text:
\begin{document}
\title[]
{Compact complement topologies and k-spaces}
\author{K. Keremedis, C. \"{O}zel, A. Pi\k{e}kosz, M. A. Al Shumrani, E. Wajch}
\address{Department of Mathematics,
University of Aegean,
Karlovassi, Samos 83200, Greece}
\email{kker@aegean.gr}
\address{Department of Mathematics, King Abdulaziz University,
P.O.Box: 80203 Jeddah 21589, Saudi Arabia}
\email{cenap.ozel@gmail.com}
\address{Institute of Mathematics, Cracow University of Technology, Warszawska 24, 31-155 Krak\'ow, Poland}
\email{pupiekos@cyfronet.pl}
\address{Department of Mathematics, King Abdulaziz University,
P.O.Box: 80203 Jeddah 21589, Saudi Arabia}
\email{maalshmrani1@kau.edu.sa}
\address{Siedlce University of Natural Sciences and Humanities, 3 Maja 54, 08-110 Siedlce, Poland}
\email{eliza.wajch@wp.pl}

\thanks{}

\subjclass{Primary: 54D50, 54D55, 54A35, 54E99. Secondary: 54D30,  54E35, 54E25}

\date{\today}

\begin{abstract}
Let $(X,\tau)$ be a Hausdorff space, where $X$ is an infinite set. The compact complement topology $\tau^{\star}$ on $X$ is defined by: $\tau^{\star}=\{\emptyset\} \cup \{X\setminus M, \text{where $M$ is compact in $(X,\tau)$}\}$. In this paper,  properties of the space $(X, \tau^{\star})$ are studied in $\mathbf{ZF}$ and applied to a characterization of $k$-spaces, to the Sorgenfrey line, to some statements independent of $\mathbf{ZF}$, as well as to partial topologies that are among Delfs-Knebusch generalized topologies. Between other results, it is proved that the axiom of countable multiple choice (\textbf{CMC}) is equivalent with each of the following two sentences: (i) every Hausdorff first countable space is a $k$-space, (ii) every metrizable space is a $k$-space. A \textbf{ZF}-example of a countable metrizable space whose compact complement topology is not first countable is given.

{\bf Keywords:} compact complement topology; countable multiple choice; $k$-space; sequential space;  Sorgenfrey line; Delfs-Knebusch generalized topological space; partial topology.

\end{abstract}

\maketitle

\section{Introduction}
The compact complement topology of the real line was considered, for instance, in Example 22 of the celebrated book by Steen and Seebach"Counterexamples in Topology" (\cite{SS}). We investigate this notion in a much wider context of Hausdorff spaces and of partially topological spaces that belong to the class of generalized topological spaces in the sense of Delfs-Knebusch (cf. \cite{DK} and \cite{Pie1}). Our results are proved in $\mathbf{ZF}$ if this is not otherwise stated. All axioms of $\mathbf{ZF}$ can be found in \cite{Ku}.

In section 2, we give elementary properties of the compact complement topology of a Hausdorff space. In particular, we show that if a Hausdorff space is locally compact and second countable, then its compact complement topology is second countable, while the compact complement topology of a non-locally compact metrizable space need not be first countable. We give an example of a countable metrizable space whose compact complement topology is not first countable. In section 3, a necessary and sufficient condition for a set to be compact with respect to the compact complement topology of a given Hausdorff space leads us to a new characterization of k-spaces. A well-known theorem of $\mathbf{ZFC}$ states that all first countable Hausdorff spaces are $k$-spaces (cf. Theorem 3.3.20 of \cite{E}). We show that this theorem may fail in $\mathbf{ZF}$. More precisely, we prove that, if $\mathcal{M}$ is a model of $\mathbf{ZF}$, then all Hausdorff first countable spaces of $\mathcal{M}$ are k-spaces if and only if all metrizable spaces in $\mathcal{M}$ are $k$-spaces which holds if and only if the axiom of countable multiple choice (Form 126 in \cite{HR}) is true in $\mathcal{M}$. In consequence, in some models of $\mathbf{ZF}$ there are metrizable spaces that are not $k$-spaces. We prove that if the Sorgenfrey line is a $k$-space, then the real line with its natural topology is sequential, so the Sorgenfrey line fails to be a $k$-space in some models of $\mathbf{ZF}$. In section 4, we introduce a notion of a compact complement partial topology corresponding to a given partial topology. Partially topological Delfs-Knebusch generalized topological spaces were introduced in Definition 2.2.67 of  \cite{Pie1}; however,  a more convenient than in \cite{Pie1} definition of a partial topology was given in \cite{OPSW}.

