text:
\begin{document}
\title{A note on mean equicontinuity}
\author[J.~Qiu]{Jiahao Qiu}
\address[J.~Qiu]{Wu Wen-Tsun Key Laboratory of Mathematics, USTC, Chinese Academy of Sciences and
School of Mathematics, University of Science and Technology of China,
Hefei, Anhui, 230026, P.R. China}
\email{qiujh@mail.ustc.edu.cn}

\author[J.~Zhao ]{Jianjie Zhao}
\address[J.~Zhao]{Wu Wen-Tsun Key Laboratory of Mathematics, USTC, Chinese Academy of Sciences and
School of Mathematics, University of Science and Technology of China,
Hefei, Anhui, 230026, P.R. China}
\email{zjianjie@mail.ustc.edu.cn}

\date{\today}

\begin{abstract}
In this note, it is shown that several results concerning mean equicontinuity proved before for minimal systems are actually held
for general topological dynamical systems. Particularly, it turns out that a dynamical system is mean equicontinuous if and only if it is equicontinuous in the mean
if and only if it is Banach (or Weyl) mean equicontinuous if and only if its regionally proximal relation is equal to the Banach  proximal relation.

Meanwhile,
a relation is introduced such that the smallest closed invariant equivalence relation containing this relation induces the maximal mean equicontinuous factor
for any system.

\end{abstract}
\keywords{Mean equicontinuity, equicontinuity in the mean}
\subjclass[2010]{54H20, 37A25}
\maketitle

\section{Introduction}
Throughout this paper, \emph{a topological dynamical system} is a pair $(X,T)$, where $X$ is a
non-empty compact metric space with a metric $d$ and $T$ is a continuous map from $X$ to
itself.

We all know that equicontinuous systems have simple dynamical behaviors.
By the well known Halmos-von Neumann theorem, a transitive equicontinuous system is conjugate to
a minimal rotation on a compact abelian metric group,
and $(X,T,\mu)$ has discrete spectrum, where $\mu$ is the unique Haar measure on $X$.
In this note, we discuss the systems with equicontinuity in the mean sense.

Recall that a dynamical system $(X,T)$ is called \emph{mean equicontinuous} if for every $\ep>0$,
there exists a $\delta>0$ such that whenever $x,y\in X$ with $d(x,y)<\delta$, we have
\[\limsup_{n\to\infty} \bar d_n(x,y)<\ep,\ \text{where}\ \bar d_n(x,y)=\frac{1}{n}\sum_{i=0}^{n-1}d(T^ix,T^iy).\]

A notion called \emph{stable in the mean in the sense of Lyapunov} or simply \emph{mean-L-stable} is introduced by Fomin~\cite{F51}.
We call a dynamical system $(X,T)$ \emph{mean-L-stable}
if for every $\ep>0$, there is a $\delta>0$ such that $d(x,y)<\delta$
implies $d(T^nx,T^ny)<\ep$ for all $n\in\Z$ except a set of
upper density less than $\ep$.
Oxtoby \cite{O52}, Auslander \cite{A59} and Scarpellini \cite{S82} also studied mean-L-stable systems.
 It is easy to see that a dynamical system
is mean-L-stable if and only if it is mean equicontinuous. Answering an open question in \cite{S82}, it was proved
by Li, Tu and Ye in \cite{YL} that a minimal mean equicontinuous system has discrete spectrum. We refer to \cite{Felipe1, Felipe2, Felipe3, GLZ17, HL, Li} for further study
on mean equicontinuity and related subjects.

In the study of a dynamical system with bounded complexity (defined by using the mean metrics), recently
Huang, Li, Thouvenot, Xu and Ye \cite{HL} introduced a notion called \emph{equicontinuity in the mean}.

We say that a dynamical system $(X,T)$ is \emph{equicontinuous in the mean}
if for every $\ep>0$, there exists a $\delta>0$ such that
$\frac{1}{n}\sum_{i=0}^{n-1}d(T^ix,T^iy)<\ep$ for all $n\in \Z$ and
all $x,y\in X$ with $d(x,y)<\delta$.
It was proved in \cite{HL} that for a minimal system the notions of mean equicontinuity and
equicontinuity in the mean are equivalent.
In this note we will show that a dynamical system is mean equicontinuous
if and only if it is equicontinuous in the mean (Theorem \ref{general}).

In \cite{YL} the notion of Banach (or Weyl) mean equicontinuity was introduced, and the authors asked if for a minimal system
Banach mean equicontinuity is equal to mean equicontinuity. This question was answered positively in \cite{DG16}.
In this note we will show that in fact for any system the two notions are equivalent (Theorem \ref{thm:Weyl-mean-equ}). Moreover,
in \cite{YL} the authors showed that if $(X,T)$ is mean equicontinuous, then its regionally proximal relation is equal to the Banach proximal relation.
In this note we will prove that the converse statement is also valid (Theorem \ref{thm:transitive-mean-eq}).

Moreover, we define a notion called
\emph{regionally proximal relation in the mean}
and we show that the mean equicontinuous structure relation
is the smallest closed invariant equivalence relation that contains regionally proximal relation in the mean (Theorem \ref{MAX ME}).

\medskip

The note is organized as follows. In Section 2, the basic notions used in the note are introduced. In Section 3,
among other things we show that mean equicontinuity is equal to equicontinuity in the mean.
In Section 4, we prove that if the regionally proximal relation is equal to Banach proximal relation then the system is mean equicontinuous. In Section 5, we prove the equivalence of mean equicontinuity and Weyl mean equicontinuity.
In the final section, we discuss the question which relation induces the maximal mean equicontinuous factor.

\section{Preliminaries}
In this section we recall some notions and aspects of the theory of topological dynamical systems.
\subsection{Subsets of non-negative integers}
Let $\Z$ ($\N$, $\mZ$, respectively) be the set of all non-negative integers (positive integers, integers, respectively).

Let $F$ be a subset of $\mathbb{Z}_+$ ($\N$, $\mZ$, respectively). Denote by $\#(F)$ the number of elements of $F$.

We say that $F$  has \emph{density} $D(F)$ if the \emph{lower density} of $F$ ($\underline{D}(F)$) is equal to the \emph{upper density} of $F$ ($\overline{D}(F)$),
that is, $D(F)=\overline{D}(F)=\underline{D}(F)$,
where
\[ \underline{D}(F)=\liminf_{n\to\infty} \frac{\#(F\cap[0,n-1])}{n}\]
and
\[ \overline{D}(F)=\limsup_{n\to\infty} \frac{\#(F\cap[0,n-1])}{n}.\]

Similarly,
we say that $F$ has \emph{Banach density} if the \emph{lower Banach density} of $F$ ($BD_*(F)$) is equal to the \emph{upper Banach density} of $F$ ($BD^*(F)$),
that is, $BD(F)=BD_{*}(F)=BD^{*}(F)$,
where,
$$
BD_{*}(F)=\liminf_{N-M \to \infty} \frac{\#(F\cap [M,N])}{N-M+1}
$$
and
$$
BD^{*}(F)=\limsup_{N-M \to \infty} \frac{\#(F\cap [M,N])}{N-M+1}.
$$

\subsection{Compact metric spaces}
Denote by $(X,d)$ a compact metric space.
For $x\in X$ and $\ep>0$, denote $B(x,\ep)=\{y\in X\colon d(x,y)<\ep\}$.
We denote by $\diam(X)$ the diameter of $X$ given by $\diam(X)=\sup_{x, y\in X}d(x,y)$,
the product space $X\times X=\{(x,y)\colon x,y\in X\}$ and the diagonal $\Delta_X=\{(x,x)\colon x\in X\}$.

Let $C(X)$ be the set of continuous complex value functions on $X$  with the supremum norm
$\Vert f\Vert=\sup_{x\in X}|f(x)|$. We denote by $C(X)^*$ the dual space of $C(X)$.

\subsection{Topological dynamics}

Let $(X,T)$ be a dynamical system.
A non-empty closed
invariant subset $Y \subset X$ (i.e., $TY \subset Y$ ) defines naturally a subsystem $(Y,T)$ of $(X,T)$.
A system $(X,T)$ is called minimal if it contains no proper subsystem.
Each point belonging
to some minimal subsystem of $(X,T)$ is called a minimal point.
The orbit of a point $x\in X$ is the set $Orb(x,T)=\{x,Tx,T^2x, \ldots,\}$.
The set of limit points
of the orbit $Orb(x,T)$ is called the $\omega$-limit set of $x$, and is denoted by $\omega(x,T)$.
For $x \in X$ and $U,V \subset X$, put $N(x,U) = \{n \in\Z: T^nx \in U\}$ and
$N(U,V) = \{n\in \Z: U\cap T^{-n}V\neq \emptyset\}$. Recall that a dynamical system $(X,T)$ is called topologically transitive
(or just transitive) if for every two non-empty open subsets $U,V$ of $X$ the set $N(U,V)$
is infinite. Any point with dense orbit is called a transitive point. Denote the set of all
transitive points by $Trans(X,T)$. It is well known that for a transitive system, $Trans(X,T)$
is a dense $G_\delta$ subset of $X$.

For two dynamical systems $(X,T)$ and $(Y,S)$. Let $\pi\colon X\to Y$ be a continuous map.
If $\pi$ is surjective with $\pi\circ T=S\circ \pi$, then we say that $\pi$ is a \emph{factor map},
the system $(Y, T)$ is a \emph{factor} of $(X, T)$ or $(X, T)$ is an \emph{extension} of $(Y, T)$.
If $\pi$ is a homeomorphism, then we say that $\pi$ is a \emph{conjugacy} and
that the dynamical systems $(X,T)$ and $(Y,S)$ are \emph{conjugate}.
By the Halmos and von Neumann theorem (see \cite[Theorem 5.18]{W82}),
a minimal system  is equicontinuous if and only if it is
conjugate to a minimal rotation on a compact abelian metric group.

A pair $(x,y)\in X\times X$ is said to be \emph{proximal}
if for any $\ep>0$, there exists a positive integer $n$ such that $d(T^nx,T^ny)<\ep$.
Let $P(X,T)$ denote the collection of all proximal pairs in $(X,T)$.
If any pair of two points in $X$ is proximal, then we say that the dynamical system $(X,T)$ is \emph{proximal}.

A pair $(x,y)\in X\times X$ is said to be \emph{Banach proximal}
if for any $\ep>0$, $d(T^nx,T^ny)<\ep$ for all $n\in \Z$ except a set of zero Banach density.
Let $BP(X,T)$ denote the collection of all Banach proximal pairs in $(X,T)$.
See \cite{LT14} for a detailed study on Banach proximality.

A pair $(x,y)$ is called \emph{regionally proximal}
if for every $\ep>0$, there exist two points  $x',y'\in X$
with $d(x,x')<\ep$ and $d(y,y')<\ep$, and a positive integer $n$ such that $d(T^nx',T^ny')<\ep$.
Let $Q(X,T)$ be the set of all regionally proximal pairs in $(X,T)$.
Clearly, $Q(X,T)\supset P(X,T)\supset BP(X,T)$.

A factor map  $\pi\colon (X,T)\to (Y,S)$ is called \emph{proximal} (\emph{Banach proximal}, respectively)
 if whenever $\pi(x)=\pi(y)$ the pair $(x,y)$ is proximal (Banach proximal, respectively ).

The factor $\pi\colon(X,T)\to (Y,S)$ is the maximal equicontinuous factor if the system $(Y,T)$
is equicontinuous and for any other factor map $\phi\colon(X,T)\to (Z,U)$,
where $(Z,U)$ is equicontinuous, there exists a factor map $\psi\colon(Y,S)\to (Z,U)$
such that $\phi=\psi \circ \pi$.
It is thus unique up to conjugacy and therefore referred to as the \emph{maximal equicontinuous factor}.
Let $\pi\colon(X,T)\to (Y,S)$ be the factor map to the maximal equicontinuous factor.
The equivalence relation $R_\pi=\{(x,y)\in X\times X: \pi(x)=\pi(y) \}$ is called the \emph{equicontinuous structure relation}.
It is shown in~\cite{EG60} that the equicontinuous structure relation is the smallest closed invariant equivalence relation
containing the regionally proximal relation.

\subsection{Invariant measures}
For a dynamical system  $(X,T)$, we denote by $M(X,T)$ the set of $T$-invariant regular Borel probability measures on $X$.
It is well known that $M(X, T)$ is always nonempty.
We say that $(X,T)$ is \emph{uniquely ergodic} if $M(X,T)$ consists a single measure.

We regard $M(X)$ as a closed convex subset of $C(X)^*$,
equipped with the weak$^*$ topology. Then $M(X)$ is a compact metric space.
An invariant measure is ergodic if and only if it is an extreme point of $M(X,T)$.

Let $\mu \in M(X, T)$. We define the \emph{support} of $\mu$ by
$\supp(\mu)$=\{$x\in X$: $\mu(U)>0$ for any neighborhood $U$ of $x\}$.
The \emph{support} of a dynamical system $(X,T)$, denoted by $\supp(X,T)$,
is the smallest closed subset $C$ of $X$ such that $\mu(C)=1$ for all $\mu\in M(X,T)$.

The action of $T$ on $X$ induces an action on
$M(X)$ in the following way: for $\mu\in M(X)$ we define $T\mu$ by
\[\int_X f(x)\ \textrm{d}T\mu(x)= \int_X f(Tx)\textrm{d}\mu(x), \quad \forall f\in C(X).\]
Hence $(M(X),T)$ is also a topological dynamical system.

Fix a measure space $(X, \mathcal{B}, \mu)$. If $f$ and $g$ are functions on $X$,
we denote by $f\otimes g$ the function on $X\times X$ given by $f\otimes g(x,x')=f(x)g(x')$
and by
$L^{\infty}(X)\otimes L^{\infty}(X) $
we denote the algebra of functions on $X\times X$ that are finite sums of
functions $f\otimes g$ with $f, g \in L^{\infty}(X, \mathcal{B}, \mu)$.
We denote by $\mu_{\Delta}$ the diagonal measure on $X\times X$ given by $\int f(x,x')\textrm{d}\mu_{\Delta}(x,x')=\int f(x,x)\textrm{d}\mu(x)$.
We notice that $\mu_\Delta (A\times B)=\mu(A\cap B)$ for any $A, B \in \mathcal{B}$.

For a dynamical system $(X,T)$, $f\in C(X)$ and $n\in\N$, let
$f_n(x)=\frac{1}{n}\sum_{i=0}^{n-1} f(T^ix).$ The following theorem is well known.

\begin{thm}\cite{O52} \label{thm:unique-ergodic}
Let $(X,T)$ be a dynamical system. Then the following conditions are equivalent:
\begin{enumerate}
  \item $(X,T)$ is uniquely ergodic;
  \item for each $f\in C(X)$, $\{f_n\}_{n=1}^\infty$ converges uniformly on $X$ to a constant;
\item for each $f\in C(X)$, there is a subsequence $\{f_{n_k}\}_{k=1}^\infty$
which converges pointwise on $X$ to a constant.
\item $(X,T)$ contains only one minimal set, and
for each $f\in C(X)$, $\{f_n\}_{n=1}^\infty$ converges uniformly on $X$.
\end{enumerate}
\end{thm}

\section{Mean equicontinuity and equicontinuity in the mean}
In this section we will show that mean equicontinuity is equal to equicontinuity in the mean.
Moreover, we will discuss what kinds of minimal sets can appear in a
transitive mean equicontinuous system.
\subsection{Mean equicontinuity and equicontinuity in the mean}
We start with the following characterizations of equicontinuous in the mean systems.
To do this, we need a simple lemma.

\begin{lem}\cite[Lemma 3.2]{YL} \label{same}
Let $(X,T)$ and $(Y,S)$ be two dynamical systems.
Then $(X\times Y,T\times S)$ is mean equicontinuous if and only if
both $(X,T)$ and $(Y,S)$ are mean equicontinuous.
\end{lem}

\begin{thm}\label{thm:unifon-conv-equi}
Let $(X,T)$ be a dynamical system. Then the following conditions are equivalent:
\begin{enumerate}
  \item $(X,T)$ is equicontinuous in the mean;
  \item for each $f\in C(X\times X)$,
  the sequence $\{\frac{1}{n}\sum_{i=0}^{n-1} f\circ(T^i\times T^i)\}_{n=1}^\infty$
  is uniformly equicontinuous;
  \item for each $f\in C(X\times X)$, the sequence
  $\{\frac{1}{n}\sum_{i=0}^{n-1} f\circ(T^i\times T^i)\}_{n=1}^\infty$ is uniformly convergent to a
  $T\times T$-invariant continuous  function $f^*\in C(X\times X)$.
\end{enumerate}
\end{thm}
\begin{proof} We only present the proof (1) implies (2) and the rest
is similar to the proof of \cite[Theorem 3.3]{YL}.

