will be considered in our future research.
\section{Concluding Remarks}
In this paper we have dealt with the notion of fuzzy $\alpha$-cut and its significance. Study of the family of fuzzy $\alpha$-cuts provides an example of graded frame which was introduced in \cite{MP3}. Moreover in this paper we generalise the notion of graded frame one step further and call it `semilinear frame'. It is to emphasise that the notion semilinearity introduced in this paper is more general than `prelinearity'; while the latter notion has been widely discussed in literature, the former notion is not. We also proposed the notion of localic frame in this work. A detailed study of G$\ddot{o}$del-like arrow provides a nice result about the relation between prelinearity and semilinearity property. The algebraic notion of semilinear frame needs to be studied in more detail. Taking a general fuzzy arrow instead of G$\ddot{o}$del arrow may also be considered as an interesting future project.

\paragraph{Acknowledgements.}
The corresponding author acknowledge Department of Science $\&$ Technology, Government of India for financial support vide reference no. SR/WOS-A/PM-1010/2014 under Women Scientist Scheme to carry out this work.

\section*{References}

\bibliography{mybibfile}

config: arxiv


text:
\section{Missing Proofs in Section \ref{sec:shadowCG}}\label{sec:appshadowCG}

\algconverg*
\begin{proof}
Recall that in the algorithm we either enter the \textsc{Trace} procedure ($\x_{t+1} := \textsc{Trace}(\x_t, \nabla f(\x_t))$, or we take a Frank-Wolfe step ($\x_{t+1} := \x_t + \gamma_t (\mathbf{v}_t - \x_t)$ for some $\gamma_t \in [0,1]$). We split the
proof of convergence into two cases depending on which case happens:
\begin{enumerate} [leftmargin = 10pt]
    \item[] \textbf{Case 1:} \textit{We enter the \textsc{Trace} procedure.} Since $\textsc{Trace}(\x_t, \nabla f(\x_t))$ traces the whole curve of $g(\lambda) =  \Pi_P(\mathbf{x}_t - \lambda \nabla f(\mathbf{x}_t))$ with exact line-search, we know that at the point $\x_{t+1}$ we have $f(\x_{t+1}) \leq f(g(\lambda))$ for all $\lambda > 0$. In particular, $f(\x_{t+1}) \leq f(g(1/L))$. Hence we get the same rate $(1 - \frac{\mu} {L})$ of decrease as PGD with fixed step size $1/L$ using Theorem \ref{PGD progress}.

    \item[] \textbf{Case 2:} \textit{We take a Frank-Wolfe step.}  Let $\gamma_t$ be the step size chosen by line-search for moving along the chosen FW direction $\bd_t := \mathbf{v}_t - \x_t$. Also, let $\gamma_t^{\max} = 1$ be the maximum step-size that one can move along $\bd_t$. Finally, recall from the previous section that $\gamma_{\bd_t} = \frac{\innerprod{-\nabla f(\mathbf{x}_t)}{\bd_t}} {L \|\mathbf{d}_{\mathbf{x}_t}\|^2}$ is the step size obtained from optimizing the smoothness inequality to obtain \eqref{progress from smoothness}. We now split the proof depending on whether $\gamma_{t} < \gamma_t^{\max}$ or not:

    \begin{enumerate}
        \item[$(i)$] \textit{First, suppose that $\gamma_{t} < \gamma_t^{\max}$}. We claim that we can use the step size from $\gamma_{\bd_t}$ to lower bound the progress even if $\gamma_{\bd_t}$ is not a feasible step size (i.e. when $\gamma_{\bd_t} > 1$). To see this, note that the optimal solution of the line-search step is in the interior of the interval $[0, \gamma_t^{\max}]$. Define $\x_\gamma := \x_t + \gamma \bd_t$. Then, because $f(\x_\gamma)$ is convex in $\gamma$, we know that $\min_{\gamma \in [0,\gamma_t^{\max}]} f(\x_\gamma) = \min_{\gamma \geq 0} f(\x_\gamma)$ and thus $\min_{\gamma \in [0,\gamma_t^{\max}]} f(\x_\gamma) = f(\x_{t+1}) \leq f(\x_\gamma)$ for all $ \gamma \geq 0$. In particular, $f(\x_{t+1}) \leq f(\x_{\gamma_{\bd_t}})$. Hence, we can use \eqref{progress from smoothness} to bound the progress per iteration as follows:
    \begin{align*}
    h(\x_t) - h(\x_{t+1}) &\geq \frac{\innerprod {-\nabla f(\x_{t})}{\bd_t^{\text{FW}}}^2} {2L \|\bd_t^{\text{FW}}\|^2}  && \text{(using \eqref{progress from smoothness})}\\
    &\geq \frac{\innerprod{-\nabla f(\x_{t})}{\frac{\bd_{\x_t}^\Pi}{\|\bd_{\x_t}^\Pi\|}}^2} {2L D^2} && \text{(choice of descent step)} \\
    &\geq  \frac{\mu} {L D^2} h(\x_t) && \text{(using \eqref{deriv performance1}).}
\end{align*}
This shows the rate stated in the theorem.
\item[$(iii)$] \textit{We have a boundary case: $\gamma_{t} = \gamma_t^{\max}$}. We further divide this case into two sub-cases:
 \begin{enumerate}
        \item[(a)] First assume that $\gamma_{\bd_t} \leq \gamma_t^{\max}$ so that the step size from smoothness is feasible. Then, using the same argument as above we also have a $(1 - \frac{\mu} {LD^2})$-geometric rate of decrease.

        \item[(b)] Finally assume that $\gamma_{\bd_t} > \gamma_t^{\max}$ and $\bd_t = \bd_t^{\mathrm{FW}}$. Observe that $\gamma_{\bd_t} = \frac{\innerprod {-\nabla f(\x_{t})}{\bd_t^{\mathrm{FW}}}}{L \|\bd_t^{\mathrm{FW}}\|^2} > \gamma_t^{\max} = 1$ implies that $\innerprod  {-\nabla f(\x_t)}{\bd_t^{\mathrm{FW}}} \geq L\|\bd_t^{\mathrm{FW}}\|_2^2$. Hence, using the fact that $\gamma_t = \gamma_t^{\max} =1$ in the smoothness inequality in \eqref{descent lemma}, we have
        \begin{align*}
            h(\x_{t}) - h(\x_{t+1}) &\geq  \innerprod  {-\nabla f(\x_{t})}{\bd_t^{\mathrm{FW}}} - \frac{L}{2} \|\bd_t^{\mathrm{FW}}\|_2^2
            & \geq \frac{h(\x_{t})}{2} && \text{(using Wolfe gap \eqref{Wolfe-gap})}
        \end{align*}
        Hence, we get a geometric rate of decrease of 1/2.
        \end{enumerate}
        \end{enumerate}
\end{enumerate}
The iteration complexity of the number of oracle calls stated in the theorem now follows using the above rate of decrease in the primal gap.
  \end{proof}