\section{Basic properties of compact complement topologies}

Throughout this article, we assume that $\tau$ is a topology on an infinite set $X$ such that $(X,\tau)$ is a Hausdorff space.

\begin{definition} \label{star}
 We denote by $\mathcal{K}(\tau)$ the collection of all $\tau$-compact sets, i.e. of all sets that are compact in the space $(X, \tau)$. The \emph{compact complement topology} of $(X, \tau)$ is the collection
$$\tau^{\star}=\{\emptyset\} \cup \{X\setminus M: M\in\mathcal{K}(\tau)\}.$$
\end{definition}

Since it is true in $\mathbf{ZF}$ that a compact subspace of a Hausdorff space is closed (see Theorem 3.1.8 of \cite{E}), it is easy to check in $\mathbf{ZF}$ that $\tau^{\star}$ is a topology on $X$.

For a subset $Y$ of $X$ and a topology $\mathcal{T}$ on $X$, let
$$\mathcal{T}\vert Y=\{V\cap Y: V\in\mathcal{T}\}.$$
Then $(Y, \mathcal{T}\vert Y)$ is a topological subspace of $(X, \mathcal{T})$.

\begin{theorem}\label{theorem1} Let $Y\subseteq X$. The following conditions hold in $\mathbf{ZF}$:
\begin{enumerate}
\item[(i)] $\tau^{\star}\vert Y$ is coarser than $\tau\vert Y$, i.e., $\tau^{\star}\vert Y \subseteq \tau\vert Y$;
\item[(ii)] if $Y$ is compact in $(X, \tau)$, then $\tau^{\star}\vert Y=\tau\vert Y$;
\item[(iii)] $\tau^{\star}\vert Y=\tau\vert Y$ if and only if there exists a $\tau$-compact set $C$ such that $Y\subseteq C$.
\end{enumerate}
\end{theorem}

\begin{proof} To prove (i), it suffices to show that $\tau^{\star}\subseteq \tau$. Let $U \in \tau^{\star}$ and $U\neq\emptyset$. Then $U=X\setminus M$ for a compact subspace $M$ of $(X,\tau)$. Since a compact subspace of a Hausdorff space is closed, $M$ is closed in $(X,\tau)$. Hence, $U \in \tau$ and, in consequence, $\tau^{\star}\subseteq \tau$.

(ii) Suppose that $Y$ is $\tau$-compact and $V\in \tau$. Since $(X, \tau)$ is Hausdorff, the set $Y$ is $\tau$-closed, so $A=Y\cap (X\setminus V)$ is a $\tau$-closed subset of the $\tau$-compact set $Y$. Hence, $A$ is $\tau$-compact. Notice that $V\cap Y=Y\cap(X\setminus A)$. This implies that $V\cap Y\in\tau^{\star}\vert Y$ and $\tau\vert Y\subseteq\tau^{\star}\vert Y$.