To make the idea of the proof clearer, when proving (1)$\Rightarrow$(2),
we assume $f\in C(X)$ instead of $f\in C(X\times X)$, because if $(X,T)$ is equicontinuous in the mean
if and only if so is $(X\times X,T\times T)$.

(1)$\Rightarrow$(2)
Fix $f\in C(X)$ and $\ep>0$.
By continuity of $f$,
there exists $\eta\in(0,\frac{\ep}{2\Vert f\Vert})$ such that if $x$, $y\in X$ with $d(x,y)<\eta$
then $|f(x)-f(y)|<\frac{\ep}{2}$.
As $(X,T)$ is equicontinuous in the mean,
there is $\delta\in(0,\eta)$ such that if $x,y\in X$ with $d(x,y)<\delta$, one has
\[\frac{1}{n}\sum_{i=0}^{n-1}d(T^ix,T^iy)<\eta^2,\ n=1,2,\dotsc.\]
For every $n\in \N$ and $x,y\in X$ with $d(x,y)<\delta$, let
\[I_n(x,y)=\{i\in [0,n-1]\colon d(T^ix,T^iy)\geq \eta\}.\]
Then $\#(I_n(x,y))\leq \eta n$.
So for every $n\in \N$ and $x,y\in X$ with $d(x,y)<\delta$, we have
\begin{align*}
\Bigl\vert\frac{1}{n}\sum_{i=0}^{n-1} f(T^ix)-\frac{1}{n}\sum_{i=0}^{n-1} f(T^iy)\Bigr\vert
&\leq \frac{1}{n} \sum_{i=0}^{n-1}\bigl\vert f(T^ix)-f(T^iy)\bigr\vert\\
&\leq \frac{1}{n}\Bigl(\sum_{i\in I_n(x,y)} 2 \Vert f\Vert+
 \sum_{i\in [0,n-1]\setminus I_n(x,y)}\bigl\vert f(T^ix)-f(T^iy)\bigr\vert\Big)\\
 &\leq 2\eta \Vert f\Vert +\frac{\ep}{2}<\ep.
\end{align*}
This shows that $\{\frac{1}{n}\sum_{i=0}^{n-1} f\circ T^i\}_{n=1}^\infty$
is uniformly equicontinuous.
\end{proof}

Before proving the main result of this section we give a proof of a result in \cite{O52} which is outlined there.
We need the following lemmas.

\begin{lem}\label{mean-eq-minimal}\cite{HL}
Let $(X,T)$ be  a minimal  dynamical system.
Then $(X,T)$ is mean equicontinuous
if and only if it is equicontinuous in the mean.
\end{lem}

\begin{lem}\label{lem:mean-eq-BP}\cite[Theorem 3.5]{YL}
Let $(X,T)$  be a dynamical system. If $(X,T)$ is mean equicontinuous, then $Q(X,T)=P(X,T)=BP(X,T)$ and
it is a closed invariant equivalence relation.
\end{lem}

\begin{lem} \cite{LT14} \label{PRBD}
Let $(X,T)$ be a dynamical system. Then the support of $(X,T)$ is the smallest closed subset $K$
of $X$ such that for every $x\in X$ and every open neighborhood $U$ of $K$, $N(x,U)$ has Banach density one.
\end{lem}

Now we are ready to show
\begin{thm}\label{transitive}
Let $(X,T)$ be a dynamical system.
If $(X,T)$ is mean equicontinuous,
then for every $x\in X$, $(\overline{\text{Orb}(x,T)},T)$ is uniquely ergodic.
In particular,
if $(X,T)$ is also transitive, then $(X,T)$ is uniquely ergodic.
\end{thm}
\begin{proof}
Without loss of generality, we can assume that $X=\overline{Orb(x,T)}$.

If  $M_1,M_2$ are two minimal sets in $(X,T)$.
By Auslanser-Ellis theorem, there exist $y_1\in M_1$ and $y_2\in M_2$
such that $(x,y_1)$ and  $(x,y_2)$ are both proximal.
For a given $\ep>0$,set
$$
A_1=\{n\in\Z: d(T^nx,T^ny_1)<\ep/2\}\ \text{and}\ A_2=\{n\in\Z: d(T^nx,T^ny_2)<\ep/2\}.
$$
By Lemma~\ref{lem:mean-eq-BP}, $A_1\cap A_2\not=\emptyset$ which implies that $(y_1,y_2)$ is proximal.
As $y_1$ and $y_2$ are minimal points, then their orbit closures
are equal which deduces that $M_1=M_2$.
So there is only one minimal set in $(X,T)$, denoted by $M$.

It is clear that $(M,T)$ is also mean equicontinuous.
By Lemma \ref{mean-eq-minimal}, $(M,T)$ is equicontinuous in the mean.
Then $(M,T)$ is  uniquely ergodic by Theorem  \ref{thm:unique-ergodic} and Theorem \ref{thm:unifon-conv-equi}.
For every $z\in X$, by Auslanser-Ellis theorem again,
there  exists a point $y\in M$ such that $(z,y)$ is proximal.
By Lemma~\ref{lem:mean-eq-BP}, $(z,y)$ is a Banach proximal piont.
So for every open neighborhood $U$ of $M$ and any $z\in X$,
$N(z,U)$ has Banach density one. Then by Lemma \ref{PRBD} we have $\supp(X,T)\subset M$.
As $(M,T)$ is  uniquely ergodic, so is $(X,T)$.
\end{proof}

Now we begin to prove the main result of this section. We need the following lemma.

\begin{lem}\label{minimal-point}
Let $(X,T)$ be mean equicontinuous system and $\nu$ be an ergodic measure on $X$,
then every point of $\text{supp} (\nu)$ is minimal.
\end{lem}

\begin{proof}
$(\text{supp}(\nu),T)$ is a transitive system since $\nu$ is an ergodic measure on $X$.
By Theorem \ref{transitive},
$(\text{supp}(\nu),T)$ is uniquely ergodic,
and so,
it is minimal.
\end{proof}

Now we are ready to show the main result. Note that our method is different from
the proof for the minimal case.

\begin{thm}\label{general}
$(X,T)$ is mean equicontinuous if and only if equicontinuous in the mean.
\end{thm}

\begin{proof}
If $(X,T)$ is equicontinuous in the mean,
it is clear that $(X,T)$ is mean equicontinuous.

Now assume that $(X,T)$ is mean equicontinuous.
If $(X,T)$ is not equicontinuous in the mean, then there are
$x_k,y_k,z\in X,n_k\in \Z,k=1,2,\cdots$ and $\ep_0>0$
such that $\lim_{k\rightarrow\infty }x_k=z=\lim_{k\rightarrow\infty }y_k$
and for every $k\in \Z$
$$
\frac{1}{n_k}\sum_{i=0}^{n_k-1}d(T^ix_k,T^iy_k)\geq \ep_0.
$$

Let $\mu_k=\frac{1}{n_k}\sum_{i=0}^{n_k-1}\delta_{(T^ix_k,T^iy_k)}$, then $\mu_k\in M(X\times X)$,
we may assume $\mu_k\rightarrow \mu$(otherwise we may consider the subsequence),
where $\mu\in M(X\times X,T\times T)$.

We claim that $\mu(\text{supp}(\mu )\setminus \Delta_X)>0$.
Actually, $d(\cdot,\cdot)$ is a continuous function on $X\times X$, then we have
$$
\int_{X\times X}d(x,y)\textrm{d}\mu_k\longrightarrow \int_{X\times X}d(x,y)\textrm{d}\mu
$$
and
$$
\int_{X\times X}d(x,y)\textrm{d}\mu_k=\frac{1}{n_k}\sum_{i=0}^{n_k-1}d(T^ix_k,T^iy_k)\geq \ep_0
$$
which implies
$$\mu(\text{supp}(\mu)\setminus \Delta_X)>0.$$
By the ergodic decomposition, we have $\nu(\text{supp}(\mu)\setminus \Delta_X)>0$
for some ergodic measure on $X\times X $. Thus, there exists a minimal point in $\text{supp}(\mu)\setminus \Delta_X$,
since $\text{supp}(\nu)$ is a minimal set by Lemma \ref{minimal-point}. Denote this minimal point by $(u,v)$.

For $l\in\N$, let $B_l=\{(x,y)\in X\times X:d((x,y),(u,v))<\frac{1}{l} \}$,
then
$$
\mu(B_l)>0  \ \text{and}\
\mu_k(B_l)=\frac{1}{n_k}\#(\{0\leq i \leq n_k:(T^ix_k,T^iy_k)\in B_l\}).
$$
There are infinitely many $k\in \Z$ with $0\leq m_k \leq n_k$ such that $(T^{m_k}x_k,T^{m_k}y_k)\in B_l$,
since $0<\mu(B_l)\leq \liminf_{k\rightarrow\infty}\mu_k(B_l)$ for every $l\in \N$.

Put $$\delta=d(\overline{Orb((z,z),T\times T)},\overline{Orb((u,v),T\times T)}).$$ Then, $\delta>0$, since $\overline{Orb((u,v),T\times T)}\cap \Delta_X=\emptyset$.
As $(X,T)$ is mean equicontinuous, so is $(X\times X,T\times T)$ by Lemma \ref{same}. Then,
for $\frac{1}{4}\delta$, there is $\eta>0$ such that if $d((x,y),(x',y'))<\eta$,
then \[\limsup_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}d((T\times T)^i(x,y),(T\times T)^i(x',y'))<\frac{\delta}{4}.\]

We can choose $k\in \mathbb{Z_{+}}$ with
 $d((x_k,y_k),(z,z))<\eta$ and $d((T^{m_k}x_k,T^{m_k}y_k),(u,v))<\eta$,
then
 \[\limsup_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}d((T\times T)^i(x_k,y_k),(T\times T)^i(z,z))<\frac{\delta}{4}\]
 and
  \[\limsup_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}d((T\times T)^i(T^{m_k}x_{k},T^{m_k}y_{k}),(T\times T)^i(u,v))<\frac{\delta}{4}\]
which implies
\[\limsup_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}d((T\times T)^i(T^{m_k}z,T^{m_k}z),(T\times T)^i(u,v))<\frac{\delta}{2}.\]
It is a contradiction, thus $(X,T)$ is equicontinuous in the mean.
\end{proof}

\subsection{Minimal sets in a transitive mean equicontinuous  system}
In Theorem \ref{transitive} we have shown that a transitive mean equicontinuous system is uniquely ergodic,
and thus it contains a unique minimal subset.
Here we will discuss what kinds of minimal sets can appear in a
transitive mean equicontinuous system.

\begin{thm} We have the following observations.
 \begin{enumerate}
 \item If $(X,T)$ is weakly mixing and mean equicontinuous, then the unique minimal set is a fixed point.
 \item If $(X,T)$ is totally transitive and mean equicontinuous, then the unique minimal set is totally minimal and mean equicontinuous.
 Moreover, any totally transitive, minimal, mean equicontinuous system can be realized in a totally transitive non-minimal
 mean equicontinuous system.
 \item If $(X,T)$ is transitive and mean equicontinuous, then the unique minimal set is mean equicontinuous.
 Moreover, any minimal, mean equicontinuous system can be realized in a transitive non-minimal
 mean equicontinuous system.
\end{enumerate}
\end{thm}
\begin{proof}
(1). If $(X,T)$ is mean equicontinuous and weakly mixing,
 then $(X\times X,T\times T)$ is transitive, thus it is uniquely ergodic by Theorem \ref{transitive}.
 As we know $\mu\times \mu$ and $\mu_\Delta$ are invariant measure on $X\times X$
 for any invariant measure $\mu$ on $X$,
 thus $\mu\times \mu=\mu_\Delta$.
 $\mu_\Delta(\Delta_X)=1$ implies $\mu$ must be $\delta_x$ for some $x\in X$.
 Hence $(X,T)$ is also uniquely ergodic and the unique fixed point is $x$.

 \medskip
 (2). If $(X,T)$ is mean equicontinuous and totally transitive,
 it has only one minimal set by Theorem \ref{transitive}, denoted by $M$.
 Then $(M,T)$ is totally minimal.
 Actually, $(X,T^n)$ is also a transitive mean equicontinuous system for every $n\in\mathbb{N}$,
 again by Theorem \ref{transitive}, there is only one $T^n$-invariant measure
 on $X$ denoted by $\mu_n$.
Let $\mu$ be the unique invariant measure on $(X,T)$ and it is also invariant on $T^n$,
hence $\mu=\mu_n$ and $M=\text{supp}(\mu)=\text{supp}(\mu_n)$,
which implies $(M,T^n)$ is minimal. It is clear that $(X,T)$ is mean equicontinuous.

Now let $(X_1,T_1)$ be a totally minimal mean equicontinuous system and $(X_2,T_2)$ be a weakly mixing
system such that the uniquely ergodic measure is supported on a fixed point $p$. Then, $(X_2,T_2)$
is mean equicontinuous and $(X_1\times X_2, T_1\times T_2)$ is the system we want.

 \medskip
(3). The first statement follows again by Theorem \ref{transitive}. Let $(X_1,T_1)$ be a minimal mean equicontinuous system
and $(X_2,T_2)$ be a weakly mixing
system such that the uniquely ergodic measure is supported on a fixed point $p$. Then, $(X_2,T_2)$
is mean equicontinuous and $(X_1\times X_2, T_1\times T_2)$ is the system we want.
\end{proof}

\section{Regionally proximal and Banach proximal relations}
Lemma \ref{lem:mean-eq-BP} shows that for a mean equicontinuous system $(X,T)$,
we have $BP(X,T)=P(X,T)=Q(X,T)$.

We will show the converse is also valid, i.e.
for a dynamical system $(X,T)$, if $BP(X,T)=P(X,T)=Q(X,T)$
then it is equicontinuous in the mean. In fact we will prove more by providing a series of equivalent statements, see Theorem \ref{thm:transitive-mean-eq} for details.

We start with some preparations.
The following lemma is just a simple observation.

\begin{lem}
Let $(X,T)$ be a dynamical system, if $(x,y)\in BP(X,T)$, then for any neighborhood $U$ of $\Delta_{X}$,
we have $BD(N((x,y),U))=1$.