TITLE: Solve a first order differential equation by substitution
QUESTION [0 upvotes]: For a given differential equation
$$xy'+2y\log y -4x^2y = 0;$$ $$ y(1)=1 $$
I want to use the substitution $v=\log y$. Which implies that $$v'=\frac{dv}{dy}\frac{dy}{dx}= \frac{1}{y} y'$$Hence: $$y'=yv'$$
That being said, solving the equation I try as follows:
$$xyv'+2yv-4x^2y = 0 $$
$$xv'+2v = 4x^2 $$
$$v'+\frac{2}{x}v= 4x $$
$$v'=4x-\frac{2}{x}v $$
How can I use seperation of variables here? Or how can I solve this thing ?
thanks

REPLY [1 votes]: You've reduced your equation to the so called Bernoulli differential equation. Using notation from wikipedia just set
$$
P(x)=\frac{2}{x}\qquad Q(x)=4x\qquad n=0
$$


TITLE: Electrons, spins, and degeneracy
QUESTION [3 upvotes]: In an atom, two electrons can have the same set of $n,\ell,m$ quantum numbers as long as they have opposite spins.  My introductory physics and chemistry courses have all introduced this as two electrons occupying the same energy state; however, the spins of the electrons cause splitting of the energy levels they occupy.  This leads me to think two electrons with opposite spins are only able to share the same set of $n,\ell,m$ quantum numbers because they actually occupy slightly different energy levels.  My question is: if there were no splitting of energy levels due to spin, would two electrons still be able to occupy the now-exactly the same energy level or do they need to have different energies, no matter how small the difference?

REPLY [2 votes]: This echoes JeffDror's (correct) answer but also includes some math/notation that might help illuminate things further.
Let's consider essentially your example.  Let $H$ denote the hamiltonian for a single electron in a central potential (like when it moves around a nucleus.  Then the state space (Hilbert space) of the system is spanned by energy eigenstates
\begin{align}
  |n,\ell,m,m_s\rangle
\end{align}
where $m_s$ is the $z$-component of spin quantum number.  Note that the energy of each of these states is determined entirely by the principal quantum number $n$, so there is a lot of degeneracy in the spectrum.  Mathematically, when $H$ acts on any of these states, it yields a value $E_n$ that only depends on $n$;
\begin{align}
  H|n,\ell,m,m_s\rangle = E_n|n,\ell,m,m_s\rangle.
\end{align}
In particular, a spin up electron, with $m_s = +1$, has the same energy as a spin down electron with $m_s=-1$;
\begin{align}
  H|n,\ell,m,+1\rangle &= E_n|n,\ell,m,+1\rangle \\
  H|n,\ell,m,-1\rangle &= E_n|n,\ell,m,-1\rangle \\
\end{align}
There is a two-fold spin degeneracy in energy levels; different spins do not cause energy splitting.  This situation would be different if, for example, there were an external magnetic field that interacted with the magnetic moment of the electron.
Now consider a system of two, non-interacting electrons in the central potential.  The state space of this system is spanned by (tensor) product states that specify the individual state of each electron;
\begin{align}
  |n_1,\ell_1,m_1,m_{s,1}\rangle|n_2,\ell_2,m_2,m_{s,2}\rangle
\end{align}
For convenience, let's suppress the orbital angular momentum quantum numbers in the notation for now, so that such a state would be written as
\begin{align}
  |n_1, m_{s,1}\rangle|n_2, m_{s,2}\rangle.
\end{align}
Let $H_2$ denote the hamiltonian of this system, then the energy of such a state is given by the sum of the energies of the individual states;
\begin{align}
  H_2|n_1, m_{s,1}\rangle|n_2, m_{s,2}\rangle = (E_{n_1}+E_{n_2})|n_1, m_{s,1}\rangle|n_2, m_{s,2}\rangle
\end{align}
However, states of this form are not allowed in this system because electrons are fermions, and the Pauli Exlusion Principle tells us that the for such a system, the only states that are allowed are those that are antisymmetric in the quantum numbers of the electrons.  This means that if we write some two-electron state, and we switch the factors in each product state that appears, then the overall effect on the state should be that is gets multiplied by $-1$.  As an example, consider the following state:
\begin{align}
  |2,+1\rangle|2,-1\rangle.
\end{align}
Is this state allowed?  Now, because when we flip the factors, we don't get negative of the same state back;
\begin{align}
  |2,+1\rangle|2,-1\rangle \neq |2,-1\rangle|2,+1\rangle.
\end{align}
But we can fix this.  Consider, instead, the following state:
\begin{align}
  |\psi\rangle = \frac{1}{\sqrt 2}\Big(|2,+1\rangle|2,-1\rangle - |2,-1\rangle|2,+1\rangle\Big)
\end{align}
Notice what happens when we flip all the terms in this state;
\begin{align}
  \mathrm{Flip}(|\psi\rangle)
&= \frac{1}{\sqrt 2}\Big(|2,-1\rangle|2,+1\rangle - |2,+1\rangle|2,-1\rangle\Big)\\
&= -\frac{1}{\sqrt 2}\Big(|2,+1\rangle|2,-1\rangle - |2,-1\rangle|2,+1\rangle\Big) \\
&= -|\psi\rangle
\end{align}
This state works!  When we exchange the quantum numbers of the two particles, the state changes by a negative sign.  This state actually has a special name; it's called the ``spin singlet state."  Notice that the two factors in each product state always have on electron having spin up, and the other electron having spin down.  This antisymmetry property cannot be satisfied by a state in which the electrons have the same spin because flipping would just given the same state back again without the minus sign!
So the upshot is this:

Regardless of the form of the hamiltonian, and in particular whether or not different spin states lead to different energies, the allowed states of electrons are determined by symmetry properties of their states.


TITLE: Integral Calculus, right or wrong?
QUESTION [5 upvotes]: I have two questions, which I decided and my feedback does not match that of my book, but do not seem to be wrong.
Calculate the volume of the solid obtained by rotating the region bounded by the curves

I) $y=e^x,\;y=0,\;x=0,\;x=1$ Around the $x$-axis

My answer was $$V=\pi\int_{0}^{1}\;\left(e^x\right)^2\;dx=\frac{\pi}{2}\int_0^1\;e^{2x}\cdot 2\;dx$$
$$\boxed{V=\frac{\pi}2\left(e^2-1\right)}$$
And my feedback says $$V=\frac{\pi}{12}\left( e^2-1 \right)$$
$$$$

II) Have the other is $y=\sec x,\; y=1,\; x=-1,\;x=1$ Around the $x$-axis

$$V=\pi\int_{-1}^1\;\left( (\sec x)^2-1^2 \right)\;dx=\left.\pi\left( \tan x-x \right)\right|_{-1}^1$$
$$V=\pi(\tan(1)-2-\tan{(-1)})$$
And my feedback says $$V=2\pi(\tan1-1)$$
$$$$
Am I wrong again?

REPLY [5 votes]: For the first one, I think someone's just accidentally written 12 instead of 2.
I can't see any flaws in your working.
For the second,
$$\tan(-1)=\frac{\sin(-1)}{\cos(-1)}=-\frac{\sin(1)}{\cos(1)}=-\tan(1)$$
$$\pi(\tan(1)−2−\tan(−1))=\pi(\tan(1)-2-(-\tan(1)))$$
$$=\pi(2\tan(1)-2)=2\pi(\tan(1)-1)$$
you were right but just needed to do a bit more :)


TITLE: Boolean representation of logical statements of more than two extreme conditions
QUESTION [1 upvotes]: I had a question to determine the highest and and lowest and the middle paid of three employees. I tried to solve the problem using logical values T or F, then getting the connection between the truth tables, but I got stuck since the first statement concerns the highest and the second concerns the lowest paid and there is a hole about the middle one that I could't represent, using truth values since not highest may be lowest or middle and vice versa. The question is below from Rosen discrete math book.