(iii) If $C$ is a $\tau$-compact set such that $Y\subseteq C$, then since it follows from (ii) that $\tau\vert C=\tau^{\star}\vert C$, we immediately deduce that $\tau\vert Y=\tau^{\star}\vert Y$. Finally,  suppose that $Y$ is a subset of $X$ such that $\tau^{\star}\vert Y=\tau\vert Y$. Let $V\in\tau$ and $\emptyset\neq V\cap Y\neq Y$. Since $V\cap Y\in\tau^{\star}\vert Y$, there exists a $\tau$-compact set $K_0$ such that $V\cap Y=Y\setminus K_0$. Fix $x_0\in V\cap Y$. Then $x_0\notin K_0$. By the $\tau$-compactness of $K_0$, there exists a pair $U_1, U_2$ of disjoint members of $\tau$ such that $x_0\in U_1$ and $K_0\subseteq U_2$. Of course, $U_2\cap Y\neq\emptyset$ because $V\cap Y\neq Y$. Since $U_1\cap Y$ and $U_2\cap Y$ are both in $\tau^{\star}\vert Y$, there exist $\tau$-compact sets $K_1, K_2$ such that $U_i\cap Y=Y\setminus K_i$ for $i\in\{1,2\}$. Let $K=K_1\cup K_2$. Then $K$ is $\tau$-compact and $Y=Y\setminus (U_1\cap U_2)=(Y\setminus U_1)\cup (Y\setminus U_2)\subseteq K_1\cup K_2=K$.
\end{proof}

\begin{corollary} $(X, \tau)$ is compact if and only if $\tau=\tau^{\star}$.
\end{corollary}

\begin{remark}  \label{2.4}
In general, $\tau^{\star}\vert Y$ is not equal to $(\tau \vert Y)^{\star}$. For instance, if $Y$ is the open interval $(0, 1)$ of $\mathbb{R}$, while $\tau_{nat}$ is the usual topology of $\mathbb{R}$, then $\tau_{nat}^{\star}\vert Y\neq(\tau_{nat} \vert Y)^{\star}$ . If this does not lead to misunderstanding, we shall denote the space $(\mathbb{R}, \tau_{nat})$ by $\mathbb{R}$ and call it the real line.
\end{remark}

From the fact that $\tau^{\star}\subseteq \tau$, we have the following obvious results:

\begin{proposition} (i) If $(X,\tau)$ is separable,  so is $(X, \tau^{\star})$.\\

(ii) If $(X, \tau)$ is hereditarily separable, so is $(X, \tau^{\star})$.\\

(iii)  If $(X, \tau)$ is hereditarily Lindel\"of, so is $(X, \tau^{\star})$.\\

(iv) If $(X,\tau)$ is connected, so is $(X, \tau^{\star})$.
\end{proposition}

The statement that every infinite set is Dedekind infinite (Form 9 in \cite{HR}) is denoted by \textbf{Fin} in Definition 2.13 of \cite{Her}.

\begin{theorem} The following sentences are equivalent in $\mathbf{ZF}$:
\begin{enumerate}
\item[(i)] \textbf{Fin}.
\item[(ii)] For every discrete space $(X, \tau)$, the space $(X, \tau^{\star})$ is hereditarily separable.
\item[(iii)] For every uncountable discrete space $(X, \tau)$, the space $(X, \tau^{\star})$ is separable.
\end{enumerate}
\end{theorem}
\begin{proof} Let $(X, \tau)$ be a discrete space, i.e. $\tau=\mathcal{P}(X)$. If $X$ is countable, then, of course, $(X, \tau^{\star})$ is hereditarily separable. Consider the case when  $Y\subseteq X$ and $Y$ is uncountable. If \textbf{Fin} holds, then $Y$ is Dedekind infinite, so $Y$ contains an infinitely countable subset. It is clear that when $D$ is an infinitely countable subset of $Y$, then $D$ is dense in $(Y, \tau^{\star}\vert Y)$. Hence, (i) implies (ii). It is obvious that (ii) implies (iii) and that (iii) implies (i).
\end{proof}