\end{lem}

\begin{lem}\label{measure}
Let $(X,T)$ be a dynamical system.
If there exists $\mu\in M(X\times X, T\times T)$ such that $\mu(BP(X,T))=1$,
then $\mu(\Delta_X)=1$.
\end{lem}
\begin{proof}
Assume that $\mu(\Delta_{X})<1$, i.e. $\mu(BP(X,T)\setminus \Delta_{X})>0$,
As $X$ is a compact metric space there exists a closed set $F\subset \text{supp}(\mu)\cap BP(X,T)\setminus \Delta_{X}$
with $\mu(F)>0$.
By the ergodic decomposition theorem, there exists an ergodic measure $\nu$ with $\nu(F)>0$.
By Birkhoff ergodic theorem there exists $z\in F$ such that
$$ \frac{1}{n} \#(N(z,F)\cap[0,n-1])=\frac{1}{n}\sum_{i=0}^{n-1} \chi_{F}(T^{i}z)\rightarrow \int \chi_F \textrm{d}\nu=\nu(F)>0,$$
then we have $D(N(z,F))>0$.
We choose neighborhoods $U$ and $V$ of $F$ and $\Delta_X$ respectively with $U\cap V=\emptyset$,
then $\underline{BD}(N(z,U))\geq D(N(z,F))>0$.
On the other hand, we have $BD(N(z,V))=1$, since $z\in BP(X,T)$.
The contradiction shows the lemma.
\end{proof}

For a minimal system the following result was known, see \cite{DG16} and \cite{YL}. We now show
it holds for a general system.

\begin{thm}\label{thm:transitive-mean-eq}
Let $(X,T)$ be a dynamical system.
Let $(Z,S)$ be the maximal equicontinuous factor of $(X,T)$
and  $\pi\colon (X,T)\to (Z,S)$ be the factor map.
Then the following conditions are equivalent:
\begin{enumerate}
	\item $(X,T)$ is mean equicontinuous;
	\item  $\pi$ is Banach proximal;
\item $BP(X,T)=P(X,T)=Q(X,T)$;
	\item $\pi\colon (X,\mu,T) \to (Z,\nu,S)$ is measure-theoretically isomorphic, where $\mu$ and $\nu$ are
	any invariant measures on $X$ and $Z$  respectively with $\pi(\mu)=\nu$;
	\item $(X,T)$ is equicontinuous in the mean.
\end{enumerate}
\end{thm}
\begin{proof}
	(1) $\Rightarrow$ (2) by Lemma \ref{lem:mean-eq-BP}. (1) $\Leftrightarrow$ (5) by Theorem \ref{general}. Moreover, it is clear
that (2) $\Leftrightarrow$ (3).

	(3) $\Rightarrow$ (4)
This is essentially proved in \cite[Theorem 3.8]{YL}.
Here we provide a proof for completeness.

Let $\mu$ be an invariant measure on $(X,T)$ and $\nu$ be the invariant measure on $(Z,S)$ with $\pi(\mu)=\nu$.
We consider the disintegration of $\mu$ over $\nu$.
That is, for a.e. $y\in Z$ we have a measure $\mu_y$ on $X$ such that $\supp(\mu_y)\subset \pi^{-1}(y)$
and $\mu=\int_{y\in Z}\mu_y \textrm{d}\nu$.
Let $W=\{(u,v)\in X\times X: \pi(u)=\pi(v)\}$.
As $\supp(\mu_y)\subset \pi^{-1}(y)$,
we have $\supp(\mu_y\times\mu_y)\subset \pi^{-1}(y)\times \pi^{-1}(y)\subset W$, a.e. $y\in Y$.
Let $\mu\times_Z\mu=\int_{y\in Z}\mu_y\times\mu_y\textrm{d}\nu$.
Then $\mu\times_Z\mu$ is an invariant measure on $(X\times X,T\times T)$.
Moreover,
\[\mu\times_Z\mu(W)=\int_{y\in Z} \mu_y\times\mu_y(W)\textrm{d}\nu=1,\]
then $\supp(\mu\times_Z\mu)\subset W$.
By Lemma \ref{measure}
we have $\mu\times_Z\mu(\Delta_X)=1$.
Since
\[\mu\times_Z\mu(\Delta_X)=\int_{y\in Y}\mu_y\times \mu_y(\Delta_X)\textrm{d}\nu=1,\]
we have $\mu_y\times\mu_y(\Delta_X)=1$ a.e. $y\in Z$.
Then for a.e. $y\in Z$,
there exists a point $c_y\in \pi^{-1}(y)$ such that $\mu_y=\delta_{c_y}$.

Let $Y_0$ be the collection of $y\in Z$ such that $\mu_y$ is not equal to $\delta_x$ for
any $x\in X$. Then $\nu(\cup_{i\in \Z}S^{-i}Y_0)=0$. Let $Z_0=Z\setminus \cup_{i\in \Z}S^{-i}Y_0$
and $X_0=\{c_y:y\in Z_0\}$.
Then $\nu(Z_0)=1$. Now we show that $X_0$ is a measurable set. In fact,
the map $y\mapsto\mu_y$ from $Z_0$ to $M(X)$ is measurable and
$x\mapsto \delta_x$ is an embedding. Since $Z_0$ is a measurable set and
maps are 1-1, it follows from Souslin's theorem that $X_0$ is a
measurable set, and it is clear that $\mu(X_0)=\mu(\pi^{-1}Z_0)=\nu(Z_0)=1$. By the same
argument, $\pi: (X_0,\mu,T)\to (Z_0,\nu,S)$ is a
measure-theoretic isomorphism.

(4) $\Rightarrow$ (5)
If $(X,T)$ is not equicontinuous in the mean, then there are
$x_k,x\in X,n_k\in \Z,k=1,2,\cdots$ and $\ep_0>0$
such that $\lim_{k\rightarrow\infty }x_k=x$
and for every $k\in \Z$
$$
\frac{1}{n_k}\sum_{i=0}^{n_k-1}d(T^ix_k,T^ix)\geq \ep_0.
$$

Let $\pi(x_k)=z_k$ and $\pi(x)=z$. We define
$$\mu_k=\frac{1}{n_k}\sum_{i=0}^{n_k-1}\delta_{(T^ix_k,T^ix)}$$
and
$$\nu_k=\frac{1}{n_k}\sum_{i=0}^{n_k-1}\delta_{(S^iz_k,S^iz)}$$
then
$$(\pi \times \pi)(\mu_k)=\nu_k.$$
By taking the subsequence,
there exists $\mu$ and $\nu$ on $M(X\times X,T\times T)$ and $M(Z\times Z,S\times S)$
respectively with $\lim_{k\rightarrow\infty}\mu_k=\mu,\lim_{k\rightarrow\infty}\nu_k=\nu$
and $(\pi \times \pi)(\mu)=\nu$.

We claim that $\mu(\text{supp}(\mu )\setminus \Delta_X)>0$.
Actually, $d(\cdot,\cdot)$ is a continuous function on $X\times X$, then we have
$$
\int_{X\times X}d(x,y)\textrm{d}\mu_k\longrightarrow \int_{X\times X}d(x,y)\textrm{d}\mu
$$
and $\int_{X\times X}d(x,y)\textrm{d}\mu_k=\frac{1}{n_k}\sum_{i=0}^{n_k-1}d(T^ix_k,T^ix)\geq \ep_0$
which implies $\mu(\text{supp}(\mu)\setminus \Delta_X)>0$.
There are open sets $U$ and $V$ of $X$ with $U\cap V=\emptyset$ and $\mu(U\times V)>0$.

Let $\mu'$ and $\nu'$ be the projection of $\mu$ and $\nu$ onto the
first component of $X$ and $Z$ respectively. It is clear that $\mu'\in M(X,T)$ and $\nu'\in M(Z,S)$.
It is easy to see $\text{supp}(\nu)\subset \overline{Orb((z,z),S\times S)}$.
Then $\nu'(\overline{Orb(z,S)})=\nu(\overline{Orb(z,S)}\times Z)=1$,
which implies $\text{supp}(\nu')\subset \overline{Orb(z,S)}$.
Furthermore,
$\nu'$ is the only invariant measure on $\overline{Orb(z,S)}$,
since $\overline{Orb(z,S)}$ is uniquely ergodic.
As for every $f,g\in C(Z)$, we have
$$\int_{Z\times Z}f(z_1)g(z_2)\textrm{d}\nu_k(z_1,z_2)\longrightarrow \int_{Z\times Z}f(z_1)g(z_2)\textrm{d}\nu(z_1,z_2)$$
and
$$\int_{Z\times Z}f(z_1)g(z_2)\textrm{d}\nu_k(z_1,z_2)=\frac{1}{n_k}\sum_{i=0}^{n_k-1}f(S^iz_{k})g(S^iz)\longrightarrow
\int_Z f(z)g(z)\textrm{d}\nu'(z),$$
thus $\nu(A\times B)=\nu'(A\cap B)$.
So $\nu'(\pi(U)\cap \pi(V))=\nu(\pi(U)\times \pi(V))\geq \mu(U\times V)>0$.

Obviously, $\pi(\mu')$ is an invariant measure on $\overline{Orb(z,S)}$,
thus we have $\pi(\mu')=\nu'$. As
$\pi\colon (X,\mu',T) \to (Z,\nu',S)$ is measure-theoretic isomorphic, we have $\nu'(\pi(U)\cap \pi(V))=\mu'(U\cap V)=0$,
it is a contradiction.
This shows $(X,T)$ is mean equicontinuous.
\end{proof}

\section{Mean equicontinuity and Weyl mean equicontinuity}

Following \cite{DG16} and \cite{YL},
a dynamical system $(X,T)$ is called \emph{Banach mean equicontinuous or Weyl mean equicontinuous}

if for every $\ep>0$, there exists a $\delta>0$ such that
\[
\limsup_{n-m\to\infty}\frac{1}{n-m}\sum_{i=m}^{n-1}d(T^ix,T^iy)<\ep
\]
for all $x,y\in X$ with $d(x,y)<\delta$.
It is clear that  each Weyl mean equicontinuous system is mean equicontinuous.
It is shown in~\cite{DG16} that if a minimal system is
mean equicontinuous then it is Weyl mean equicontinuous.
In this section we show that for a general dynamical system mean equicontinuity is equivalent to
Weyl mean equicontinuity. That is, we have

\begin{thm}\label{thm:Weyl-mean-equ}
A dynamical system $(X,T)$ is mean equicontinuous if and only if
it is Weyl mean equicontinuous.
\end{thm}

Before proving the Theorem, we need the following lemma.

\begin{lem}
If a dynamical system $(X,T)$ is uniquely ergodic, then
for any $f\in C(X, \mathbb{R})$ and $x\in X$,
\[\lim_{n-m\to\infty}\frac{1}{n-m}\sum_{i=m}^{n-1}
f(T^ix)=\lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}
f(T^ix).\]
\end{lem}
\begin{proof}
Let $\mu$ be the unique invariant measure on $(X,T)$.
Then for any $f\in C(X, \mathbb{R})$ and $x\in X$,
\[ \lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}
f(T^ix)=\int f \textrm{d}\mu.\]
If the conclusion does not hold, then there exist $f\in C(X, \mathbb{R})$, $x\in X$ and two sequences $\{n_k\}$ and $\{m_k\}$ with
$n_k- m_k\to\infty $ such that
\[\lim_{n_k-m_k\to\infty}\frac{1}{n_k-m_k}\sum_{i=m_k}^{n_k-1}
f(T^ix) =c \neq \int f \textrm{d}\mu.\]
As $M(X)$ is compact, passing to a subsequence if necessary
we may assume that
\[ \lim_{k\to\infty}
\frac{1}{n_k-m_k}\sum_{i=m_k}^{n_k-1}\delta_{T^ix}
=\nu.\]
It is easy to see that $\nu$ is an invariant measure.
As $(X,T)$ is unqiuely ergodic then $\nu=\mu$.
So
\[\lim_{n_k-m_k\to\infty}\frac{1}{n_k-m_k}\sum_{i=m_k}^{n_k-1}
f(T^ix) = \int f\textrm{d}\mu.\]
This is a contradiction.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:Weyl-mean-equ}]
As $(X,T)$ is mean equicontinuous, then so is $(X\times X,T\times T)$.
Fix $(x,y)\in X\times X$.
By Theorem \ref{transitive}, $(\overline{Orb((x,y),T\times,T)},T\times T)$ is uniquely ergodic.
Now applying the above theorem to the distance function $d(\cdot,\cdot)$ and $(x,y)$, we get
\[
\lim_{n-m\to\infty}\frac{1}{n-m}\sum_{i=m}^{n-1}d(T^ix,T^iy)=
\lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}d(T^ix,T^iy).
\]
Then the result follows from the definition.
\end{proof}

We now give the following conclusion to end this section.

\begin{thm}
Let $(X,T)$ be a mean equicontinuous system, then
for every
$\varepsilon >0$, there are $\delta>0$ and $N>0$, such that whenever $d(x,y)<\delta$, we have
$$\frac{1}{n}\sum_{i=j}^{n+j-1} d(T^ix,T^iy)<\varepsilon$$
 for all $ n\geq N$ and $j\geq 0$.
\end{thm}

\begin{proof}
Assume that there are
$x_k,y_k,z\in X,n_k,j_k\in \Z,k=1,2,\cdots$ and $\ep_0>0$
such that $\lim_{k\rightarrow\infty }x_k=z=\lim_{k\rightarrow\infty }y_k$
and for every $k\in \Z$
$$
\frac{1}{n_k}\sum_{i=j_k}^{n_k+j_k-1}d(T^ix_k,T^iy_k)\geq \ep_0.
$$

Let $\mu_k=\frac{1}{n_k}\sum_{i=j_k}^{n_k+j_k-1}\delta_{(T^ix_k,T^iy_k)}$, and then $\mu_k\in M(X\times X)$.
We may assume $\mu_k\rightarrow \mu$(otherwise we may consider the subsequence),
where $\mu\in M(X\times X,T\times T)$.

We claim that $\mu(\text{supp}(\mu )\setminus \Delta_X)>0$.
In fact, $d(\cdot,\cdot)$ is a continuous function on $X\times X$, then we have
$$
\int_{X\times X}d(x,y)\textrm{d}\mu_k\longrightarrow \int_{X\times X}d(x,y)\textrm{d}\mu
$$
and
$$
\int_{X\times X}d(x,y)\textrm{d}\mu_k=\frac{1}{n_k}\sum_{i=j_k}^{n_k+j_k-1}d(T^ix_k,T^iy_k)\geq \ep_0
$$
which implies
$$\mu(\text{supp}(\mu)\setminus \Delta_X)>0.$$
By ergodic decomposition theorem, we have $\nu(\text{supp}(\mu)\setminus \Delta_X)>0$
for some ergodic measure $\nu$ on $X\times X $, thus there exists a minimal point $(u,v)$ in $\text{supp}(\mu)\setminus \Delta_X$
since $\text{supp}(\nu)$ is a minimal set by Lemma \ref{minimal-point}.

Let $B_l=\{(x,y)\in X\times X:d((x,y),(u,v))<\frac{1}{l} \}$.
Then we have
$$\mu(B_l)>0 \ \text{and}\
\mu_k(B_l)=\frac{1}{n_k}\#(\{j_k\leq i \leq n_k+j_k-1:(T^ix_k,T^iy_k)\in B_l\}).$$
Thus for any $l\in \Z$ there exist infinte $k\in \Z$ with $j_k\leq m_k \leq n_k+j_k-1$ such that $(T^{m_k}x_k,T^{m_k}y_k)\in B_l$,
since $0<\mu(B_l)\leq \liminf_{k\rightarrow\infty}\mu_k(B_l)$.