Steve would like to determine the relative salaries of three
  coworkers using two facts. First, he knows that if Fred
  is not the highest paid of the three, then Janice is. Second, he knows that if Janice is not the lowest paid, then
  Maggie is paid the most. Is it possible to determine the
  relative salaries of Fred, Maggie, and Janice from what
  Steve knows? If so, who is paid the most and who the
  least? Explain your reasoning.

REPLY [0 votes]: Steve would like to determine the relative salaries of three coworkers using two facts. First, he knows that if Fred is not the highest paid of the three, then Janice is.

$$(F{\lt}J\vee F{\lt}M)\to (F{<}M{<}J\vee M{<}F{<}J)\tag 1$$
Clearly that means that: $$F{<}J\to (F{<}M{<}J\vee M{<}F{<}J) \tag{1.1}$$ ... and, less obviously, that: $$F{<}M\to (F{<}M{<}J)\tag {1.2}$$

Second, he knows that if Janice is not the lowest paid, then Maggie is paid the most.

$$(F{<}J\vee M{<}J)\to (F{<}J{<}M\vee J{<}F{<}M) \tag 2$$
Likewise, this infers that: $$F{<}J\to F{<}J{<}M \tag{2.1}$$ ... and also that: $$M{<}J\to \bot\tag{2.2}$$

Is it possible to determine the relative salaries of Fred, Maggie, and Janice from what Steve knows? If so, who is paid the most and who the least? Explain your reasoning.

Fred cannot be paid less than Jannice (1.1 and 2.1 contradict), and Fred cannot be paid less than Maggie (1.2 and 2.1 contradict), and Maggie cannot be paid less than Jannice (2.2).   Which means Fred is paid the most, and Jannice the least. $$J{<}M{<}F\tag{done}$$


TITLE: equivalent norm to uniform norm on $C[a,b]$
QUESTION [0 upvotes]: I know that on $C[a,b]$ the uniform norm (sup norm) is not equivalent to $L_p$-norm. Can anyone help me find a norm which is actually equivalent to the uniform norm on $C[a,b]$?

REPLY [1 votes]: The main difficulty is to avoid writing a norm which is isometric to the sup norm.  You didn't actually state that the norm shouldn't be isometric, but it's a natural nontriviality condition (which is not satisfied by such tricks as scaling the sup norm, or multiplying by some fixed function which is bounded away from zero, as I suggested in an answer I've since deleted).
Here's an approach which produces an equivalent but nonisometric norm: Take the unit ball of the sup norm, add a suitable$^\dagger$ point $f_0$ (and its negative), take convex hull, and consider the result to be the unit ball of a new norm.  The new norm is equivalent to the sup norm with the formal identity map (with isomorphism constant $\le\|f_0\|$), but the unit ball of the sup norm has only two extreme points$^\ast$ (the constants $\pm 1$) while the unit ball of the new norm has at least four (the constants $\pm 1$, and $\pm f_0$).
$^\dagger$ suitable point $f_0$: I think it's sufficient to take any function for which there exist $x$ with $|f_0(x)|>1$ (so that $f_0$ will be an extreme point of the new unit ball) and there exist $x$ with $|f_0(x)|<1$ (so that the constants $\pm 1$ remain extreme points as well).  But I haven't checked this very carefully.
$^\ast$ unit ball of the sup norm has only two extreme points: Any function with sup norm 1, other than the constants $\pm 1$, can be perturbed up and down in a region away from where it attains its norm, and then written as an average of those perturbations.


TITLE: Determine whether or not the following sequence converges. If so, find the limit.
QUESTION [1 upvotes]: Im not sure if I am evaluating the convergence of the following sequence correctly, and I am unsure of how to determine the limit. All help is greatly appreciated.
$a_n = {1 \over (\sqrt{n^2-1}-\sqrt{n^2 +n})}$
My attempt at the convergence:
$ {1 \over (\sqrt{n^2-1}-\sqrt{n^2 +n})} < {1 \over n} $
$ \lim_{x\to \infty} {1 \over n} = 0$
Therefore, ${1 \over (\sqrt{n^2-1}-\sqrt{n^2 +n})}$ converges to as $ {n \to \infty} $

REPLY [1 votes]: You wrote
$$\frac1{\sqrt{n^2+1}-\sqrt{n^2+n}}\le\frac1n$$
which is true...but perhaps not in the way you meant: your secuence is negative, so if you want to use the squeeze theorem this is not the way.
Kola's answer shows you the way to reach the actual limit.

REPLY [0 votes]: We have: $\sqrt{n^2-1} - \sqrt{n^2+n} = (\sqrt{n^2-1} - n) - (-n +\sqrt{n^2+n}) = -\dfrac{1}{\sqrt{n^2-1} + n} - \dfrac{1}{\sqrt{1+\dfrac{1}{n}} + 1} \to -\dfrac{1}{2}$. Thus $a_n \to -2$.


TITLE: Setting the volume of a torus with integral
QUESTION [0 upvotes]: The disk in the $xz$ plane with center at $(x,z) = (R,0)$ and radius $r$ $(0< r < R)$ is rotated about the $z$-axis to generate a torus in $R^{3}$. Setup the integral to determine the volum of the generated torus. Slice the circle in narrow band and rotate those about the $z$-axis
the band are of height $2\sqrt{r^{2}-(x-R)^{2}}$ and thus a volume of
$$\Delta V = 2\pi x \cdot2\sqrt{r^{2}-(x-R)^{2}}\, \Delta x$$
This leads to the total volume
$$V = \int_{R-r}^{R+r} 2\pi x \cdot 2 \sqrt{r^{2}-(x-R)^{2}}\,\mathrm dx = 4\pi \int_{R-r}^{R+r} x \cdot2 \sqrt{r^{2}-(x-R)^{2}}\,\mathrm dx$$
using $u = x-R$ we get :
$$V = 4\pi \int_{-r}^{+r} (u + R) \sqrt{r^{2}-(u)^{2}}\,\mathrm du \tag{1}\label{eq1}$$
And then we get
$$V = 2\pi \int_{-r}^{+r} R\cdot2 \sqrt{r^{2}-(u)^{2}}\,\mathrm du \tag{2}\label{eq2}$$
$$V = 2\pi R \cdot\pi r^{2} \tag{3}\label{eq3}$$
What I don't understand is how we get from $(1)$ to $(2)$ and $(2)$ to $(3)$.