\begin{corollary} It is consistent with $\mathbf{ZF}$ that there exists an uncountable discrete space $(X, \tau)$ such that $(X, \tau^{\star})$ is separable and not hereditarily separable.
\end{corollary}
\begin{proof} Let $\mathcal{M}$ be any model of $\mathbf{ZF}$ in which $\mathbb{R}$ contains an uncountable Dedekind finite set. For instance, $\mathcal{M}$ can be Cohen's original model $\mathcal{M}$1 of \cite{HR}. Then if $\tau$ is the discrete topology on $\mathbb{R}$, the space $(\mathbb{R}, \tau^{\star})$ is not hereditarily separable because for each uncountable Dedekind finite subset $Y$ of $\mathbb{R}$, the space $(Y, \tau^{\star}\vert Y)$ is not separable. Of course, $(\mathbb{R}, \tau^{\star})$ is separable because $\mathbb{R}$ is Dedekind infinite.
\end{proof}

\begin{proposition}
For every Hausdorff space $(X, \tau)$, the space $(X, \tau^{\star})$ is $T_{1}$.
\end{proposition}

\begin{proof}
Let $x \in X$.  Since finite sets are compact, we have that $X\setminus \{x\}$ is open in $(X, \tau^{\star})$. Hence, $(X, \tau^{\star})$ is a $T_{1}$-space.
\end{proof}

\begin{proposition}
$(X,\tau)$ is not compact if and only if $(X, \tau^{\star})$ is not Hausdorff. Moreover, if $(X,\tau)$ is not compact, then any two non-empty $\tau^{\star}$-open sets have a non-empty intersection.
\end{proposition}

\begin{proof}
Assume that $(X,\tau)$ is not compact. Let $U$ and $V$ be any two non-empty open sets in $(X, \tau^{\star})$. Then $X\setminus U$ and $X\setminus V$ are compact  in $(X,\tau)$, so $(X\setminus U)\cup (X\setminus V)$ is compact in $(X, \tau)$. Hence, $X\neq (X\setminus U)\cup (X\setminus V)= X\setminus (U\cap V)$. This implies that $U\cap V\neq\emptyset$; thus, $(X, \tau^{\star})$ is not Hausdorff. On the other hand, if we assume that $(X, \tau^{\star})$ is not Hausdorff, then, since $(X, \tau)$ is Hausdorff, we have $\tau\neq\tau^{\star}$, so $(X, \tau)$ is not compact by Corollary 2.3.
\end{proof}

\begin{corollary}
If $(X,\tau)$ is not compact, then the following conditions are satisfied:
\begin{enumerate}
\item[(i)]  every set $V\in\tau^{\star}$ is connected in $(X, \tau^{\star})$;
\item[(ii)]  $(X, \tau^{\star})$  is connected and locally connected.
\end{enumerate}
\end{corollary}

\begin{proof} Suppose that $(X, \tau)$ is not compact and that $\emptyset\neq V\in\tau^{\star}$. If $V$ were disconnected in $(X, \tau^{\star})$, there would exist a pair $U, W$ of non-empty disjoint members of $\tau^{\star}$ which would contradict Proposition 2.9. Hence, $V$ is connected in $(X, \tau^{\star})$. This is why (i) holds. Of course, (ii) follows from (i).
\end{proof}

\begin{remark} Some authors call a topological space \emph{hyperconnected} or \emph{irreducible} if all open sets of this space are connected. In the light of Corollary 2.10, if $(X, \tau)$ is not compact, the space $(X, \tau^{\star})$ is hyperconnected.
\end{remark}

\begin{theorem} If $(X, \tau)$ is locally compact and second countable, then $(X, \tau^{\star})$ is second countable.
\end{theorem}