Put $$\delta=d(\overline{Orb((z,z),T\times T)},\overline{Orb((u,v),T\times T)})$$
then $\delta>0$, since $\overline{Orb((u,v),T\times T)}\cap \Delta_X=\emptyset$.
As $(X,T)$ is mean equicontinuous,
so is $(X\times X,T\times T)$ by Lemma \ref{same}. Thus,
for $\frac{1}{4}\delta$, there is $\eta>0$ such that if $d((x,y),(x',y'))<\eta$
then \[\limsup_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}d((T\times T)^i(x,y),(T\times T)^i(x',y'))<\frac{\delta}{4}.\]

There are infinitely many $k\in \Z$ with
 $d((x_k,y_k),(z,z))<\eta$ and $d((T^{m_k}x_k,T^{m_k}y_k),(u,v))<\eta$,
then
 \[\limsup_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}d((T\times T)^i(x_k,y_k),(T\times T)^i(z,z))<\frac{\delta}{4}\]
 and
  \[\limsup_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}d((T\times T)^i(T^{m_k}x_{k},T^{m_k}y_{k}),(T\times T)^i(u,v))<\frac{\delta}{4}\]
which implies
\[\limsup_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}d((T\times T)^i(T^{m_k}z,T^{m_k}z),(T\times T)^i(u,v))<\frac{\delta}{2}.\]
It is a contradiction which shows the theorem.
\end{proof}

\section{Mean equicontinuous relation}

It is well known that the equicontinuous structure relation
is the smallest closed invariant equivalence relation containing the regionally proximal relation.
In \cite{YL} the authors showed that each topological dynamical system admits a maximal mean equicontinuous factor.
Inspired by the above ideas, we now define a new notation called \emph{a pair sensitive in the mean}
and introduce the mean equicontinuous structure relation. We show that the maximal mean equicontinuous factor
is induced by the smallest invariant closed equivalence relation containing the relation of sensitivity in the mean.

\begin{defn}
Let $(X,T)$ be a dynamical system.
We say $(x,y)$ is \emph{a pair sensitive in the mean},
if $x=y$ or
for any $\tau>0$, there exists $c=c(\tau)>0$ such that
for every $\epsilon>0$,
there exist $x',y'\in X$ and $n\in \N$ such that
$d(x',y')<\ep$ and
$$ \frac{1}{n}\#(\{0\leq i\leq n-1:d(T^ix',x)<\tau,d(T^iy',y)<\tau\})>c$$
Let $Q_{me}(X,T)$ be the set of all pairs sensitive in the mean  in $(X,T)$,
and we call that  \emph{the relation of sensitivity in the mean}.
\end{defn}
Clearly, if $T$ is a homeomorphism, then $Q_{me}(X,T)\subset Q(X,T)$.
Let $S_{me}(X, T)$ be the smallest closed $T\times T$ invariant equivalence relation
such that $X/S_{me}(X, T)$ is a mean equicontinuous system.
We will show that $S_{me}(X, T)$ is the smallest closed $T\times T$ invariant equivalent relation that contains
the relation of sensitivity in the mean.
This will be done through the following lemmas.

First we observe that

\begin{lem} \label{NOT ME}
Let $(X,T)$ be a dynamical system.
Then $(X,T)$ is not mean equicontinuous system if and only if
there exists $x, x_{k}\in X, n_{k}\in \mathbb{N}$ and $\varepsilon_{0}>0$,
such that $x_{k}\rightarrow x, \frac{1}{n_{k}}\sum_{i=0}^{n_{k}-1}d(T^{i}x,T^{i}x_{k})\geq \varepsilon_{0}$.
\end{lem}

It is easy to check:

\begin{lem}
 Let $\pi:(X,T)\rightarrow (Y,S)$ be a factor map.
 If $(x,y)\in Q_{me}(X, T)$, then we have $(\pi(x),\pi(y))\in Q_{me}(Y,S)$.
\end{lem}

\begin{lem} \label{ME-Q}
 Let $(X,T)$ be a dynamical system, then $(X,T)$ is mean equicontinuous if and only if
 $Q_{me}(X, T)=\Delta_X.$
\end{lem}
\begin{proof}
If $(X,T)$ is mean equicontinuous, it is clear that $Q_{me}(X, T)=\Delta_X$.

Conversely, assume that $Q_{me}(X, T)=\Delta_X$. Suppose that $(X,T)$ is not mean equicontinuous.
By Lemma ~\ref{NOT ME}, there are $x_k,x \in X,n_k\in\Z$ and $\ep_0>0$
such that $\lim_{k\rightarrow\infty}d(x_k,x)=0$
 and for every $k\in \N$, we have
 $$
 \frac{1}{n_k}\sum_{i=0}^{n_k-1}d(T^ix_k,T^ix)\geq\ep_0.
 $$
Let $\mu_k=\frac{1}{n_k}\sum_{i=0}^{n_k-1}\delta_{(T^ix_k,T^ix)}$, then $\mu_k\in M(X\times X)$,
we may assume $\mu_k\rightarrow \mu$(otherwise we may consider the subsequence),
where $\mu\in M(X\times X,T\times T)$.
 From the proof of Theorem ~\ref{general}, it follows that $\mu(\text{supp}(\mu)\setminus \Delta_X)>0$.

Let $(y,z)\in \text{supp}(\mu)\setminus \Delta_X$.
Fix $\tau>0$, choose $l\in\N$ such that $\frac{1}{l}<\tau$.
Let $B_l=\{(u,v)\in X\times X:d((y,z),(u,v))<\frac{1}{l} \}$,
where $d((y,z),(u,v))=d(y,u)+d(z,v)$
then
$$
\mu(B_l)>0  \ \text{and}\
\mu_k(B_l)=\frac{1}{n_k}\#(\{0\leq i \leq n_k-1:d(T^ix_k,y)+d(T^ix,z)<\frac{1}{l}\}).
$$
There exist infinite $k\in \Z$ such that $\mu_k(B_l)>\frac{\mu(B_l)}{2}>0$
since $0<\mu(B_l)\leq \liminf_{k\rightarrow\infty}\mu_k(B_l)$.
For $\ep>0$, we can choose $k\in \Z$ satisfying above proposition with $d(x_k,x)<\ep$,
hence
$$
\frac{1}{n_k}\#(\{0\leq i \leq n_k-1:d(T^ix_k,y)<\tau,d(T^ix,z)<\tau\})>\frac{\mu(B_l)}{2}.
$$
It follows that $(y,z)\in Q_{me}(X, T)$.
It is a contradiction which implies the lemma.
\end{proof}

\begin{lem}\label{mean equicontinuous factor}
Let $(X,T)$ be a dynamical system and $\mathcal{A}(Q_{me}(X, T))$ be the
smallest closed $T\times T$ invariant equivalence relation containing $Q_{me}(X, T)$,
then $X/\mathcal{A}(Q_{me}(X, T))$ is the maximal mean equicontinuous factor.

\end{lem}

\begin{proof}
Let $Y=X/\mathcal{A}(Q_{me}(X, T))$ and $\pi:X\rightarrow Y$ be the factor map.
As $\pi\colon X\to Y$ is a continuous surjective,
we can choose a metric on $X$ (also denoted by $d$)
such that $d(x,y)\geq d(\pi(x),\pi(y))$  for all $x,y\in X$.

Assume that $(Y,T)$ is not mean equicontinuous.
By Lemma ~\ref{ME-Q}, there exist $x,y\in Y$ with $x\neq y$ and $(x,y)\in Q_{me}(Y,T)$.
Let $\tau<\frac{1}{4}d(x,y)$.
For $k\in \N$, there are $x_k,y_k\in Y$ and $n_k\in \Z$ with $d(x_k,y_k)<\frac{1}{k}$ such that
$$
\frac{1}{n_k}\#(\{0\leq i \leq n_k-1:d(T^ix_k,x)<\tau,d(T^iy_k,y)<\tau\})>c
$$
for some $c>0$.
For every $k\in \N$,  choose $u_k,v_k\in X$ such that $\pi(u_k)=x_k,\pi(v_k)=y_k$.
Without loss of generality, we can assume that
$\lim_{k\rightarrow\infty}x_k=z=\lim_{k\rightarrow\infty}y_k$
and
$\lim_{k\rightarrow\infty}u_k=w=\lim_{k\rightarrow\infty}v_k$,
then $\pi(w)=z$.
$$\frac{1}{n_k}\sum_{i=0}^{n_k-1}d(T^iu_k,T^iv_k)\geq
\frac{1}{n_k}\sum_{i=0}^{n_k-1}d(T^ix_k,T^iy_k)\geq
c\cdot(d(x,y)-2\tau)>\frac{c}{2}\cdot d(x,y)>0.$$
Let $\mu_k=\frac{1}{n_k}\sum_{i=0}^{n_k-1}\delta_{(T^iu_k,T^iv_k)}$
and  $\nu_k=\frac{1}{n_k}\sum_{i=0}^{n_k-1}\delta_{(T^ix_k,T^iy_k)}$.
Without loss of generality, assume that $\mu_k\rightarrow \mu$
and $\nu_k\rightarrow \nu$, where $\mu\in M(X\times X,T\times T)$
and $\nu\in M(Y\times Y,T\times T)$.
From the proof of Theorem ~\ref{general}, we have
$\mu(\text{supp}(\mu)\setminus \Delta_X)>0$ and $\nu(\text{supp}(\nu)\setminus \Delta_Y)>0$.
Moreover $\pi\times \pi(\mu)=\nu$ since $\pi\times \pi(\mu_k)=\nu_k$.

If $(a,b)\in \text{supp}(\mu) $, by the proof of Theorem ~\ref{general}, we have
$(a,b)\in Q_{me}(X,T)$ which implies $\pi(a)=\pi(b)$, hence $\pi\times \pi (\text{supp}(\mu))=\Delta_Y$.

Therefore $\nu(\Delta_Y)=\pi\times \pi(\mu)(\Delta_Y)=\mu((\pi\times \pi)^{-1}\Delta_Y)=
\mu(\text{supp}(\mu))=1$.

It is a contradiction which implies the lemma.
\end{proof}

By Lemma \ref{ME-Q} and Lemma \ref{mean equicontinuous factor}, we have the main result in this section.

\begin{thm} \label{MAX ME}
Let $(X,T)$ be a dynamical system. Then
 $S_{me}(X, T)$ is  the smallest closed $T\times T$ invariant equivalence relation containing $Q_{me}(X, T)$.
\end{thm}

\medskip
\noindent {\bf Acknowledgments.}
The authors would like to thank Jian Li and Xiangdong Ye for bringing us the questions and for useful discussions when doing the research.
We also thank Jie Li for the careful reading which help the writing of the paper. Finally the authors thank the referee for his/her careful reading.

The authors were supported by NNSF of China (11431012).

\bibliographystyle{amsplain}


TITLE: Trying to define an abstract notion of a function that turns sets of sentences into the set of models satisfying those sentences.
QUESTION [4 upvotes]: Let $\mathsf{Sen}$ denote a Boolean algebra, thought of as a collection of sentences, and let $\mathsf{Mod}$ denote a set without any additional structure, thought of as a collection of models. I'm trying to define (abstractly) the notion that a function $f : \mathcal{P}(\mathsf{Sen}) \rightarrow \mathcal{P}(\mathsf{Mod})$ turns a set of sentences $A \in \mathcal{P}(\mathsf{Sen})$ into the set of all models satisfying those sentences $f(A) \in \mathcal{P}(\mathsf{Mod})$.
Now its clear that $f$ ought to turn arbitrary unions into intersections. That is, for all $\mathcal{A} \subseteq \mathcal{P}(\mathsf{Sen})$ we require that $$f(\bigcup_{A \in \mathcal{A}} A) = \bigcap_{A \in \mathcal{A}}f(A).$$
So my first question is: going on our intuition involving first/second-order logic and first/second-order models, ought we require that $f$ turn arbitrary intersections into unions? And should we require that $f$ interact nicely with complements?
Now furthermore, $f$ ought to be compatible with the Boolean Algebra structure on $\mathsf{Sen}$. For instance, if $\phi,\psi \in \mathsf{Sen},$ then we want $$f(\{\phi \wedge \psi\}) = f(\{\phi\}) \cap f(\{\psi\}),\quad f(\{\phi \vee \psi\}) = f(\{\phi\}) \cup f(\{\psi\}),\quad f(\{\phi^c\}) = \mathsf{Mod} \setminus f(\{\phi\})$$
My second question is: Are the requirements of the above paragraph sufficient, given that they're only written in terms of singleton sets, when after all $f$ can potentially operate on infinite collections of sentences? Again, let us go with the intuition that comes from thinking about first/second-order logic and first/second-order models.

REPLY [4 votes]: $f$ is a Galois connection between the poset of sets of sentences and the poset of sets of models. Here is a general way to induce Galois connections between such posets: let $X, Y$ be sets and let $R \subseteq X \times Y$ be a relation, which we will write $xRy$. Then we get a Galois connection between the poset of subsets of $X$ and the poset of subsets of $Y$ given by the functions
$$f(S) = \{ y : xRy \forall x \in S \}$$
where $S \subseteq X$ and $f(S) \subseteq Y$, and
$$g(T) = \{ x : xRy \forall y \in T \}$$
where $T \subseteq Y$ and $g(T) \subseteq X$. We recover several classical examples of Galois connections this way.

The motivating example behind the name: let $L/K$ be a Galois extension, let $X = K$, and let $Y = \text{Gal}(L/K)$. The relation is $xRy$ iff the group element $y$ fixes the element $x$. This induces the Galois connection between subextensions of $L/K$ and subgroups of $Y$ which is the subject of the fundamental theorem of Galois theory
Let $X = k[x_1, ... x_n]$ ($k$ an algebraically closed field) and let $Y = k^n$. The relation is $xRy$ iff the function $x$ vanishes on the point $y$. This induces the Galois connection between ideals of $k[x_1, ... x_n]$ and subvarieties of $k^n$ which is the subject of the Nullstellensatz.
In this case, let $X$ be the set of sentences and let $Y$ be the set of models. The relation is $xRy$ iff the sentence $x$ is true in the model $y$. This induces the Galois connection between sets of sentences and sets of models which is the subject of the completeness theorem (in first-order logic).

I think it was Lawvere who first noticed this. The slogan is "syntax is adjoint to semantics." Connecting the last two examples together, you can think of sentences as functions on the space of models and taking the set of models satisfying a collection of sentences as taking the subspace on which a collection of functions vanishes.


TITLE: On a proof regarding the sigma algebra generated by a single random variable.
QUESTION [0 upvotes]: I left (b) and (c) for the sake of the curious.
What I am trying to do is Exercise (a) except that I recall that $\sigma(Y):= ( \{ w : Y(w) \in B \} : B \in \mathcal{B} )$ is the definition of sigma algebra of a random variable so I do not see what is there to prove in the first request.

REPLY [2 votes]: Take a look further to the front of the book, maybe you can find the definition of the generated sigma-algebra that the author uses.
Typically, the generated sigma-algebra is defined just in the way you described. In this case you only need to prove the second assertion of this exercise.
However, there are also other definitions. One possible definition is that $\sigma(Y)$ is the intersection of all sigma-algebras with respect to which $Y$ is measurable. If the author defines $\sigma(Y)$ in this way, you need to prove that it coincides with $Y^{-1}(\mathcal{B})$.


TITLE: Properties of the category of smooth vector bundles over a smooth manifold
QUESTION [16 upvotes]: I am wondering if there are any sources that discuss the properties of the category of vector bundles over a smooth manifold. It seems that most differential geometry texts I've looked at avoid explicitly discussing the properties of this category in a systematic fashion.
I know that the category is additive, but are there are any other useful things to know about this category? For instance:

Is it a closed monoidal category with respect to the tensor product?
Is there a sharp result that states the conditions under which kernels and quotients exist?
Under what conditions do functorial constructions on vector spaces induce corresponding constructions on smooth vector bundles?
How much of homological algebra applies to this category?