REPLY [1 votes]: $(1)\to(2):$ $u\sqrt{r^2-u^2}$ is an odd function, which means
$$\int_{-r}^r u\sqrt{r^2-u^2}\,\mathrm du=0.$$
Therefore,
\begin{align*}
V=4\pi\int_{-r}^r (u+R)\sqrt{r^2-u^2}\,\mathrm du&=\underbrace{4\pi\int_{-r}^r u\sqrt{r^2-u^2}\,\mathrm du}_{=0}+4\pi\int_{-r}^r R\sqrt{r^2-u^2}\,\mathrm du\\
&=2\pi R\int_{-r}^r 2\sqrt{r^2-u^2}\,\mathrm du.
\end{align*}
$(2)\to(3):$ Notice $y=\sqrt{r^2-x^2}$ is a semi-circle above the $x$-axis with radius $r$. If we consider the geometric meaning of the integral, the following integral is exactly the area of a circle with radius $r$:
$$\int_{-r}^r 2y\,\mathrm dx=\int_{-r}^r 2\sqrt{r^2-x^2}\,\mathrm dx=\pi r^2.$$
Therefore,
$$2\pi R\int_{-r}^r 2\sqrt{r^2-u^2}\,\mathrm du=2\pi R\int_{-r}^r 2\sqrt{r^2-x^2}\,\mathrm dx=2\pi R\cdot\pi r^2.$$


TITLE: Prove the limit of the following complex sequence equals $ 1 $ or $ -1$
QUESTION [0 upvotes]: Let $z_{n+1}=0.5 \cdot (z_n + \frac{1}{z_n} ) $ for all natural $n$. Assume that $z_0$ is non-zero complex numer. Show, that limit of such sequence equals $ 1$ if $z_0$ Has positive real part and $-1$ if its real part is negative. I'm not even sure if it is true, Banach contraction principle doesn't apply in this particular case... I'm literally clueless.

REPLY [1 votes]: Hint
$$z_{n+1}=\frac 12 \left(z_n + \frac{1}{z_n} \right)\implies z_{n+1}=z_n-\frac{z_n^2-1}{2z_n} $$ This is the iterative scheme for the solution of $z^2-1=0$ using Newton method.


TITLE: Prove that to each $\epsilon >0$, there exists a $\delta >0$ so that the Lebesgue integral...
QUESTION [14 upvotes]: Suppose $f$ is in $L^1$ space of $\mu$, where $\mu$ is the Lebesgue measure. Prove that to each $\epsilon >0$, there exists a $\delta >0$ so that the Lebesgue integral of the absolute value of $f$ is less than $\epsilon$ (over the set $E$) whenever the measure of $E$ is less than $\delta$.
I know that the integral of $f$ is finite. So for arbitrary $f$ we can find the integral of $f$ to be less than some $\epsilon$, but how would I connect that with the measure of $E$ less than $\delta$. Any help would be welcome.

REPLY [16 votes]: The first thing you should understand is that "less than some $\epsilon$" will not cut it. You don't get to choose $\epsilon$ — it will be given to you, and it can be  any positive number whatsoever. You must be prepared  to produce $\delta>0$ (based on the $\epsilon$) such that the implication
$$\mu(E)<\delta \implies \int_E |f|\,d\mu<\epsilon \tag{goal}$$
will hold.
Let's begin with something easy.
Easy case: $f$ is bounded, that is, there is a number $M$  such that $|f|\le M$ (almost everywhere). Then
$$\int_E |f|\,d\mu \le \int_E M\,d\mu = M\,\mu(E) < M\delta \tag1$$
Since we want to conclude with "$\int_E |f|\,d\mu<\epsilon$", the right choice of $\delta$ is $\delta=\epsilon/M$. Now you are prepared for the challenge: no matter what $\epsilon$ someone throws at you, you choose $\delta=\epsilon/M$ and (goal) is achieved.
General case. If $f$ is not bounded, we can still introduce a bounded function into consideration, namely $|f|_M(x)=\min(|f(x)|,M)$ (here $M$ is ours to choose). This is called the truncation of $|f|$ at level $M$: if you don't see why, picture the graphs of $|f|$ and of $|f|_M$.
How to deal with $|f|_M$ is more or less clear (the easy case).  What to do with the rest, namely $|f|-|f|_M$? Turns out that by choosing $M$  sufficiently large, we can make $\int_{\mathbb R} (|f|-|f|_M)\,d\mu$ is as small as we wish (I give a hint below). So, the strategy is:

Pick $M$ so that $\int_{\mathbb R} (|f|-|f|_M)\,d\mu<\epsilon/2$.
Pick $\delta = (\epsilon/2)/M$; this ensures that $\int_E |f|_M\,d\mu<\epsilon/2$.

Promised hint: since $|f(x)|=\lim_{M\to\infty} |f(x)|_M$ for almost every $x$, by the monotone convergence theorem $$\int_{\mathbb R} |f| = \lim_{M\to\infty}\int_{\mathbb R} |f|_M\,d\mu $$


TITLE: Find ALL solutions to $23x + 13y = 275$ when $x,y \ge 0$
QUESTION [1 upvotes]: Find ALL solutions to $23x + 13y = 275$ when $x,y \in \Bbb N$.
How can I approach this exercise? do I need to use GCD?
I got to $y = 23k + 16, x = 13t + 4$, when $k,t \in \Bbb Z$

REPLY [0 votes]: In some cases it is possible to have an easy brute force search.
Here for instance the divisibility by $5$ brings great thnigs.
$23x+13y\equiv3x+3y\equiv3(x+y)\equiv275\equiv0\pmod5$ and since $3\land5=1$ we get $x+y\equiv0\pmod 5\iff x+y=5k$.
So $23x+13y=65k+10x=275$, obviously $k$ is also odd for $10$ to divide $275-65k$
$
\begin{array}{c|cccc}
k & 65k & 275-65k & x & y\\
\hline
1 & 65 & 210 & 21 & -16\\
3 & 195 & 80 & 8 & 7\\
\end{array}
$
Brute force is reduced to only 2 choices!