\begin{proof} Assume that $\mathcal{B}$ is a countable open base of a locally compact Hausdorff space $(X, \tau)$. Let $\mathcal{A}$ be the collection of all sets $U\in\mathcal{B}$ which have compact closures $\text{cl}_{\tau}U$ in $(X, \tau)$. By the local compactness of $(X, \tau)$, the collection $\mathcal{A}$ is an open base of $(X, \tau)$. Let $[\mathcal{A}]^{<\omega}$ be the collection of all finite subcollections of $\mathcal{A}$.  We put
$$\mathcal{B}^{\star}=\{ X\setminus\text{cl}_{\tau}(\bigcup\mathcal{C}): \mathcal{C}\in [\mathcal{A}]^{<\omega}\}.$$
Then $\mathcal{B}^{\star}$ is a countable subcollection of $\tau^{\star}$. To check that $\mathcal{B}^{\star}$ is an open base of $(X, \tau^{\star})$, let us consider any non-empty set $V\in\tau^{\star}$ and a point $x\in V$. Let $K=X\setminus V$ and let $\mathcal{U}$ be the collection of all $U\in\mathcal{A}$ such that $x\notin\text{cl}_{\tau} U$. Since $(X, \tau)$ is Hausdorff, we have $K\subseteq\bigcup\mathcal{U}$. By the compactness of $K$, there exists a finite $\mathcal{U}_K\subseteq\mathcal{U}$ such that $K\subseteq\bigcup\mathcal{U}_K$. Let $W=X\setminus\text{cl}_{\tau}(\bigcup\mathcal{U}_K)$. Then $W\in\mathcal{B}^{\star}, x\in W$ and $W\subseteq V$.
\end{proof}

The axiom of countable choice, denoted by $\mathbf{CC}$ in \cite{Her}, states that every non-empty countable collection of non-empty sets has a choice function (see Form 8 in \cite{HR}). The axiom of countable choice for $\mathbb{R}$, denoted by $\mathbf{CC}(\mathbb{R})$ in \cite{Her}, states that every non-empty countable collection of non-empty subsets of $\mathbb{R}$ has a choice function (see Form 94 in \cite{HR}).

\begin{remark} In view of Exercise E3 to Section 4.6 of \cite{Her},  $\mathbf{CC}(\mathbb{R})$ is equivalent to the statement: for every second countable topological space $Z$,  every open base of $Z$ contains a countable open base of $Z$. Let  us notice that, if $\mathcal{M}$ is a model of $\mathbf{ZF}$ in which there exists a dense infinite Dedekind finite subset $D$ of $\mathbb{R}$, then it holds true in $\mathcal{M}$ that the collection $\mathcal{B}^{\star}$ of all sets of the form $\mathbb{R}\setminus\bigcup_{i\in n+1}[a_i, b_i]$ with $n\in\omega, a_i,b_i\in D$ and $a_i<b_i$ for each $i\in n+1$ is an open base of $(\mathbb{R}, \tau_{nat}^{\star})$ which does not contain a countable open base of $(\mathbb{R}, \tau_{nat}^{\star})$.
\end{remark}

We recall that a topological space $(Z, \mathcal{T})$ is \emph{submetrizable} if there exists a metrizable topology $\mathcal{T}'$ on $Z$ such that $\mathcal{T}'\subseteq\mathcal{T}$.

\begin{theorem} The following conditions are equivalent:
\begin{enumerate}
\item[(i)] $(X, \tau^{\star})$ is metrizable;
\item[(ii)] $(X, \tau^{\star})$ is submetrizable;
\item[(iii)] $(X, \tau)$ is a compact metrizable space.
\end{enumerate}
\end{theorem}

\begin{proof} Of course, (i) implies (ii). Assume (ii). If $(X, \tau)$ is not compact, then $(X, \tau^{\star})$ is not Hausdorff by Proposition 2.9. Hence, (ii) implies that $(X, \tau)$ is compact. In this case, $\tau=\tau^{\star}$ by Corollary 2.3. In consequence, $(X, \tau)$ is both compact and submetrizable. Since every compact submetrizable space is metrizable, (ii) implies (iii). That (iii) implies (i) follows from Corollary 2.3.
\end{proof}

\begin{proposition} Let $x_0\in X$. Then $\{x_0\}$ is of type $G_{\delta}$ in $(X, \tau^{\star})$ if and only if $X\setminus\{x_0\}$ is a $\sigma$-compact subspace of $(X, \tau)$.
\end{proposition}
\begin{proof} \emph{Necessity.} Suppose that $\{U_n:n\in\omega\}\subseteq \tau^{\star}$ and $\{x_0\}=\bigcap_{n\in\omega} U_n$. Then the sets $K_n= X\setminus U_n$ are all compact in $(X, \tau)$ and $X\setminus \{x_0\}=\bigcup_{n\in\omega} K_n$, so $X\setminus\{x_0\}$ is $\sigma$-compact in $(X, \tau)$.