REPLY [5 votes]: The book Characteristic Classes (This may be behind a paywall) by Milnor & Stasheff answers some of your questions; in particular, they prove a theorem about "continuous functors": any functor on the category of vector spaces which is continuous on the space of isomorphisms between two vector spaces gives a functor on the category of vector bundles on some topological space.
One reason this category is not talked about very often is it is not that nice for exactly the reasons you mention. As Andreas Cap points out above, kernels and quotients do not always exist, and thus, homological algebra does not always work. The problem is you may have some morphisms of vector bundles whose rank changes on a submanifold of smaller dimension; the quotient or kernel of such a morphism won't be a vector bundle, since the dimension is not locally constant, so the local triviality condition is violated. (Some examples here).
Instead, one may work in the category of "coherent sheaves" - these were invented to be a category containing the category of vector bundles on a space, but including the missing kernels/quotients. This is common in algebraic geometry and related fields; the idea is to change your local picture from being some vector space over $\mathbb{R}$ or $\mathbb{C}$ to being a finitely generated module over the ring of smooth functions on your open sets. The main difference is that this allows things like "torsion" - maybe you have an element in your local module that is killed by the ideal of the ring of functions consisting of those functions that vanish on a submanifold of dimension one; this element can be thought of as being supported on that submanifold. This way, if the dimension of a kernel goes up near a point, you can model this behavior with such elements.
Unfortunately, coherent sheaves do not immediately give as nice of a geometric picture as vector bundles, and are fundamentally algebraic objects. The algebra is harder, and [to my knowledge - someone please correct me if I'm mistaken], things don't work so well over real manifolds - you need complex manifolds, where everything is basically algebraic already.


TITLE: Factoring a polynomium with no pattern
QUESTION [3 upvotes]: Along the way of deriving the formula for a sum of integer squares on this webpage (it is about proving the pyramid volume formula), I noticed this step:
$$2n^2+7n+6=(n+2)(2n+3)$$
From right-to-left, you simply multiply the terms into each other's brackets.
But from left-to-right, as is done in the proof, I can't figure out the method.
Is there are formula for such factoring or is it simply a "good guess"?

REPLY [2 votes]: The polynomial $2n^2+7n+6$ is a simple quadratic with discriminant $\Delta = 49-48=1$ and roots $\cfrac{-7 \pm \sqrt{1}}{4}=-2,-\frac{3}{2}\,$, therefore it is divisible by $(n+2)(n+\frac{3}{2})$. The dominant coefficient is $2\,$, so the full factorization is $\,2(n+2)(n+\frac{3}{2}) = (n+2)(2n+3)\,$.


TITLE: Sequence of solutions of an ODE
QUESTION [0 upvotes]: I'm trying to solve the following problem, which says
Let $f : E \to E$ be continuos; suppose $f(x) \leq M$. For each $n=1,2,...,$ let $x_n: [0,1] \to E$ be a solution to $x'=f(x)$. If $x_n(0)$ converges, show that a subsequence of $\{x_n\}$ converges uniformly to a solution.
Using the fact that $f(x) \leq M$ i can fulfill the hypothesis of Arzelà–Ascoli theorem and prove that exist a subsequence of $\{x_n\}$ that converges uniformly, but i believe that i only have convergence and i need to prove that such subsequence converges to a solution, in order to do that i get stuck since i really don't know how to approach it and don't know how to use the fact that $x_n(0)$ converges.
If someone could helping me with that, thank you in advance

REPLY [0 votes]: You need to know that $x_n(0)$ converges in order to prove that $x_n$ is uniformly bounded (and you need to know that, if you want to apply Ascoli-Arzelà!).
This is clear if you apply the fundamental theorem of calculus to your sequence of solution:
$$
x_n(t) = x_n(0) + \int_0^t f(x_n(s)) ds.
$$


TITLE: Why is the sky of the moon always dark?
QUESTION [9 upvotes]: Why is the sky of the moon always dark compared to the sky of the earth, doesn't it have day and night like earth?

REPLY [0 votes]: I would like to expand here on the mechanisms of scattering. Light is scattered by particles in the air which act like dipoles, and oscillate because of the electromagnetic frequency of light. Oscillating dipoles may emit a different frequency. There are two mechanisms of scattering.

Rayleigh scattering:
This occurs when particle sizes are smaller than 10% of the wavelength of light. This causes the blue color of the sky and yellow color of the sun. The scattering is inversely proportional to fourth power of wavelength. When light passes through thicker layers of atmosphere at sunrise or sunset, only longer wavelengths(red and yellow) survive and the rest are scattered. So, the sky which scatters blue wavelengths appears blue and red/yellow wavelengths of sun survive to reach our eyes to give it its color.
Mie scattering:
For larger sized particles, all wavelengths are scattered identically. It is caused by larger particles like water droplets in clouds. That's why they appear white or grey.

As you can see here, Sun from the space looks white!

(source: stanford.edu)


TITLE: For any point inside a triangle
QUESTION [1 upvotes]: Let $P$ be an interior point of the triangle $\triangle ABC$. Assume that $AP$, $BP$ and $CP$ meet the opposite sides $BC$, $CA$ and $AB$ at $D$, $E$ and $F$, respectively. Show that
$\frac{AF}{FB} +\frac{AE}{EC}= \frac{AP}{PD}$.

REPLY [1 votes]: The statement is known as the Van-Aubel theorem.
Here is a simple proof:
Apply Menelaus theorem for triangles $\triangle ACD$ and $\triangle ABD$. You get:
$$\frac {AP}{PD} * \frac {DB}{BC} * \frac {CE} {EA} = 1$$
$$\frac {AF}{FB} * \frac {BC}{CD} * \frac {DP} {AP} = 1$$
Now $$\frac {AE} {CE} = \frac {AP}{PD} * \frac {DB}{BC} $$ and
$$\frac {AF}{FB} = \frac {CD}{BC} * \frac {AP} {PD} $$
Sum these equalities and you obtain what you wanted.


TITLE: Determining relativistic velocities of two moving bodies at an angle
QUESTION [0 upvotes]: I'm currently doing a practice problem from a special relativity textbook and wanted to see if my methodology works. The Question reads:
"Two rocketships leave the earth at equal speeds of c/2 and make equal angles of 60 degrees with the x-direction. find the relative velocity of the two rocketships".
The question doesn't specify what frame to find the velocity from so I'll just assume you find it for the earth's perspective and one of the rocket's perspective.
Using some simple trig, I found the rockets to be moving apart at 0.866c in the earth's frame. To find the velocity from one of the rocket's perspectives I used the equation $U_y'=\frac{U_y\sqrt{1-v^2/c^2}}{ 1-\frac{vU_x}{c^2}}$ but I'm pretty sure I did something wrong because I attempted the calculation a couple times and I got a value greater than c on one attempt.
Very interested to see what you all get or to see if I'm even using the right equation. All feedback is welcome and thanks in advance!

REPLY [0 votes]: The velocity addition formula in this case is:
$V_{x} = \frac {V_{x}^{'} + v}{1+\frac{vV_{x}^{'}}{c^{2}}}$
along $x$-axis.


TITLE: Two ultrafilters converging to different points
QUESTION [2 upvotes]: Suppose $U_1$ and $U_2$  be two different ultrafilters on $\mathbb Z$. Is it always possible to find a $f:\mathbb Z\to \mathbb C$ in $\ell^\infty(\mathbb Z)$ such that $\lim_{U_1}f\neq \lim_{U_2}f$ ?
If the ultrafilers happens to be principal then one can easily find such a $f$.

REPLY [2 votes]: As you point out, if one of them is principal then clearly there exists such an $f$.
More generally, assume $A\in U_1\setminus U_2$. Then in particular the complement of $A$, $A^c \in U_2$. Then we may define $f$ to be $1$ on $A$, $-1$ on $A^c$. It is clearly bounded and $\lim_{U_1}f = 1, \lim_{U_2}f = -1$


TITLE: Vehicle routing with unit demand
QUESTION [1 upvotes]: Hi i am looking at a special type of the VRP where the vehicle isn't restricted by the capacity, but can at most visit a given number of costumers $Q$ in a single route. As I see it, it can be be reprecented as a CVRP where the demand for each location is equal to 1 and the capacity of the vehicle is a number $Q$.
My assumption is that the number of routes needed $M$ is equal to, where $N$ is the total number of costumers:
$M = ceil \left(\frac{N}{Q} \right)$
Is this assumtion true, I have not found any litterature on this topic, and can not myself come up with a proof to the statement... Hope you guys can help.

REPLY [2 votes]: Yes, you can model it as demand $1$ and capacity $Q$.  If $M$ is the number of vehicles, then the total capacity is $M Q$, and this must cover the total demand $N$, so $M Q\ge N$, which implies that $M\ge N/Q$. Because $M$ is integer, you can strengthen the lower bound to $M\ge\lceil N/Q\rceil$. Whether this lower bound is attainable depends on the underlying graph.


TITLE: Is the position of an electron part of his quantum state? Pauli exclusion principle
QUESTION [3 upvotes]: I have a problem regarding the Pauli exclusion principle which as far as I understand states that two or more identical fermions cannot occupy the same quantum state.
So is the position ($r_1$) of an electron in an atom part of its quantum state ?

If yes, why do two electrons in  a $ 1s$ orbital have opposite spins despite having different positions?

I am guessing that its related to Heisenberg uncertainty principle as follows: since the electrons' position's  aren't known with complete precision, they may coincide.

REPLY [1 votes]: If you look at the periodic table, which has ordered the elements according to their behaviour, you can see that there are 2, 2+8 and 2+8+8 electrons in the first three periods. The amazing thing is that you can imagine (please only for yourself) the electrons as bar magnets. In helium they are standing head to foot opposite each other and in 8 electrons they are arranged in the edges of a cube, 4 with the north pole to the nucleus and 4 with the south pole.
This image is helpful for me for understanding the founded empirically Pauli principle or for example the symmetrical behavior of methan CH4.


TITLE: How can I optimize a rational function?
QUESTION [0 upvotes]: I know I need to take the derivative and set it equal to zero but what equations can I create?

REPLY [0 votes]: This is one of many kinds of optimization problems. In general, you should ask yourself, "What is the quantity I want to optimize?", and then "How can I represent this quantity mathematically?".

What is the quantity I want to optimize?
This would be the area of the triangle, and in this case the type of optimization you are seeking is minimizing its area.

How can I represent this quantity mathematically?
The area of a triangle has a well known formula, in particular for $b$ and $h$ the base and height lengths of the triangle:
$$A = \frac{1}{2}bh$$

Put it in a formula of one variable.

The base can be seen to be of length $x$, where $x$ is the point where the triangle intersects the $x$ axis. Similarly, we can see that we can create a formula for the line, namely:
$$y-2 = m(x-3)$$
for $m$ the slope of the line. At $x = 0 $, this gives you the height $h$ of your triangle, and so we can substitute this into our original equation:
$$A(x) = \frac{1}{2}x(-3m+2)$$
Keep in mind though that $m$ is a function of $x$, so you should calculate this prior to differentiating your function. This should be enough to help you finish the rest of your problem.


TITLE: Not Equivalent Interpretation
QUESTION [4 upvotes]: Can someone provide an interpretation to show that the following are not equivalent:
$$\forall x \in D, P(x) \vee Q(x)\;\;\text{vs.}\;\;(\forall x \in D, P(x)) \vee (\forall x \in D, Q(x))$$
They seem equivalent.

REPLY [1 votes]: The first statement asserts that for all $x \in D, P$ is true or $Q$ is true. Perhaps both may be true, assuming the statement uses the inclusive OR. However, this statement does not claim that both $P \text{ and } Q$ must be true for any given $x$. It also does not claim that either (or both) must be true for all $x$.
The second statement on the other hand, claims that either $P$ must be true for all $x$ or $Q$ must be true for all $x$ or both are simultaneously true for all $x$. So for example, with this statement it may not be the case that $P$ is true for all $x$ and $Q$ is true for some $x$ but not others. However, this possibility is included by the first statement.


TITLE: Find a probability of $n$ event happening from $m$ types
QUESTION [1 upvotes]: The question is: to find a sum
$$
S=\sum\limits_{n_1+n_2+\ldots+n_m = n,\ n_i=0,1,\ldots,n} p_1^{n_1}p_2^{n_2}\cdots p_m^{n_m},
$$
where $p_i\in[0,1]$.
UPDATE. This issue has no probabalistic meaning, sorry for misleading you.

REPLY [1 votes]: It seems the following.
Without loss of generality we may assume that all numbers $p_i$ are mutually different, because in the opposite case we simply decrease $m$. Denote $S$ as $S(n;p_1,\dots,p_m)$.
If $m=1$ then $S(n;p_1)=p_1^n$.
If $m=2$ then $S(n;p_1,p_2)=\frac{p_1^{n+1}- p_2^{n+1}}{p_1-p_2}$.
Let $m\ge 3$. Then
$$S(n;p_1,\dots,p_m)=\sum_{k=0}^n \sum_{n_{m-1}+n_m=k} S(n-k;p_1,\dots,p_{m-2})p_{m-1}^{n_{m-1}} p_m^{n_m}=$$ $$\sum_{k=0}^n \sum_{n_{m-1}+n_m=k} S(n-k;p_1,\dots,p_{m-2}) \frac{p_m^{k+1} - p_{m-1}^{k+1}}{p_m -p_{m-1}}=$$ $$\frac{1}{p_m -p_{m-1}}\left(S(n+1;p_1,p_2,\dots,p_{m-2},p_m)- S(n+1;p_1,\dots,p_{m-1}) \right).$$
then by recursion we obtain
$$S(n;p_1,\dots,p_m)=\sum_{i=1}^m p_i^{n+m-1}\prod_{j\ne i} (p_i-p_j)^{-1}.$$
Comparing this expression with Lagrange interpolation formula, we can conclude that $S(n;p_1,\dots,p_m)$ is equal to the highest coefficient of an unique $(m-1)$-degree polynomial $P$ such that $P(x)=x^{n+m+2}$ for each $x\in\{p_1,\dots,p_m\}$. :-)


TITLE: Trouble with newton's method.
QUESTION [2 upvotes]: This is the problem I'm doing and the start of the explanation of how to do it:

Use Newton's Method to approximate to $4$ decimal place: $$\sqrt[3]{11}$$
This would be the equivalence of finding the zero for $x^3-14=0$. So that would make our function $f(x)=x^3-14$. $x_1=3$ will be our starting point since it is closest perfect cube root.

Based on other example problems, I thought it should be $x^3-11 = f(x)$. Trying that, using the formula $x_{n+1} = x_{1} - \frac{x^3-11}{3x^3}$ I just kept getting negative numbers that just bounced back and forth, eventually, between -1.something and -2.something. So with my understanding of how I get f(x) is wrong, or the example problem is wrong again I somehow either screwed up the overly simple math of setting the cube root of a 11 to zero, or (despite checking and adding plenty of parentheses) I'm failing at my calculator input. Or some combination of those. I need held figuring what.

REPLY [2 votes]: It looks like there's a typo in the explanation, it should be $x^3 - 11$ not $x^3 - 14$.  Then the approach you've described attempting is correct except that you've got the wrong $f'$ in your correction term.  It should be:
$$x_{n+1} = x_n - \frac{x_n^3-11}{3x_n^2}$$


TITLE: Convergence of square root operators
QUESTION [2 upvotes]: Let $Q_n$ and $Q$ be compact positive and symmetric operators. Let $A_n = {Q_n}^{\frac12}$ and $A=Q^{\frac12}$. Given $Q_n$ converges to $Q$ w.r.t. operator norm. Does $A_n$ converges to $A$? Thanks.

REPLY [0 votes]: Sketch: The square root is operator monotone. Now, if $Q_n\rightarrow Q$, then, given $\epsilon>0$, there is an $N$ such that for all $n\geq N$,
$$
(Q-\epsilon I)_+\leq Q_n\leq Q+\epsilon I.
$$
It follows that
$$
(Q-\epsilon I)_+^{1/2}\leq Q_n^{1/2}\leq (Q+\epsilon I)^{1/2}.
$$
It therefore suffices to show that
$$
(Q\pm\epsilon I)_+^{1/2}\rightarrow Q^{1/2},
$$
in norm, as $\epsilon\rightarrow 0$. But this convergence we can show by working in an orthonormal eigenbasis for $Q$!
Here is a sketch of a more elementary proof, with wider applicability: Let $f\in C(\mathbb R)$ be any continuous function (for the square root, put $f(x)=0$ for $x\leq 0$). Since $\|Q_n\|\leq \|Q-Q_n\|+\|Q\|$, it suffices to consider $f$ restricted to $[0,2\|Q\|]$ for $n$ large enough. Then we can pick a polynomial $p$ which approximates $f$ uniformly, say $\|p-f\|_{[0,2\|Q\|]}<\epsilon$, and thus
\begin{align*}
\|f(Q)-f(Q_n)\|&\leq\|f(Q)-p(Q)\|+\|p(Q)-p(Q_n)\|+\|p(Q_n)-f(Q_n)\|\\
&\leq 2\epsilon+ \|p(Q)-p(Q_n)\|,
\end{align*}
whenever $n$ is large enough, say $n\geq N$. But it is not hard to show that $p(Q_n)\rightarrow p(Q)$ in norm!