text:
\section{Potential well $\boldsymbol{PW}$}\label{08/07/02/23:51}
In this section, we shall discuss fundamental properties of the sets $PW$, $PW_{-}$ and $PW_{+}$. In particular, we will prove that these sets are invariant under the flow defined by the equation (\ref{08/05/13/8:50}) (see Propositions \ref{09/06/21/18:53}, \ref{09/06/21/19:28} and \ref{08/05/26/10:57}). Moreover, we  prove  Theorem \ref{08/10/20/5:21} here.
\par
We begin with the following fact:
\begin{lemma}
\label{09/06/21/15:50}
The set $PW$ does not contain any function $f$ with $\mathcal{K}(f)=0$, i.e.,   \begin{equation}\label{10/02/16/20:47}
\left\{ f \in H^{1}(\mathbb{R}^{d}) \bigm| \mathcal{K}(f)=0 \right\}
\cap PW =\emptyset ,
\end{equation}
so that
\begin{equation}\label{10/02/16/20:48}
PW=PW_{+}\cup PW_{-}, \qquad PW_{+}\cap PW_{-}=\emptyset.
\end{equation}
\end{lemma}
\begin{proof}[Proof of Lemma \ref{09/06/21/15:50}]
Let $f$ be a function in $H^{1}(\mathbb{R}^{d})$ with $\mathcal{K}(f)=0$. Then,  the condition $\mathcal{K}(f)=0$ leads to that
\begin{equation}\label{09/12/15/8:28}
\mathcal{H}(f)=\frac{d(p-1)-4}{d(p-1)}\left\| \nabla f \right\|_{L^{2}}^{2}.
\end{equation}
Using (\ref{09/12/15/8:28}) and the definitions of $N_{2}$ (see (\ref{08/07/02/23:24})) and $\widetilde{N}_{2}$  (see (\ref{09/04/29/14:31})), we obtain that
\begin{equation}\label{10/01/06/18:01}
\begin{split}
\widetilde{\mathcal{N}}_{2}(f)&=
\sqrt{ \frac{d(p-1)-4}{d(p-1)}
}^{\frac{d}{2}(p-1)-2}
\mathcal{N}_{2}(f)
\\
& \ge
\sqrt{ \frac{d(p-1)-4}{d(p-1)} }^{\frac{d}{2}(p-1)-2}N_{2}=\widetilde{N}_{2}.
\end{split}
\end{equation}
Hence, it follows from (\ref{08/06/15/14:38}) that $f \not \in PW$.
\par
We immediately obtain (\ref{10/02/16/20:48}) from (\ref{10/02/16/20:47}) and the definitions of $PW_{+}$ and $PW_{-}$ (see (\ref{10/02/13/14:44}) and (\ref{10/02/13/14:45})).
\end{proof}
In the next lemma, we consider a path constructed from the ground state:
\begin{lemma}\label{08/06/14/23:37}
Let $Q_{\omega}$ be the ground state of the equation (\ref{08/05/13/11:22}) for  $w>0$. We consider a path $\Gamma_{\omega} \colon [0,\infty) \to H^{1}(\mathbb{R}^{d})$ given by $\Gamma_{\omega}(s)=sQ_{\omega}$ for $s\ge 0$. Then, $\Gamma_{\omega}$ is continuous and satisfies that
\begin{align}\label{10/02/16/20:56}
& \Gamma_{\omega}(s)\in PW_{+}
\quad
\mbox{for all $s \in (0,1)$},
\\[6pt]
\label{10/02/16/21:03}
&\Gamma_{\omega}(1)=Q_{\omega} \notin PW=PW_{+}\cup PW_{-},
\\[6pt]
\label{10/02/16/20:58}
&\Gamma_{\omega}(s) \in \left\{f \in H^{1}(\mathbb{R}^{d}) \bigm| \mathcal{H}(f) > 0\right\} \cap PW_{-}
\quad
\mbox{for all  $s \in \left(1, \ \left\{ \frac{d(p-1)}{4}\right\}^{\frac{1}{p-1}}\right)$},
\\[6pt]
\label{10/02/16/20:57}
&\Gamma_{\omega}(s)
 \in \left\{f \in H^{1}(\mathbb{R}^{d})\setminus \{0\} \bigm| \mathcal{H}(f) \le 0\right\} \subset PW_{-}
\quad
\mbox{for all $s \in \left[ \left\{ \frac{d(p-1)}{4}\right\}^{\frac{1}{p-1}}, \ \infty\right)$}.
\end{align}
In particular, $PW_{-}\neq \emptyset$ and $PW_{+} \neq \emptyset$.
\end{lemma}
\begin{proof}[Proof of Lemma \ref{08/06/14/23:37}] The continuity of $\Gamma_{\omega}\colon (0,\infty)\to H^{1}(\mathbb{R}^{d})$ is obvious from its definition. We shall prove the properties (\ref{10/02/16/20:56})--(\ref{10/02/16/20:57}). As stated in (\ref{10/03/27/10:23}) (see also (\ref{09/12/16/20:17})), the ground state $Q_{\omega}$ satisfies that $\mathcal{K}(Q_{\omega})=0$, which immediately yields that
\begin{equation}\label{10/03/27/10:02}
\left\| \nabla Q_{\omega} \right\|_{L^{2}}^{2}
=
\frac{d(p-1)}{2(p+1)}\left\| Q_{\omega} \right\|_{L^{p+1}}^{p+1}.
\end{equation}
Using (\ref{10/03/27/10:02}), we obtain that
\begin{equation}\label{09/09/14/0:33}
\mathcal{H}(\Gamma_{\omega}(s))=s^{2}\left( 1-\frac{4}{d(p-1)}s^{p-1} \right)\|\nabla Q_{\omega} \|_{L^{2}}^{2}.
\end{equation}
This formula gives us that
\begin{align}
\label{10/03/15/21:47}
&\mathcal{H}(\Gamma_{\omega}(s))>0
\quad
\mbox{for all  $s \in \left(0, \ \left\{ \frac{d(p-1)}{4}\right\}^{\frac{1}{p-1}}\right)$},
\\[6pt]
\label{10/03/15/21:35}
&\mathcal{H}(\Gamma_{\omega}(s)) \le 0
\quad
\mbox{for all $s \in \left[ \left\{ \frac{d(p-1)}{4}\right\}^{\frac{1}{p-1}}, \ \infty\right)$}.
\end{align}
Similarly, we can verify that
\begin{align}\label{10/03/15/21:20}
&\mathcal{K}(\Gamma_{\omega}(s))>0 \quad \mbox{for all $s \in (0,1)$},
\\[6pt]
\label{10/03/15/21:26}
&\mathcal{K}(\Gamma_{\omega}(1))=\mathcal{K}(Q_{\omega})=0,
\\[6pt]
\label{10/03/15/21:27}
&\mathcal{K}(\Gamma_{\omega}(s))<0 \quad \mbox{for all $s\in (1,\infty)$}.
\end{align}
Now, it follows from (\ref{09/09/14/0:33}) and (\ref{10/03/27/10:23}) that
\begin{equation}\label{10/03/15/21:11}
\widetilde{\mathcal{N}}_{2}(\Gamma_{\omega}(s))=s^{p-1}\left( 1-\frac{4}{d(p-1)}s^{p-1} \right)^{\frac{d}{4}(p-1)-1}
\hspace{-7pt}
N_{2}
\qquad
\mbox{for all $s \in \left(0,\  \left\{ \frac{d(p-1)}{4}\right\}^{\frac{1}{p-1}}\right)$}.
\end{equation}
This relation (\ref{10/03/15/21:11}) shows that
\begin{equation}\label{10/03/15/21:22}
\widetilde{\mathcal{N}}_{2}(\Gamma_{\omega}(s))< \widetilde{\mathcal{N}}_{2}(\Gamma_{\omega}(1))=\widetilde{N}_{2}
\qquad
\mbox{for all $s \in \left(0,\  \left\{ \frac{d(p-1)}{4}\right\}^{\frac{1}{p-1}}\right)\setminus\{1\}$}.
\end{equation}
Hence, (\ref{10/03/15/21:47}) and (\ref{10/03/15/21:22}), with the help of (\ref{08/06/15/14:38}), lead us to the conclusion that
\begin{equation}\label{10/03/15/22:13}
\Gamma_{\omega}(s) \in PW
\qquad
\mbox{for all $s \in \left(0,\  \left\{ \frac{d(p-1)}{4}\right\}^{\frac{1}{p-1}}\right) \setminus \{1\}$}
.
\end{equation}
Then, (\ref{10/02/16/20:56}) follows from (\ref{10/03/15/21:20}).
Moreover, (\ref{10/02/16/20:58}) follows from  (\ref{10/03/15/21:47}) and (\ref{10/03/15/21:27}). The second claim (\ref{10/02/16/21:03}) is a direct consequence of (\ref{10/03/15/21:26}) and Lemma \ref{09/06/21/15:50}.
\par
It remains to prove (\ref{10/02/16/20:57}). Since $\mathcal{B}(f)\ge 0$ and $\mathcal{K}(f)<\mathcal{H}(f)$ for all $f \in H^{1}(\mathbb{R}^{d})\setminus \{0\}$, we have a relation
\begin{equation}\label{10/03/15/21:34}
\left\{f \in H^{1}(\mathbb{R}^{d})\setminus \{0\} \bigm| \mathcal{H}(f) \le 0\right\} \subset PW_{-}.
\end{equation}
Then, (\ref{10/02/16/20:57}) immediately follows from (\ref{10/03/15/21:35}) and this relation (\ref{10/03/15/21:34}).
\end{proof}