\emph{Sufficiency.} Suppose that $X\setminus\{x_0\}=\bigcup_{n\in\omega} C_n$ where all the sets $C_n$ are compact in $(X, \tau)$. Then the sets $V_n= X\setminus C_n$ are all open in $(X, \tau^{\star})$ and $\{x_0\}=\bigcap_{n\in\omega} V_n$.
\end{proof}

\begin{corollary} If $(X, \tau)$ is not $\sigma$-compact, then the following conditions are satisfied:
\begin{enumerate}
\item[(i)] there does not exist a one-point set of type $G_{\delta}$ in $(X, \tau^{\star})$;
\item[(ii)] $(X, \tau^{\star})$ is not first countable;
\item[(iii)] $(X, \tau^{\star})$ is not second countable;
\item[(iv)] $(X, \tau^{\star})$ is not quasi-metrizable.
\end{enumerate}
\end{corollary}

\begin{remark} We denote by $\mathbb{S}$ the Sorgenfrey line, i.e the topological space $(\mathbb{R}, \tau_{\mathbb{S}})$ where $\tau_{\mathbb{S}}$ is the topology on $\mathbb{R}$ which has as an open base the collection of all half-open intervals $[a, b)$ where $a, b\in\mathbb{R}$ and $a<b$.  The Sorgenfrey line is one of the most frequently used examples of a submetrizable, quasi-metrizable but not metrizable space, so we shall pay a special attention to it.

The countable union theorem (Form 31 in \cite{HR}, abbreviated to $\mathbf{CUT}$ in \cite{Her}) states that countable unions of countable sets are countable sets. Let $\mathbf{CUT}(\mathbb{R})$ be the statement: for every family $\{A_n: n\in\omega\}$ of countable subsets of $\mathbb{R}$, the union $\bigcup_{n\in\omega}A_n$ is countable (see Form 6 in \cite{HR})). It is easy to prove in $\mathbf{[ZF+CUT(\mathbb{R})]}$ that the Sorgenfrey line is not $\sigma$-compact by using the following simple argument: since all compact subsets of $\mathbb{S}$ are countable, if $\mathbb{S}$ were $\sigma$-compact, $\mathbb{R}$ would be a countable union of countable sets; however, $\mathbb{R}$  cannot be a countable union of countable sets because $\mathbb{R}$ is uncountable. This is not a proof in $\mathbf{ZF}$ that the Sorgenfrey line is not $\sigma$-compact because $\mathbf{CUT}(\mathbb{R})$ fails in some models of $\mathbf{ZF}$ (see Theorem 10.6 of \cite{J}).
\end{remark}

\begin{proposition} In every model of $\mathbf{ZF}$,  the Sorgenfrey line is not $\sigma$-compact.
\end{proposition}
\begin{proof} Consider any countable collection $\{K_n: n\in\omega\}$ of compact sets of the Sorgenfrey line. Then all  the sets $K_n$ are countable, closed in $\mathbb{R}$ and they do not have left accumulation points in $\mathbb{R}$. Therefore, each $K_n$ is nowhere dense in $\mathbb{R}$. Since $\mathbb{R}$ is a separable completely metrizable space, by Theorem 4.102 of \cite{Her}, the interior in $\mathbb{R}$ of the set $\bigcup_{n\in\omega} K_n$ is empty. Hence, $\mathbb{R}\neq\bigcup_{n\in\omega} K_n$.
\end{proof}