TITLE: Convergence of $a_0 \cdot a_1 \cdot ... \cdot a_n$
QUESTION [4 upvotes]: Let $(a_n)_{n \geq 0}$ be a sequence with $a_0,a_1 \in (0,1)$ and $$a_{n+1}=a_n^2a_{n-1}-a_na_{n-1}+1$$
  Find $$\lim_{n \to \infty} (a_0 \cdot a_1 \cdot ... \cdot a_n)$$

I managed to prove that $a_n \in (0,1)$ and that the sequence is increasing and convergent to $1$. Then I denoted $b_n=a_0 \cdot a_1 \cdot ... \cdot a_n$ and since it is strictly decreasing and bounded, it must be convergent to a number $l$.
Using the relation given I obtained $$b_{n+1}=\frac{b_n^3}{b_{n-1}b_{n-2}}-\frac{b_n^2}{b_{n-2}}+b_n \iff b_{n+1}-b_n=\frac{b_n^2}{b_{n-1}b_{n-2}}(b_n-b_{n-1})$$ and thus got $$b_{n+1}-b_n=\frac{b_n^2 \cdot b_{n-1}}{b_1\cdot b_0}(b_2-b_1)$$
Applying limits yields $0=\frac{l^3(b_2-b_1)}{b_1b_0}$, hence $l=0$.
My question is whether this problem could have been done differently, maybe without the notation of $b_n$. At first I tried to write $$a_0\cdot a_1 \cdot ... \cdot a_n <a_n^{n+1}$$ and then $\lim_{n \to \infty}a_n^{n+1}=e^{\lim_{n \to \infty} (a_n-1)(n+1)}$ and tried to prove that $$\lim_{n \to \infty}n(a_n-1)=-\infty$$ using Cesaro-Stolz, but didn't get anything...

REPLY [5 votes]: $$a_{n+1}=a_n^2a_{n-1}-a_na_{n-1}+1\Longleftrightarrow\frac{1-a_{n+1}}{1-a_n}=a_na_{n-1}$$
By telescopic product we have, $$\frac{a_{n+1}-1}{a_1-1}= \prod_{k=1}^{n}\frac{1-a_{k+1}}{1-a_k}=\prod_{k=1}^{n}(a_ka_{k-1}) = \prod_{k=1}^{n}a_k\prod_{k=1}^{n}a_{k-1}  = a_0\prod_{k=0}^{n}a_k\prod_{k=0}^{n-1}a_{k} $$
Taking limit on both side give,
$$\lim_{n\to \infty} (1-a_n) = a_0(1-a_1)\left(\prod_{k=0}^{\infty}a_k\right)^2$$
Thus, $$\prod_{k=0}^{\infty}a_k = \sqrt{\frac{\lim\limits_{n\to \infty} (1-a_n)}{a_0(1-a_1)}}=0$$
since you proved that $$\lim\limits_{n\to \infty} a_n=1$$


text:
\section{Unused for now}

\subsubsection{Decaying eigenvalues}
The expressions~\eqref{eq:mdl_scaling} and \eqref{eq:adj_scaling} were computed assuming all eigenvalues were constant.
We now provide analytical expressions when the eigenvalues decay, and show that \mdltag\ in such cases can grow
slowly than $\mc O(d/n)$ for $d<n$.
To keep the analysis tractable, we make an assumption that the eigenvalues are arranged such that the product $\evali \myvec_i^2$ decays as $A/i^{\decay}$. In such cases, standard calculus shows that
\begin{align}
    \mdltag(\decay) \precsim \begin{cases}
     \displaystyle\mc{O}\parenth{\frac{C\sqrt {\min\braces{d, n}}}{n}}, \quad&\text{if } \decay= \frac12 \\[4mm]
    \displaystyle\mc{O}\parenth{\frac{C (1+\log(\min\braces{d, n}) }{n}},\quad&\text{if } \decay= 1,

    \end{cases}
\end{align}
for some choices of constant $C$ (which may differ from line-to-line).
In other words, when $\decay=0.5$, \mdltag\ increases sub-linearly with $d$ until $d=n$, and then remains bounded for $d>n$ by $C/\sqrt n$.
When the terms decay faster, i.e., $\decay=1$, the growth of \mdltag\ is even slower ($\log d$) with $d$, and for high-dimensions \mdltag\ remains bounded as $\mc{O}((\log n)/n)$.
One can derive similar expressions for \textrm{Adj-MDL-COMP} under a variety of assumptions, we omit the
derivation for brevity.

We provide more detailed expressions for these cases below:
\begin{align*}
    \decay &= \frac{1}{2} \quad\Longrightarrow \quad
    \mdltag = \begin{cases} \frac{C\sqrt d}{2n} + \frac{d}{2n} \log\parenth{\frac{C}{\sqrt d} + 1}
     = \mc{O}\parenth{\frac{C\sqrt d}{n}}, \quad\text{if } d \leq n\\
    \frac{C}{2\sqrt n} + \frac12 \log\parenth{\frac{C}{\sqrt n} + 1}
    = \mc{O}\parenth{\frac{C}{\sqrt n}} \quad\text{if } d > n.
    \end{cases} \\
    \decay &= 1 \quad\Longrightarrow \quad
    \mdltag = \begin{cases}
    \frac{d}{2n} \log\parenth{\frac{C}{d}+1} + \frac{C}{2n} \log\parenth{C+d}
     = \mc{O}\parenth{\frac{C(1+\log d)}{n}}, \quad\text{if } d \leq n\\
    \frac{1}{2} \log\parenth{\frac{C}{n}+1} + \frac{C}{2n}\log(C+n)
    = \mc{O}\parenth{\frac{C(1+\log n)}{\sqrt n}} \quad\text{if } d > n.
    \end{cases}
    \\
    \decay &= 2 \quad\Longrightarrow \quad
    \mdltag = \begin{cases}
    \frac{d}{2n} \log\parenth{\frac{C}{d^2}+1} + \frac{\sqrt C}{n} \tan\inv\parenth{\frac {d}{\sqrt C} }
     = \mc{O}\parenth{\frac{ \sqrt C}{n}}, \quad\text{if } d \leq n\\
     \frac{1}{2} \log\parenth{\frac{C}{n^2}+1} + \frac{\sqrt C}{n} \tan\inv\parenth{\frac {n}{\sqrt C} }
     = \mc{O}\parenth{\frac{ \sqrt C}{n}}, \quad\text{if } d > n

    \end{cases}
\end{align*}

\subsection{Other variants of MDL-COMP}
\begin{align*}
    \ropt &:= \min_{\lammat}\frac1n \kull{\ptrue}{\qdist_{\lammat}} \\ &= \frac1n\log\det((\mat I_n - \xmat (\xtx+\lammatopt)\inv \xmat \tp )\inv)
    = \frac1n \sum_{i=1}^{\minnd} \log(1+ \frac{\evali}{ \lambda_{i, \mrm{opt} }}) \\
    \lambda_{i, \mrm{opt}}  &= \frac{\sigma_i^2}{w_i^2}, \qtext{and} w = \mat U\tp \tstar \\
    \mc L(\lammatopt) &= \frac{1}{n} \sum_{i=1}^{\minnd} \log( \lambda_{i, \mrm{opt}}/\Delta) \qtext{and} \\
    \mdltag &:= \ropt + \mc L(\lammatopt)
    =  \frac1n \sum_{i=1}^{\minnd} \log(1+ \frac{\evali}{\lambda_{i, \mrm{opt}}})  +  \frac{1}{n} \sum_{i=1}^{\minnd} \log(\lambda_{i, \mrm{opt}}/\Delta) \\
    &= \frac1n\sum_{i=1}^{\minnd} \log( \evali + \lambda_{i, \mrm{opt}}) - \min\braces{1, \frac{d}{n}} \log\Delta
\end{align*}
OR
\begin{align*}
    \mdltag_2 &:= \min_{\lammat}\frac1n \kull{\ptrue}{\qdist_{\lammat}} + \mc L(\lammat)  \\
    &= -\frac{\minnd}{2n} + \min_{\lammat}\frac{1}{2n} \sumn[\minnd] \parenth{
    \frac{
    (\evali\myvec_i^2/\noisevar
    + 1)\lambda_i}{\lambda_i+\evali}
    + \log\parenth{\evali+\lambda_i}
    } \\
    &= -\frac{\minnd}{2n} +\frac{1}{2n} \sumn[\minnd] \log\parenth{\evali} \\
    \lambda_{i, \mrm{opt}} &= 0
\end{align*}
OR
\begin{align*}
    \mdltag_3 &:= \min_{\lambda}\frac1n \kull{\ptrue}{\qdist_{\lambda}} + \frac1n \log (\lambda/\Delta)  \\
    &= -\frac{\minnd}{2n} + \min_{\lambda} \brackets{\frac1n \log (\lambda/\Delta)  + \frac{1}{2n} \sumn[\minnd] \parenth{
    \frac{
    (\evali\myvec_i^2/\noisevar
    + 1)\lambda}{\lambda+\evali}
    + \log\parenth{1+\frac{\evali}{\lambda}}
    }} \\
    \lambda_{\mrm{opt}} &= \text{no closed form}
\end{align*}

config: arxiv


TITLE: Trying to understand a part of the RSA algorithm...
QUESTION [1 upvotes]: The original paper published mentions this...
D(E(M)) ≡ (E(M))^d ≡ (M^e)^d (mod n) = M^(ed) (mod n)

E(D(M)) ≡ (D(M))^e ≡ (M^d)^e (mod n) = M^(ed) (mod n)

and

M^(ed) ≡ M^(k·φ(n)+1) (mod n) (for some integer k).

It's not clear to me how they arrived at the exponent for M in the last line...
k·φ(n)+1
Would anyone be able to help identify what I'm failing to grasp?

REPLY [1 votes]: During key generation $e$ and $d$ were explicitly chosen such that $ed\equiv 1\pmod{\varphi(n)}$, which means exactly that $ed=k\varphi(n)+1$ for some $k$.

REPLY [1 votes]: $d$ is $e$'s inverse modulo $\varphi(n)$. so $d*e = 1(mod(\varphi(n))$.
so $d*e = k*\varphi(n) + 1$ for some integer $k$..


TITLE: Codimension inequality
QUESTION [2 upvotes]: Let $N$ and $R$ be subspace of vector space $X$(not necessary finite dimension)$\DeclareMathOperator{\codim}{codim}$
Assume $N\subset R \subset X$, prove the codimension inequality:
$$\codim R \le \codim N$$
This is very simple if we use the dimension formula: $\dim X = \dim N + \codim N = \dim R + \codim  R\:$ (where codimension is defined as $\codim  N = \dim(X/N)$ )
How to prove it without this formula?

REPLY [0 votes]: Perhaps easier (though really similar under the hood to the various quotient-based arguments, above), using Extension of linearly independent set to a basis in an infinite dimensional vector space .
A basis of $N$, $\mathcal{N}$, can be extended to a basis of $R$, $\mathcal{R}$, which can be extended to a basis of $X$, $\mathcal{X}$.  Then $\mathcal{X} \smallsetminus \mathcal{R} \subseteq \mathcal{X} \smallsetminus \mathcal{N}$, as was to be shown.


TITLE: Eliminating $\theta$ from the system $x\sin\theta-y\cos\theta=-\sin4\theta$, $x\cos\theta+y\sin\theta=\frac52-\frac32\cos4\theta$
QUESTION [2 upvotes]: Eliminate $\theta$ from the system of equations.
  $$\begin{align}
x\sin\theta-y\cos\theta&=\phantom{\frac52\frac32}-\sin4\theta \\
x\cos\theta+y\sin\theta&=\frac52-\frac32\cos4\theta
\end{align}$$

I am stuck at this question after squaring and adding.

REPLY [0 votes]: There is a catch,  a trap sort of. Posting nevertheless..
If we make change of variable by substituting $( x=r \cos  \theta, y= r \sin \theta ) $ it changes in a strange way.
$$ r (\sin \theta \cos \theta - \sin \theta \cos \theta)  = 0 = f(\theta),$$
$$ r( \sin ^2\theta +\cos^2 \theta ) = r = \sqrt{x^2+y^2} = g(\theta) = 5/2-3/2 \cos 4 \theta $$
The first equation says $r=0$ .
The second figure is a 4 leaved rose configured to be never  at the origin.
So problem is a designed contradiction..
$(x,y) $ are determinate or indeterminate constants including null solutions  or so it seems for now to me..


text:
\begin{document}
\preprint{}

\title{Quantum cluster algebra structure on the finite dimensional representations of $U_q(\widehat{sl_{2}})$}

\author{Hai-Tao Ma, Yan-Min Yang, Zhu-Jun Zheng}

\email{zhengzj@scut.edu.cn}
\affiliation{Department of Mathematics\\
 South China University of
 Technology\\ Guangzhou 510641, P. R. China.}

\begin{abstract}
In this paper,
we give a quantum cluster algebra structure on the deformed Grothendieck ring of $\CC_{n}$, where $\CC_{n}$ is  a full subcategory of finite dimensional representations of $U_q(\widehat{sl_{2}})$ defined in section II.

\end{abstract}

\keywords{ Quantum cluster algebra, Deformed Grothendieck ring}

\maketitle

\section{Introduction}

Cluster algebras were introduced by Fomin and Zelevinsky\cite{FZ0}, and they played more and more important role in representation theory and other areas. In 2010, Hernandez and Leclerc \cite{HL0} posted a conjecture on the monoidal categorification of cluster algebras \cite{FZ0} and gave a model of monoidal category for certain cluster algebras. They  proved the case $n = 1$ for the Lie algebras of type $AD$\cite{HL0,HL1}. Almost at the same time, Nakajima\cite{Na1} used the theory of perverse sheaves and q,t-characters to prove the case $n = 1$ for the Lie algebras of type $ADE$. In 2012, Qin \cite{Qin} generalized Nakajima's geometric approach and obtained monoidal categorifications of the cluster algebras associated to any acyclic quiver. Recently, Yang and Zheng\cite{YZ1} proved the conjecture for the case $A_3, n = 2$.

In 2010, Nakajima\cite{Na1} asked that if there is a quantum cluster algebra structure on the deformed Grothendieck ring of full subcategory of the finite dimensional representation of the quantum loop algebra. In 2012, Qin and Yoshiyuki Kimura \cite{QY1} obtained a deformed monoidal categorifications of acyclic quantum cluster algebras with specific coefficients through geometric approach.

    In this paper, by the algebraic definition of q, t-characters given by Hernandez\cite{H0}, we want to give an answer of Nakajima's problem of $U_q(\widehat{sl_{2}})$ case. That is , we proved that for any $n \in \N$, there is a quantum cluster algebra structure on the deformed Grothendieck ring of the full subcategory $\CC_{n}$.

    The paper is organized as follows. In Section 2, we present necessary definitions and
facts of quantum cluster algebras.
In Section 3, we give the main result and the detail of the proof.