Now, we are in a position to prove Theorem \ref{08/10/20/5:21}.

\begin{proof}[Proof of Theorem \ref{08/10/20/5:21}]
We consider the path $\Gamma_{\omega}$ given in Lemma \ref{08/06/14/23:37}.
Then, (\ref{10/02/16/20:56}) and the continuity of $\Gamma_{\omega}$ yield that\begin{equation}\label{10/03/28/9:20}
\Gamma_{\omega}(s) \in PW_{+}
\quad
\mbox{for all $s \in (0,1)$},
\qquad
\lim_{s\uparrow 1}\Gamma_{\omega}(s) =\Gamma_{\omega}(1)=Q_{\omega}
\quad
\mbox{strongly in $H^{1}(\mathbb{R}^{d})$}.
\end{equation}
Hence, we have $Q_{\omega} \in \overline{PW_{+}}$. Similarly, (\ref{10/02/16/20:58}) and the continuity of $\Gamma_{\omega}$ show that $Q_{\omega} \in \overline{PW_{-}}$.
\end{proof}

We find from the following lemma that $PW_{+}$ has a ``foliate structure''.

\begin{lemma}\label{08/09/01/23:45}
For all $\eta \in (0,\widetilde{N}_{2})$ and $\alpha>0$, there exists $f \in PW_{+}$ such that
 $\widetilde{\mathcal{N}}_{2}(f)=\eta$
and
 $\left\| f \right\|_{L^{2}}=\alpha$.
\end{lemma}

\begin{proof}[Proof of Lemma \ref{08/09/01/23:45}]
We construct a desired function from the continuous path $\Gamma_{\omega}\colon [0,\infty) \to H^{1}(\mathbb{R}^{d})$ ($\omega>0$) given in Lemma \ref{08/06/14/23:37}. Let us remind you that
\begin{align}
\label{10/03/27/15:09}
&\Gamma_{\omega}(s) \in PW_{+}
\quad
\mbox{for all $s \in (0,1)$},
\\[6pt]
&
\label{10/03/27/15:19}
\widetilde{\mathcal{N}}_{2}(\Gamma_{\omega}(s))=s^{p-1}\left( 1-\frac{4}{d(p-1)}s^{p-1} \right)^{\frac{d}{4}(p-1)-1}
N_{2}
\quad
\mbox{for all $s \in (0, 1)$},
\\[6pt]
&
\label{10/03/27/15:37}
\widetilde{\mathcal{N}}_{2}(\Gamma_{\omega}(0))=0,
\qquad
\widetilde{\mathcal{N}}_{2}(\Gamma_{\omega}(1))=\widetilde{N}_{2}
\end{align}
(see (\ref{10/03/15/21:11}) for (\ref{10/03/27/15:19})). Using (\ref{10/03/27/15:19}), we find  that $\widetilde{\mathcal{N}}_{2}(\Gamma_{\omega}(s))$ is continuous and monotone increasing with respect to $s$ on $[0,1]$. Hence, the intermediate value theorem, together with (\ref{10/03/27/15:37}), shows that for any $\eta \in (0, \widetilde{N}_{2})$, there exists $s_{\eta} \in (0,1)$ such that $\widetilde{\mathcal{N}}_{2}(\Gamma_{\omega}(s_{\eta}))=\eta$. Moreover, (\ref{10/03/27/15:09}) gives us that $\Gamma_{\omega}(s_{\eta}) \in PW_{+}$. We put $f=\Gamma_{\omega}(s_{\eta})$ and consider the scaled function $f_{\lambda}:=\lambda^{\frac{2}{p-1}}f(\lambda \cdot)$ for $\lambda>0$. Then, it is easy to see that
\begin{equation}
\label{10/03/27/16:10}
f_{\lambda} \in PW_{+},
\quad
\widetilde{\mathcal{N}}_{2}(f_{\lambda})=\widetilde{\mathcal{N}}_{2}(f)=
\eta,
\quad
\left\| f_{\lambda} \right\|_{L^{2}}=\lambda^{-\frac{d(p-1)-4}{2(p-1)}}\left\| f\right\|_{L^{2}}
\quad
\mbox{for all $\lambda>0$}.
\end{equation}
Hence, for any $\alpha>0$, $f_{\lambda}$ with $\lambda =\left( \frac{ \left\| f \right\|_{L^{2}}}{\alpha}\right)^{\frac{2(p-1)}{d(p-1)-4}}$ is what we want.
\end{proof}

Finally, we give the invariance results of the sets $PW$, $PW_{+}$ and $PW_{-}$  under the flow defined by the equation (\ref{08/05/13/8:50}):

\begin{lemma}[Invariance of $PW$]\label{09/06/21/18:53}
Let $\psi_{0} \in PW$ and $\psi$ be the corresponding solution to
the equation (\ref{08/05/13/8:50}). Then, we have that
\[
\psi(t) \in PW
\quad
\mbox{for all $t \in I_{\max}$}.
\]
\end{lemma}
\begin{proof}[Proof of Lemma \ref{09/06/21/18:53}]
This lemma immediately follows from the mass and energy conservation laws (\ref{08/05/13/8:59}) and (\ref{08/05/13/9:03}).
\end{proof}

\begin{proposition}[Invariance of $PW_{+}$]
\label{09/06/21/19:28}
Let $\psi_{0} \in PW_{+}$ and $\psi$ be the corresponding solution to the equation (\ref{08/05/13/8:50}). Then, $\psi$ exists globally in time and satisfies the followings:
\begin{equation}\label{10/01/06/21:29}
 \psi(t) \in PW_{+}
\quad
\mbox{for all  $t \in \mathbb{R}$},
\end{equation}
\begin{equation}\label{09/06/21/21:58}
\left\| \nabla \psi (t) \right\|_{L^{2}}^{2}
< \frac{d(p-1)}{d(p-1)-4}\mathcal{H}(\psi_{0})
\quad
\mbox{for all  $t \in \mathbb{R}$},
\end{equation}
\begin{equation}\label{09/06/21/19:30}
\mathcal{K}(\psi(t)) >
\left( 1-\frac{\widetilde{N}_{2}(\psi_{0})}{\widetilde{N}_{2}}\right)
\mathcal{H}(\psi_{0})
\quad
\mbox{for all  $t \in \mathbb{R}$}.
\end{equation}
\end{proposition}

\begin{proof}[Proof of Proposition \ref{09/06/21/19:28}]
We first prove the invariance of $PW_{+}$ under the flow defined by (\ref{08/05/13/8:50}). With the help of Lemma \ref{09/06/21/18:53}, it suffices to show that
\begin{equation}\label{09/08/15/18:45}
\mathcal{K}(\psi(t))>0
\quad
\mbox{for all $t \in I_{\max}$}.
\end{equation}
Supposing the contrary that (\ref{09/08/15/18:45}) fails, we can take $t_{0}\in I_{\max}$ such that
\begin{equation}\label{10/03/27/16:44}
\mathcal{K}(\psi(t_{0}))= 0.
\end{equation}
Then, (\ref{10/03/27/16:44}) and the energy conservation law (\ref{08/05/13/9:03}) yield that
\begin{equation}\label{10/03/27/17:02}
\begin{split}
0&=\mathcal{K}(\psi(t_{0}))
\\[6pt]
&=\mathcal{H}(\psi(t_{0}))
- \frac{2}{p+1}\left\{ \frac{d}{4}(p-1)-1 \right\}\|\psi(t_{0})\|_{L^{p+1}}^{p+1}
\\[6pt]
&=
\mathcal{H}(\psi_{0}) - \frac{2}{p+1}\left\{ \frac{d}{4}(p-1)-1 \right\}
\frac{2(p+1)}{d(p-1)} \|\nabla \psi(t_{0})\|_{L^{2}}^{2}.
\end{split}
\end{equation}
Since $\psi_{0}\in PW_{+}\subset PW$, we have $\mathcal{H}(\psi_{0})<\mathcal{B}(\psi_{0})$. This inequality and (\ref{10/03/27/17:02}) lead us to that
\begin{equation}\label{10/03/27/17:07}
0 < \mathcal{B}(\psi_{0})-\frac{d(p-1)-4}{d(p-1)} \|\nabla \psi(t_{0})\|_{L^{2}}^{2},
\end{equation}
which is  equivalent to
\begin{equation}\label{10/03/27/17:18}
\|\nabla \psi(t_{0})\|_{L^{2}}^{2}
<
\frac{d(p-1)}{d(p-1)-4}\mathcal{B}(\psi_{0})
=
\left( \frac{ N_{2}}{\|\psi(t_{0})\|_{L^{2}}^{p+1-\frac{d}{2}(p-1)}}\right)^{\frac{4}{d(p-1)-4}}
.
\end{equation}
Dividing the both sides of (\ref{10/03/27/17:18}) by $\|\nabla \psi(t_{0})\|_{L^{2}}^{2}$,
we obtain that
\begin{equation}\label{10/03/27/17:24}
1< \left( \frac{N_{2}}{\mathcal{N}_{2}(\psi(t_{0}))}\right)^{\frac{4}{d(p-1)-4}}.
\end{equation}
On the other hand, (\ref{10/03/27/16:44}), together with the definition of $N_{2}$, leads us to that $\mathcal{N}_{2}(\psi(t_{0}))\ge N_{2}$, so that
\begin{equation}\label{10/03/27/17:26}
\left( \frac{N_{2}}{\mathcal{N}_{2}(\psi(t_{0}))}\right)^{\frac{4}{d(p-1)-4}}
\le 1.
\end{equation}
This inequality (\ref{10/03/27/17:26}) contradicts (\ref{10/03/27/17:24}). Thus, (\ref{09/08/15/18:45}) must hold.
\par
Once we obtain (\ref{09/08/15/18:45}), we can easily obtain the following uniform bound:
\begin{equation}\label{10/01/06/23:07}
\left\| \nabla \psi (t) \right\|_{L^{2}}^{2}
< \frac{d(p-1)}{d(p-1)-4}\mathcal{H}(\psi_{0})
\quad
\mbox{for all $t \in I_{\max}$}.
\end{equation}
Indeed, it follows from the energy conservation law (\ref{08/05/13/9:03}) and (\ref{09/08/15/18:45}) that
\begin{equation}\label{10/03/28/10:15}
\begin{split}
\mathcal{H}(\psi_{0})&=\mathcal{H}(\psi(t))=
\left\| \nabla \psi(t)\right\|_{L^{2}}^{2}-
\frac{2}{p+1}\left\| \psi(t) \right\|_{L^{p+1}}^{p+1}
\\[6pt]
&>\frac{d(p-1)-4}{d(p-1)}\left\| \nabla \psi(t)\right\|_{L^{2}}^{2}
\quad
\mbox{for all $t \in I_{\max}$}.
\end{split}
\end{equation}

The estimate (\ref{10/01/06/23:07}), together with the sufficient condition for the blowup (\ref{10/01/27/11:26}), leads us to that $I_{\max}=\mathbb{R}$. Hence,  (\ref{10/01/06/21:29}) and (\ref{09/06/21/21:58}) follow from (\ref{09/08/15/18:45}) and (\ref{10/01/06/23:07}), respectively.
\par
It remains to prove (\ref{09/06/21/19:30}). The Gagliardo-Nirenberg inequality (\ref{08/05/13/15:45}) gives us that
\begin{equation}\label{10/03/28/10:21}
\begin{split}
\mathcal{K}(\psi(t))
&
=\left\| \nabla \psi(t) \right\|_{L^{2}}^{2}
-\frac{d(p-1)}{2(p+1)}\left\| \psi(t) \right\|_{L^{p+1}}^{p+1}
\\[6pt]
&\ge \left\| \nabla \psi(t) \right\|_{L^{2}}^{2}
-\frac{d(p-1)}{2(p+1)}\frac{1}{N_{3}}\mathcal{N}_{2}(\psi(t))
\left\| \nabla \psi(t)\right\|_{L^{2}}^{2}.
\end{split}
\end{equation}
Moreover, this inequality (\ref{10/03/28/10:21}) and the relation $N_{3}=\frac{d(p-1)}{2(p+1)}N_{2}$ (see (\ref{08/05/13/11:25})) yield that
\begin{equation}\label{08/12/16/18:09}
\begin{split}
\mathcal{K}(\psi(t))
&
\ge \left\| \nabla \psi(t) \right\|_{L^{2}}^{2}
-\frac{\mathcal{N}_{2}(\psi(t))}{N_{2}}
\left\| \nabla \psi(t) \right\|_{L^{2}}^{2}
.
\end{split}
\end{equation}
Here, it follows from (\ref{09/06/21/21:58}) that
\begin{equation}
\label{09/08/15/19:08}
\begin{split}
\mathcal{N}_{2}(\psi(t))&<
\sqrt{\frac{d(p-1)}{d(p-1)-4}}^{\frac{d}{2}(p-1)-2}\hspace{-1cm}
\left\|\psi_{0} \right\|_{L^{2}}^{p+1-\frac{d}{2}(p-1)}
\sqrt{\mathcal{H}(\psi_{0})}^{\frac{d}{2}(p-1)-2}
\\
&= \sqrt{\frac{d(p-1)}{d(p-1)-4}}^{\frac{d}{2}(p-1)-2}\widetilde{\mathcal{N}}_{2}(\psi_{0}).
\end{split}
\end{equation}
Hence, combining (\ref{08/12/16/18:09}), (\ref{09/08/15/19:08}) and $\mathcal{H}(\psi(t))>\mathcal{K}(\psi(t))>0$, we obtain that\begin{equation}\label{10/03/10:27}
\begin{split}
\mathcal{K}(\psi(t))&>
\left\| \nabla \psi(t) \right\|_{L^{2}}^{2}
-
\sqrt{\frac{d(p-1)}{d(p-1)-4}}^{\frac{d}{2}(p-1)-2}
\frac{\mathcal{\widetilde{\mathcal{N}}}_{2}(\psi_{0})}{N_{2}}
\left\| \nabla \psi(t) \right\|_{L^{2}}^{2}
\\[6pt]
&
=
\left( 1-\frac{\widetilde{\mathcal{N}}_{2}(\psi_{0})}{\widetilde{N}_{2}}\right)
\left\| \nabla \psi(t)\right\|_{L^{2}}^{2}
\\[6pt]
&\ge
\left( 1-\frac{\widetilde{\mathcal{N}}_{2}(\psi_{0})}{\widetilde{N}_{2}}\right) \mathcal{H}(\psi(t))
=
\left( 1-\frac{\widetilde{\mathcal{N}}_{2}(\psi_{0})}{\widetilde{N}_{2}}\right) \mathcal{H}(\psi_{0})
.
\end{split}
\end{equation}
\end{proof}