\section{Definitions and Notations}\label{s2}
In this section, we give a simplified introduction to the theory of quantum cluster algebras \cite{BZ0} and finite dimensional representations of quantum affine algebras \cite{CP0,CP1,H0}.
\subsection{Quantum cluster algebra}

\begin{definition}
Let $\widetilde{B}$ be an $m \times n$ integer matrix with rows labeled by $[1,m]$ and columns labeled by an n-element subset $\textbf{ex} \subset [1, m]$. Let $\Lambda$ be a skew-symmetric $m \times m$ integer matrix with rows and columns labeled by $[1, m]$. We say that a pair $(\Lambda, \widetilde{B})$ is compatible if, for every $j \in \textbf{ex}$ and $i \in [1, m]$, we have
$$\sum_{ k = 1}^{m}b_{kj}\lambda_{ki} = \delta_{ij}d_{j}$$
for some integers $d_{j}(j \in \textbf{ex}).$
\end{definition}

Fix an index $k \in \textbf{ex}$ and a sign $\varepsilon \in \{\pm 1\},$ the matrix $\widetilde{B}^{'} = \mu_{k}(\widetilde{B})$ is defined as
$$\widetilde{B}^{'} = E_{\varepsilon}\widetilde{B}F_{\varepsilon},$$
where $E_{\varepsilon}$ is the $m \times m$ matrix with entries

$$ e_{ij} =\left\{\begin{array}{ll}
\delta_{ij} & \text{if $j \neq  k$,} \\[.05in]
-1
 & \text{if $i = j = k$,
 }  \\[.05in]
 max(0, -\varepsilon b_{ik}) &\text{if $i\neq j = k$,}
 \end{array}   \right.
$$
$F_{\varepsilon}$ is the $n \times n$ matrix with rows and columns labeled by
$\textbf{ex}$ with  entries

$$ f_{ij} =\left\{\begin{array}{ll}
\delta_{ij} & \text{if $i \neq  k$,} \\[.05in]
-1
 & \text{if $i = j = k$,
 }  \\[.05in]
 max(0, \varepsilon b_{ik}) &\text{if $i = k \neq j$.}
 \end{array}   \right.
$$

Set $\Lambda^{'} = E_{\varepsilon}^{T} \Lambda E_{\varepsilon}.$
\begin{definition}
Let $(\Lambda, \widetilde{B})$ be a compatible pair, and $k \in \textbf{ex}$. We say that the compatible pair $(\Lambda^{'}, \widetilde{B}^{'})$ is obtained from $(\Lambda, \widetilde{B})$ by the mutation in direction k.
\end{definition}

Let $L$ be a lattice of rank m with a skew-symmetric bilinear form $\Lambda : L \times L \rightarrow \Z.$
The based quantum torus associated with $L$ is the $\Z[q^{\pm1/2}]$-algebra $T = T(\Lambda)$ with a distinguished $\Z[q^{\pm 1/2}]$-basis $\{X^{e}| e \in L\}$
and the multiplication given by
$X^{e}X^{f} = q^{\Lambda(e, f)/2}X^{e + f},$
$X^{0} = 1, (X^{e})^{-1} = X^{-e}.$
For a skew-field $F$ of $T$, a toric frame in $F$ is a map $M : \Z^{m} \rightarrow F \backslash \{0\}$ defined as
$M(c) = \varphi(X^{\eta(c)}),$
where $\varphi$ is an automorphism of $F$, and $\eta : \Z^{m} \rightarrow L$ is an isomorphism of lattices.

\begin{definition}
A quantum seed is a pair $(M, \widetilde{B})$, where
M is a toric frame in F; $\widetilde{B}$ is an $m \times n$ integer matrix with rows labeled by $[1, m]$ and columns labeled by an n-element subset $\textbf{ex} \subset [1, m];$ the pair$(\Lambda_{M}, \widetilde{B})$ is a compatible pair.
\end{definition}

Let $(M,\widetilde{B})$ be a quantum seed, and $(M^{'}, \widetilde{B}^{'})$ be the quantum seed obtained by the mutation in direction $ k \in \textbf{ex}$. For $i \in [1, m]$, let $X_{i} = M(e_{i})$ and $X_{i}^{'} = M^{'}(e_{i})$. Then $X_{i}^{'} = X_{i}$ for $i \neq k$, and $X_{k}^{'}$ is given by the following quantum analog of the exchange relation:
$$X_{k}^{'} = M(-e_{k} + \sum_{b_{ik} > 0}b_{ik}e_{i}) + M(- e_{k} + \sum_{b_{ik} < 0}b_{ik}e_{i}).$$

Two quantum seeds are mutation-equivalent if they can be obtained from each other by a sequence of quantum seed mutations. For a quantum seed $(M,\widetilde{B}),$ we denote by $\widetilde{\textbf{X}} = (X_{1}, X_{2},\cdots,X_{m})$ the corresponding "free generating set" in $F$ given by $X_{i} = M(e_{i})$. We call the subset$\textbf{X} = \{X_{j} | j \in \textbf{ex}\} \subset \widetilde{\textbf{X}}$ the cluster of the quantum seed $(M, \widetilde{B}),$
and set $\textbf{C} = \widetilde{\textbf{X}} - \textbf{X}$.

\begin{definition}
Let $S$ be a mutation-equivalence class of quantum seeds in $F$. The quantum cluster algebra associated with $S$ is the $\Z[q^{\pm1/2}]$-subalgebra of the ambient skew-field F, generated by the union of clusters of all seeds in $S$, together with the element of $\textbf{C}$.
\end{definition}

\subsection{Quantum loop algebra}
$U_q(\widehat{sl_{2}})$ is the corresponding quantum affine algebra with parameter $q\in C^*$ not a root of unity. $\CC$ is the category of finite-dimensional $U_q(\widehat{sl_{2}})$-representations of type 1.
For $\ell\in \Z_{\geq0}$,   $\CC_\ell$ be the full subcategory of $\CC$:
 for any $V$ of $\CC_{\ell}$, the roots of the Drinfeld polynomials of
every composition factor of $V$ belong to $\{q^{-2k}\mid 0\le k \le \ell\}$.

Denote the Grothendieck ring of $\CC_{\mathbb{\ell}}$ by $R_{\mathbb{\ell}}$, then $R_{\mathbb{\ell}}=\mathbb{Z}\big[[W_{1,2k}]_{i\in I,0\leq k\leq \ell}\big]$, where  $W_{1,2k}$ are Kirillov-Reshetikhin modules with the highest l-weights
$m_{1,2k} = Y_{q^{2k}}.$

\begin{theorem}\cite{H0}
In the $sl_{2}$ case, the deformed Grothendieck ring $Rep_{t,n}$ is a $\Z[q^{\pm1/2}]$-algebra generated by $[W_{1,0}], [W_{1,2}], \cdot\cdot\cdot, [W_{1,2n}]$ with relations
$$[W_{1,l_{1}}]\ast[W_{1,l_{2}}] = q^{r}[W_{1,l_{2}}]\ast[W_{1,l_{1}}], if \ l_{1} \geq l_{2} \ and \ l_{1} \neq l_{1} + 2, where \ r = (-1)^{(l_{1} - l_{2})/2}.$$
$$[W_{1,l}]\ast[W_{1,l - 2}] = q^{-1}[W_{1,l-2}]\ast[W_{1,l}] + (1 - q^{-1}). $$
\end{theorem}

\section{Main Results}\label{s3}

In this section, we construct a quantum cluster algebra structure on the deformed Grothendieck ring  of $\CC_{n}$  for the Lie algebra $sl_2$. To be easily reading, we give an example for a special case.

The $\Gamma_{n}$ is the quiver
\[
\begin{matrix} n+1& \rightarrow & n &\rightarrow & \cdots&\rightarrow & 1
\end{matrix},
\]

where
$n+1$ is the frozen point.

 Let $\widetilde{B_{n}}$ is a matrix correspond to $\Gamma_{n}$ , i.e
$$\widetilde{B_{n}} = \left(
                        \begin{array}{ccccc}
                          0 & -1 & 0 & \cdot\cdot\cdot & 0 \\
                          1 &0 & -1 & \cdot\cdot\cdot & 0 \\
                          0 & 1 & 0 & \cdot\cdot\cdot & 0 \\
                          0 & 0& 1 & \cdot\cdot\cdot & 0 \\
                          \vdots &\vdots & \vdots  & \vdots  & \vdots  \\
                          0 & 0 & 0 & \cdot\cdot\cdot & 1 \\
                        \end{array}
                      \right)_{(n+1,n)}
$$

Let $\Lambda_{n} = (\lambda_{ij})_{(n+1,n+1)}$  be the skew matrix  defined by

$$ \lambda_{ij} =\left\{\begin{array}{ll}
(-1)^{i+j+1} & \text{if $j > i$ and i is odd,} \\[.05in]
0
 & \text{if $j > i$ and i is even ,
 }  \\[.05in]
 - \lambda_{ji} &\text{if $j < i$.}
 \end{array}   \right.
$$
That is
$$\Lambda = \left(
    \begin{array}{ccccccc}
      0 & -1 & 0 & -1 & 0 & -1 & \cdot\cdot\cdot \\
      1 & 0 & 0 & 0 & 0 & 0 & \cdot\cdot\cdot \\
      0 & 0 & 0 & -1 & 0 & -1 & \cdot\cdot\cdot \\
      1 & 0 & 1 & 0 & 0 & 0 & \cdot\cdot\cdot \\
      \vdots& \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\
    \end{array}
  \right)_{(n+1,n+1)}.
$$
It is easy to see $(\Lambda_{n},\widetilde{B}_{n})$ is compatible.

Let $\A_{n} = \A(\Gamma_{n})$  be the quantum cluster algebra associated with a  pair$(\Lambda_{n},\widetilde{B}_{n})$. The ambient field $F$ of fractions of the quantum torus with generators $Y_{1}, Y_{2}, \cdots, Y_{n+1}$ satisfying relations $Y_{i}Y_{j} = q^{\lambda_{ij}}Y_{j}Y_{i}$.
For $0 \leq i \leq n-1$, let $X_{2i}=\mu_{n-i}(Y_{n-i})$, and $X_{2n} = Y_{1}$.

\begin{proposition}\label{equithm}
 $\A_n $ as a  $\Z[q^{\pm1/2}]$ algebra can be generated by $\{X_{2i} \ | \ 0 \leq i \leq n\}.$
\end{proposition}

\begin{proof}
Since quantum clusters are mutation equivalence if and only if the correspondence clusters are mutation equivalence,
the number of the quantum cluster variables is equal to the number of the cluster variables. The cluster variable is one to one correspondence to the set $\{[W_{i,j}] | 1\leq i \leq n+1 , 0 \leq j \leq 2n - 2i +2, {\rm where }~ j~ {\rm is ~ even} \}$\cite{L1}.

By the definition of the quantum cluster algebra, we have the following relations:
$$X_{2n-2}X_{2n} = q^{1/2}Y_{2} + 1;$$
$$X_{2n-2i}Y_{i} = q^{-1/2}Y_{i-1} + Y_{i+1},{\rm if~ }i~ {\rm is~ even};$$
$$X_{2n-2i}Y_{i} = Y_{i-1} + q^{1/2}Y_{i+1},{\rm if} ~ i~ {\rm is~ odd};$$

By the above relations, it is easy to see that $Y_{i}(1 \leq i \leq n) \in \Z[X_{2i}(0 \leq i \leq n+1),q^{\pm1/2}]$.
For example, \ $Y_{3} = X_{2n-4}Y_{2} + q^{-1/2}Y_{1} = q^{-1/2}(X_{2n-4}X_{2n-2}X_{2n} - X_{2n-4} + X_{2n}).$

 Set $\mu = \mu_{n}\mu_{n-1}\cdots\mu_{1}$, where $\mu_{i}$ is the mutation at the direction i, $(Y_{1}^{'}, Y_{2}^{'},\cdot\cdot\cdot,Y_{n}^{'},Y_{n+1}) = \mu(Y_{1}, Y_2, \cdots,Y_{n+1})$ , $Y_{i}^{''} = \mu_{i}(Y_i^{'})$ . By the mentioned above and the T-system, $(Y_{1}^{'}, Y_{2}^{'},\cdot\cdot\cdot,Y_{n}^{'},Y_{n+1})$ is correspondence to  $(W_{1,2n-2}, W_{2,2n-4}, \cdot\cdot\cdot, W_{n,0}, W_{n+1,0})$.  So we have $Y_{i}^{''} = X_{2n-2-2i}(1 \leq i \leq n-1)$, \ $Y_{1}^{'} = X_{2n-2}$. Similarly, we have
$Y_{i}^{'}(1 \leq i \leq n) \in \Z[X_{2i}(0 \leq i < n),q^{\pm1/2}].$

By the induction,\ we have that any quantum cluster variable belongs to $Z[X_{2i}(0 \leq i \leq n),q^{\pm1/2}].$
\end{proof}

\begin{theorem}\emph{}\label{thm}
The map
\[
\begin{array}{ccc}
  X_{i} \mapsto [W_{1,i}]
\end{array}
\]
extends to a ring isomorphism $\iota_{n}$  from the quantum cluster algebra $\A_{n}$
to the deformed Grothendieck ring $Rep_{t,n}$ of $\CC_n$.
\end{theorem}
\begin{proof}
First, if i is odd,
\[
\begin{array}{ll}
  & X_{2(n - i)}X_{2(n - i - 1)} \\
  = & (Y_{i - 1} + q^{1/2}Y_{i+1})Y_{i}^{-1}(q^{-1/2}Y_{i} + Y_{i + 2})Y_{i+1}^{-1} \\
  = & q^{-1/2}Y_{i - 1}Y_{i+1}^{-1} + 1 + Y_{i-1}Y_{i}^{-1}Y_{i+2}Y_{i+1}^{-1} + q^{-1/2}Y_{i}^{-1}Y_{i+2}.
\end{array}
\]

\[
\begin{array}{ll}
   & X_{2(n - i - 1)}X_{2(n - i)} \\
  = & (q^{-1/2}Y_{i} + Y_{i + 2})Y_{i+1}^{-1}(Y_{i - 1} + q^{1/2}Y_{i+1})Y_{i}^{-1} \\
  = & q^{1/2}Y_{i - 1}Y_{i+1}^{-1} + 1 + qY_{i-1}Y_{i}^{-1}Y_{i+2}Y_{i+1}^{-1} + q^{1/2}Y_{i}^{-1}Y_{i+2}.
\end{array}
\]

So we have$X_{2(n - i)}X_{2(n - i - 1)} = q^{-1}X_{2(n - i - 1)}X_{2(n - i)} + (1 - q^{-1})$.

If i is even,
\[
\begin{array}{ll}
   & X_{2(n - i)}X_{2(n - i - 1)}\\
  = & (q^{-1/2}Y_{i - 1} + Y_{i+1})Y_{i}^{-1}(Y_{i} + q^{1/2}Y_{i + 2})Y_{i+1}^{-1} \\
  = & q^{-1/2}Y_{i - 1}Y_{i+1}^{-1} + 1 + Y_{i-1}Y_{i}^{-1}Y_{i+2}Y_{i+1}^{-1} + q^{-1/2}Y_{i}^{-1}Y_{i+2}.
\end{array}
\]

\[
\begin{array}{ll}
   & X_{2(n - i - 1)}X_{2(n - i)} \\
  = & (Y_{i} + q^{1/2}Y_{i + 2})Y_{i+1}^{-1}(q^{-1/2}Y_{i - 1} + Y_{i+1})Y_{i}^{-1} \\
  = &  q^{1/2}Y_{i - 1}Y_{i+1}^{-1} + 1 + qY_{i-1}Y_{i}^{-1}Y_{i+2}Y_{i+1}^{-1} + q^{1/2}Y_{i}^{-1}Y_{i+2}.
\end{array}
\]

So we have$X_{2(n - i)}X_{2(n - i - 1)} = q^{-1}X_{2(n - i - 1)}X_{2(n - i)} + (1 - q^{-1})$.