\begin{proposition}[Invariance of $PW_{-}$]\label{08/05/26/10:57}
Let $\psi_{0} \in PW_{-}$ and $\psi$ be the corresponding solution to the equation (\ref{08/05/13/8:50}). Then, we have
\begin{equation}\label{09/06/21/18:52}
\psi(t) \in PW{-},
\quad
\mathcal{K}(\psi(t)) < -\varepsilon_{0}
\qquad
\mbox{for all $t \in I_{\max}$},
\end{equation}
where $\varepsilon_{0}=\mathcal{B}(\psi_{0})-\mathcal{H}(\psi_{0})>0$.
\end{proposition}
\begin{proof}[Proof of Proposition \ref{08/05/26/10:57}]
By the virtue of Lemma \ref{09/06/21/18:53}, it suffices to show that
\begin{equation}\label{10/03/28/10:48}
\mathcal{K}(\psi(t)) < -\varepsilon_{0}
\qquad
\mbox{for all $t \in I_{\max}$}.
\end{equation}
We first prove that
\begin{equation}\label{10/03/28/11:09}
\mathcal{K}(\psi_{0})<-\varepsilon_{0}.
\end{equation}
Supposing the contrary that $-\varepsilon_{0}\le \mathcal{K}(\psi_{0})$,
 we have
\begin{equation}\label{10/03/28/10:54}
\begin{split}
0&\le \mathcal{K}(\psi_{0})+\varepsilon_{0}
\\[6pt]
&=\mathcal{H}(\psi_{0})+\varepsilon_{0}
- \frac{2}{p+1}\left\{ \frac{d}{4}(p-1)-1 \right\}\|\psi_{0}\|_{L^{p+1}}^{p+1}
\\[6pt]
&=
\mathcal{B}(\psi_{0})
-
\frac{2}{p+1}\left\{ \frac{d}{4}(p-1)-1 \right\}\|\psi_{0}\|_{L^{p+1}}^{p+1}
.
\end{split}
\end{equation}
Moreover, it follows from (\ref{10/03/28/10:54}) and $\mathcal{K}(\psi_{0})<0$ that
\begin{equation}\label{10/03/28/10:55}
\begin{split}
0&
< \mathcal{B}(\psi_{0}) - \frac{2}{p+1}\left\{ \frac{d}{4}(p-1)-1 \right\}
\frac{2(p+1)}{d(p-1)}  \|\nabla \psi_{0}\|_{L^{2}}^{2}
\\[6pt]
&=
\mathcal{B}(\psi_{0})-\frac{d(p-1)-4}{d(p-1)} \|\nabla \psi_{0}\|_{L^{2}}^{2}
,
\end{split}
\end{equation}
so that
\begin{equation}\label{10/03/28/11:02}
\|\nabla \psi_{0}\|_{L^{2}}^{2}
< \frac{d(p-1)}{d(p-1)-4}\mathcal{B}(\psi_{0})
= \left( \frac{N_{2}}{\|\psi_{0}\|_{L^{2}}^{p+1-\frac{d}{2}(p-1)}}\right)^{\frac{4}{d(p-1)-4}}
.
\end{equation}
Dividing the both sides of (\ref{10/03/28/11:02}) by $\|\nabla \psi_{0}\|_{L^{2}}^{2}$, we obtain that
\begin{equation}\label{10/03/28/11:05}
1< \left( \frac{ N_{2}}{\mathcal{N}_{2}(\psi_{0})}\right)^{\frac{4}{d(p-1)-4}}
.
\end{equation}
On the other hand, since $\mathcal{K}(\psi_{0})<0$, the definition of $N_{2}$ (see (\ref{08/07/02/23:24})) implies that
\begin{equation}\label{10/03/28/11:06}
\left( \frac{ N_{2}}{\mathcal{N}_{2}(\psi_{0})}\right)^{\frac{4}{d(p-1)-4}}
\le 1 ,
\end{equation}
which contradicts (\ref{10/03/28/11:05}). Hence, we have proved (\ref{10/03/28/11:09}).
\par
Next, we prove (\ref{10/03/28/10:48}). Suppose the contrary that (\ref{10/03/28/10:48}) fails. Then, it follows from (\ref{10/03/28/11:09}) and $\psi \in C(I_{\max};H^{1}(\mathbb{R}^{d}))$
 that there exists $t_{1}\in I_{\max} \setminus \{0\}$ such that $-\varepsilon_{0}= \mathcal{K}(\psi(t_{1}))$. This relation and the energy conservation law (\ref{08/05/13/9:03}) lead us to that
\begin{equation}\label{10/03/28/11:17}
\begin{split}
0&= \mathcal{K}(\psi(t_{1}))+\varepsilon_{0}
\\[6pt]
&=\mathcal{H}(\psi_{0})+\varepsilon_{0}
- \frac{2}{p+1}\left\{ \frac{d}{4}(p-1)-1 \right\}\|\psi(t_{1})\|_{L^{p+1}}^{p+1}
\\[6pt]
&= \mathcal{B}(\psi_{0}) - \frac{2}{p+1}\left\{ \frac{d}{4}(p-1)-1 \right\}
\frac{2(p+1)}{d(p-1)}  \left( \|\nabla \psi(t_{1})\|_{L^{2}}^{2}+\varepsilon_{0}\right)
\\[6pt]
&< \mathcal{B}(\psi_{0})-\frac{d(p-1)-4}{d(p-1)} \|\nabla \psi(t_{1})\|_{L^{2}}^{2}
.
\end{split}
\end{equation}
Hence, we have that
\begin{equation}\label{10/03/28/11:20}
\begin{split}
\|\nabla \psi(t_{1})\|_{L^{2}}^{2}
&
<
\frac{d(p-1)}{d(p-1)-4}\mathcal{B}(\psi_{0})
=\left( \frac{N_{2}}{\|\psi_{0}\|_{L^{2}}^{p+1-\frac{d}{2}(p-1)}}\right)^{\frac{4}{d(p-1)-4}}
\\[6pt]
&=
\left( \frac{N_{2}}{\|\psi(t_{1})\|_{L^{2}}^{p+1-\frac{d}{2}(p-1)}}\right)^{\frac{4}{d(p-1)-4}}
,
\end{split}
\end{equation}
where we have used the mass conservation law (\ref{08/05/13/8:59}) to derive the last equality.  Then, an argument similar to the above yields a contradiction: Thus, we completed the proof.
\end{proof}