Secondly,
\[
\begin{array}{ll}
   & X_{2(n - i)}X_{2(n - j)}\\
  = & (Y_{i - 1} + q^{1/2}Y_{i+1})Y_{i}^{-1}(Y_{j} + q^{1/2}Y_{j + 1})Y_{j}^{-1}\\
  = & Y_{i - 1}Y_{i}^{-1}Y_{j-1}Y_{j}^{-1} + q^{1/2}Y_{i+1}Y_{i}^{-1}Y_{j-1}Y_{j}^{-1}
 + q^{1/2}Y_{i - 1}Y_{i}^{-1}Y_{j+1}Y_{j}^{-1} + qY_{i+1}Y_{i}^{-1}Y_{j+1}Y_{j}^{-1}.
\end{array}
\]

if i and j is odd and $i < j$;
\[
\begin{array}{ll}
   & X_{2(n - j)}X_{2(n - i)} \\
 = & Y_{j - 1} + q^{1/2}Y_{j+1})Y_{j}^{-1}(Y_{i} + q^{1/2}Y_{i + 1})Y_{i}^{-1} \\
  = & Y_{j-1}Y_{j}^{-1}Y_{i - 1}Y_{i}^{-1} + q^{1/2}Y_{j-1}Y_{j}^{-1}Y_{i+1}Y_{i}^{-1}
 + q^{1/2}Y_{j+1}Y_{j}^{-1}Y_{i - 1}Y_{i}^{-1} + qY_{j+1}Y_{j}^{-1}Y_{i+1}Y_{i}^{-1} \\
  = & q^{-1}(Y_{i - 1}Y_{i}^{-1}Y_{j-1}Y_{j}^{-1} + q^{1/2}Y_{i+1}Y_{i}^{-1}Y_{j-1}Y_{j}^{-1}
 + q^{1/2}Y_{i - 1}Y_{i}^{-1}Y_{j+1}Y_{j}^{-1} + qY_{i+1}Y_{i}^{-1}Y_{j+1}Y_{j}^{-1}) \\
  = & q^{-1}X_{2(n - i)}X_{2(n - j)} \\
  = & q^{(-1)^{j - i + 1}}X_{2(n - i)}X_{2(n - j)}.
\end{array}
\]
 If i is odd,j is even and $i +1 < j$;
 \[
 \begin{array}{ll}
    & X_{2(n - j)}X_{2(n - i)} \\
   = & (Y_{j - 1} + q^{1/2}Y_{j+1})Y_{j}^{-1}(Y_{i} + q^{1/2}Y_{i + 1})Y_{i}^{-1} \\
   = & Y_{j-1}Y_{j}^{-1}Y_{i - 1}Y_{i}^{-1} + q^{1/2}Y_{j-1}Y_{j}^{-1}Y_{i+1}Y_{i}^{-1}
 + q^{1/2}Y_{j+1}Y_{j}^{-1}Y_{i - 1}Y_{i}^{-1} + qY_{j+1}Y_{j}^{-1}Y_{i+1}Y_{i}^{-1} \\
   = & q(Y_{i - 1}Y_{i}^{-1}Y_{j-1}Y_{j}^{-1} + q^{1/2}Y_{i+1}Y_{i}^{-1}Y_{j-1}Y_{j}^{-1}
 + q^{1/2}Y_{i - 1}Y_{i}^{-1}Y_{j+1}Y_{j}^{-1} + qY_{i+1}Y_{i}^{-1}Y_{j+1}Y_{j}^{-1}) \\
   = & qX_{2(n - i)}X_{2(n - j)} \\
   = &q^{(-1)^{j - i + 1}}X_{2(n - i)}X_{2(n - j)}
 \end{array}
 \]

 If i and j are even,and $i < j$;

 \[
 \begin{array}{ll}
    & X_{2(n - j)}X_{2(n - i)} \\
   = & (Y_{j - 1} + q^{1/2}Y_{j+1})Y_{j}^{-1}(Y_{i} + q^{1/2}Y_{i + 1})Y_{i}^{-1} \\
   = & Y_{j-1}Y_{j}^{-1}Y_{i - 1}Y_{i}^{-1} + q^{1/2}Y_{j-1}Y_{j}^{-1}Y_{i+1}Y_{i}^{-1}
 + q^{1/2}Y_{j+1}Y_{j}^{-1}Y_{i - 1}Y_{i}^{-1} + qY_{j+1}Y_{j}^{-1}Y_{i+1}Y_{i}^{-1} \\
   = & q^{-1}(Y_{i - 1}Y_{i}^{-1}Y_{j-1}Y_{j}^{-1} + q^{1/2}Y_{i+1}Y_{i}^{-1}Y_{j-1}Y_{j}^{-1}
 + q^{1/2}Y_{i - 1}Y_{i}^{-1}Y_{j+1}Y_{j}^{-1} + qY_{i+1}Y_{i}^{-1}Y_{j+1}Y_{j}^{-1}) \\
   = & q^{-1}X_{2(n - i)}X_{2(n - j)} \\
   = & q^{(-1)^{j - i + 1}}X_{2(n - i)}X_{2(n - j)}
 \end{array}
 \]

 If i is even,j is odd,and $i+1 < j$;

 \[
 \begin{array}{ll}
    & X_{2(n - j)}X_{2(n - i)} \\
   = & (Y_{j - 1} + q^{1/2}Y_{j+1})Y_{j}^{-1}(Y_{i} + q^{1/2}Y_{i + 1})Y_{i}^{-1} \\
   = & Y_{j-1}Y_{j}^{-1}Y_{i - 1}Y_{i}^{-1} + q^{1/2}Y_{j-1}Y_{j}^{-1}Y_{i+1}Y_{i}^{-1}
 + q^{1/2}Y_{j+1}Y_{j}^{-1}Y_{i - 1}Y_{i}^{-1} + qY_{j+1}Y_{j}^{-1}Y_{i+1}Y_{i}^{-1} \\
   = & q(Y_{i - 1}Y_{i}^{-1}Y_{j-1}Y_{j}^{-1} + q^{1/2}Y_{i+1}Y_{i}^{-1}Y_{j-1}Y_{j}^{-1}
 + q^{1/2}Y_{i - 1}Y_{i}^{-1}Y_{j+1}Y_{j}^{-1} + qY_{i+1}Y_{i}^{-1}Y_{j+1}Y_{j}^{-1}) \\
   = & qX_{2(n - i)}X_{2(n - j)} \\
   = & q^{(-1)^{j - i + 1}}X_{2(n - i)}X_{2(n - j)}
 \end{array}
 \]

 So we have $X_{2(n - i)}X_{2(n - j)} = q^{j - i}X_{2(n - j)}X_{2(n - i)}$.

Thus,we get the conclusion.
\end{proof}

\subsection{Example}
The quiver $\Gamma_{2}$ are defined by
\[
\begin{matrix} 1& \leftarrow & 2 &\leftarrow & 3
\end{matrix}.
\]

where
$3$ is the frozen point.

$$\widetilde{B}_{2}=\left(
                      \begin{array}{cc}
                        0 & -1 \\
                        1 & 0 \\
                        0 & 1 \\
                      \end{array}
                    \right)
, \qquad
\Lambda_{2}=\left(
              \begin{array}{ccc}
                0 & -1 & 0 \\
                1 & 0 & 0 \\
                0 & 0 & 0 \\
              \end{array}
            \right)
$$

Since
$$\widetilde{B}_{2}^{T}\Lambda_{2} = \left(
                                      \begin{array}{ccc}
                                        0 & 1 & 0 \\
                                        -1 & 0 & 1 \\
                                      \end{array}
                                    \right)
              \left(
              \begin{array}{ccc}
                0 & -1 & 0 \\
                1 & 0 & 0 \\
                0 & 0 & 0 \\
              \end{array}
            \right)
            =\left(
               \begin{array}{ccc}
                 1 & 0 & 0 \\
                 0 & 1 & 0 \\
               \end{array}
             \right),
$$
  that is, $(\Lambda_{2},\widetilde{B}_{2})$ is compatible.

Let $\A_{n} = \A(\Gamma_{2})$  be the quantum cluster algebra associated with a  pair$(\Lambda_{2},\widetilde{B}_{2})$. By the definition of the quantum cluster algebra,we have $X_{0} = (q^{-1/2}Y_{1} + Y_{3})Y_{2}^{-1}$, \ $X_{2} = Y_{1}^{-1} + q^{1/2}Y_{2}Y_{1}^{-1}$, \ $X_{4} = Y_{1}$, where $(Y_{1}, Y_{2}, Y_{3})$ is the initial quantum cluster.
\begin{proposition}\label{equithm}
 $\A_2$ as a  $\Z[q^{\pm1/2}]$ algebra can be generated by $X_{0}, X_{2}, X_{4}.$
\end{proposition}

\begin{proof}
The detailed expressions of quantum cluster variables are shown as follows:
$Y_{1}, ~Y_{2},~ Y_{3},$~
$Y_{1}^{'}=Y_{1}^{-1} + q^{-1/2}Y_{1}^{-1}Y_{2} = X_{2},~$
$Y_{2}^{'}=q^{1/2}Y_{3}Y_{1}^{-1}Y_{2}^{-1} + Y_{2}^{-1} + Y_{1}^{-1}Y_{3}$,~
$ Y_{1}^{''}=Y_{3}Y_{2}^{-1} + q^{-1/2}Y_{1}Y_{2}^{-1} = X_{0}$.

Because the relation $Y_{1}Y_{2}=q^{-1}Y_{2}Y_{1}$, we have:
$$Y_{1}Y_{1}^{'} = 1+q^{-1/2}Y_{2},$$
$$Y_{1}^{''}Y_{1}^{'} = 1+q^{-1/2}Y_{2}^{'},$$
$$X_{0}Y_{2} = q^{-1/2}Y_{1} + Y_{3}.$$
That is:
$$Y_{2}=q^{1/2}(Y_{1}Y_{1}^{'}-1) = q^{1/2}(X_{4}X_{2} - 1),$$
$$Y_{2}^{'} = q^{1/2}(Y_{1}^{''}Y_{1}^{'} - 1) = q^{1/2}(X_{0}X_{2} - 1),$$
$$Y_{3} = X_{0}Y_{2} - q^{-1/2}X_{4}.$$
So we get the conclusion.

\end{proof}

\begin{theorem}\emph{}\label{thm}
The map
\[
\begin{array}{ccc}
  X_{4} \mapsto [W_{1,4}], & X_{0} \mapsto [W_{1,0}], & X_{2} \mapsto [W_{1,2}]
\end{array}
\]
extends to a ring isomorphism $\iota_{2}$ from the quantum cluster algebra $\A_2$
to the deformed Grothendieck ring $Rep_{t,2}$ of $\CC_2$.
\end{theorem}
\begin{proof}
Since

\[
\begin{array}{ll}
   & Y_{1}^{'}Y_{1}^{''} \\
  = & Y_{1}^{-1}Y_{3}Y_{2}^{-1} + q^{-1/2}Y_{1}^{-1}Y_{2}Y_{3}Y_{2}^{-1}
+ q^{-1}Y_{1}^{-1}Y_{2}Y_{1}Y_{2}^{-1} + q^{-1/2}Y_{2}^{-1}\\
  = & Y_{1}^{-1}Y_{2}^{-1}Y_{3} + q^{-1/2}Y_{1}^{-1}Y_{3} + q^{-1/2}Y_{2}^{-1} +  1
\end{array}
\]
and

\[
\begin{array}{ll}
   & Y_{1}^{''}Y_{1}^{'} \\
  = & Y_{3}Y_{2}^{1}Y_{1}^{-1} + q^{-1/2}Y_{1}Y_{2}^{-1}Y_{1}^{-1}
+ q^{-1/2}Y_{3}Y_{2}^{-1}Y_{1}^{-1}Y_{2} + q^{-1}Y_{1}Y_{2}^{-1}Y_{1}^{-1}Y_{2} \\
  = &  qY_{1}^{-1}Y_{2}^{-1}Y_{3} + q^{1/2}Y_{1}^{-1}Y_{3} + q^{1/2}Y_{2}^{-1} +  1 ,
\end{array}
\]
then we have
$$Y_{1}^{'}Y_{1}^{''} = q^{-1}Y_{1}^{''}Y_{1}^{'} + (1 - q^{-1}).$$

Since
$$Y_{1}^{'}Y_{1} = 1 + q_{1/2}Y_{1}^{-1}Y_{2}Y_{1} = 1 + q^{1/2}Y_{2}$$
and
$$Y_{1}Y_{1}^{'} = 1 + q^{-1/2}Y_{1},$$
then we have
$$Y_{1}Y_{1}^{'} = q^{-1}Y_{1}^{'}Y_{1} + (1 - q^{-1}).$$

Since
$$Y_{1}Y_{1}^{''} = Y_{1}Y_{3}Y_{2}^{-1} + q^{-1/2}Y_{1}^{2}Y_{2}^{-1} = Y_{1}Y_{2}^{-1}Y_{3} + q^{-1/2}Y_{1}^{2}Y_{2}^{-1}$$
and
$$Y_{1}^{''}Y_{1} = Y_{3}Y_{2}^{-1}Y_{1} + q^{-1/2}Y_{1}Y_{2}^{-1}Y_{2} = q^{-1}Y_{1}Y_{2}^{-1}Y_{3} + q^{-3/2}Y_{1}^{2}Y_{2}^{-1},$$
then we have
$$Y_{1}Y_{1}^{''} = qY_{1}^{''}Y_{1}.$$
so we get the conclusion.
\end{proof}


TITLE: Does nesscery for every $V\subseteq A$ open relative to $A$ $f(V)$ is open relative to $B$
QUESTION [0 upvotes]: Let $f:A\to B$ continuous function and prove or disprove the following statments.
a) Does nesscery for every $V\subseteq A$ open relative to $A$ $f(V)$ is open relative to $B$
b) Does nesscery for every $V\subseteq A$ close relative to $A$ $f(V)$ is close relative to $B$

For b) I think I can choose $A=\mathbb{R}^n$ and setting $f(x)=arctan(x)$ and the image is an open set.
for a) I think it's not true but I don't have an example any hint?

REPLY [1 votes]: a). Let $A=B=\Bbb R$ and $f(x)=x^2$ and $V=(-1,1).$ Then $V$ is open in $A$ but $f[V]=[0,1)$ is not open in $B.$
b). Let $A=\Bbb R^2$ and $B=\Bbb R.$ For $(x,y)\in A$ let $f(\,(x,y)\,)=x.$ Then $V=\{(x,1/x):0\ne x\in \Bbb R\}$ is closed in $A$ but $f[V]=\Bbb R \setminus \{0\}$ is not closed in B.
A function that maps open (closed) sets to open (closed) sets is called an open (closed) function. Continuous functions need not be open nor closed functions.
If $f:A\to B$ is continuous and $V$ is a compact subset of $A$ then $f[V]$ is a compact subset of $B$. And a compact subset of a Hausdorff space must be closed. So in the example for b) above, we choose a closed non-compact $V$.


TITLE: Milnor's octahedron
QUESTION [5 upvotes]: If $V$ is a vector space, then a curvaturelike tensor in $V$ is a quadrilinear map $F:V \times V \times V \times V \to \Bbb R$ satisfying
1) $F(x,y,z,w) = -F(y,x,z,w) = -F(x,y,w,z)$;
2) $F(x,y,z,w) = F(z,w,x,y)$;
3) $F(x,y,z,w) + F(y,z,x,w) + F(z,x,y,w) = 0$,
for all $x,y,z,w \in V$. It turns out that the Bianchi identity (item 3) actually implies item 2, assuming 1. This implication is called "Milnor's octahedron argument".
I'd like to know where this appeared for the first time. Thanks!

REPLY [4 votes]: I don't know if it's truly the first appearance, but it seems likely the name originates from Milnor's classic 1963 book Morse Theory, in which you can find the following illustrated argument:
Here (1), (2) and (3) are your 1) and 3), with (4) being your conclusion 2).