\begin{corollary} The compact complement topology of the Sorgenfrey line is not first countable.
\end{corollary}

\begin{corollary} The compact complement topology of the Sorgenfrey line is not quasi-metrizable.
\end{corollary}

\begin{proposition} The compact complement topology of the real line $\mathbb{R}$ is quasi-metrizable.
\end{proposition}

\begin{proof} For $x\in\mathbb{R}$, let $m(x)=\min\{n\in\omega: \vert x\vert<n\}$. For each $x\in\mathbb{R}$ and $n\in\omega$, we define a set $G(n, x)$ by putting:
$$G(n, x)=(x-\frac{1}{2^{n+1}}, x+\frac{1}{2^{n+1}})\cup (-\infty, -m(x)-n-2)\cup(m(x)+n+2, +\infty).$$
It is clear that, for each $x\in\mathbb{R}$,  the collection $\mathcal{B}(x)=\{ G(n, x): n\in\omega\}$ is a base of neighbourhoods of $x$ in $(\mathbb{R}, \tau_{nat}^{\star})$. One can check by a simple calculation that the following condition  satisfied: for all $x, y\in\mathbb{R}$ and $n\in\omega$, if $y\in G(n+1, x)$, then $G(n+1, y)\subseteq G(n, x)$. Let us notice that Theorem 10.2 of \cite{Gr} (Chapter 10 of \cite{KV}) holds true in $\mathbf{ZF}$, so we can infer from it that $(\mathbb{R}, \tau_{nat}^{\star})$ is quasi-metrizable in $\mathbf{ZF}$.
\end{proof}

We are going to give a simple $\mathbf{ZF}$-example of a countable  metrizable space whose compact complement topology is not first countable. We shall use the following lemma in this and in the third section:

\begin{lemma} Let us assume that  $\{A_n: n\in\omega\}$ is a collection of non-empty pairwise disjoint sets, $A=\bigcup_{n\in\omega}A_n$ and $Z=A\cup\{\infty\}$ where $\infty\notin A$. For $x, y\in Z$ let  $d(x,y)=d(y,x)$ and $d(x,x)=0$; for each pair $x,y$ of distinct points of $Z$, let $d(x,y)=\max\{\frac{1}{2^n}, \frac{1}{2^m}\}$ if $x\in A_n$ and $y\in A_m$; moreover, let $d(x, \infty)=\frac{1}{2^n}$ if $x\in A_n$. Then the function $d: Z\times Z\to\mathbb{R}$ is a metric on $Z$ such that $A$ is not closed in $(Z, \tau(d))$, while each $A_n$ is a clopen discrete subspace of $(Z, \tau(d))$ where $\tau(d)$ is the topology on $Z$ induced by $d$.
\end{lemma}
\begin{proof} Using the fact that  $\max\{a,b\}\leq\max\{a, c\}+\max\{c, b\}$ for all non-negative real numbers $a,b,c$, one can easily check that $d$ is a metric on $Z$. Since $\infty\in\text{cl}_{\tau(d)} A$, the set $A$ is not closed in $(Z,\tau(d))$. It is obvious that each $A_n$  is a clopen discrete subspace of $(Z, \tau(d))$.
\end{proof}

\begin{example} Let  $A_n=\{n\}\times \omega$ for each $n\in\omega$ and let $A=\bigcup_{n\in\omega}A_n$ Take a point $\infty\notin A$ and put $Z=A\cup\{\infty\}$. Consider the metric $d$ on $Z$ defined in Lemma 2.22. Suppose that the point $(0, 0)$ has a countable base $\{V_n: n\in\omega\}$ of open neighbourhoods in $(Z, \tau({d})^{\star})$ where $\tau(d)$ is as in Lemma 2.22. We may assume that $V_n\subseteq V_0$ for each $n\in\omega$.  The sets $Z\setminus V_n$ are all $\tau(d)$-compact, while the sets $A_n$ are not $\tau(d)$-compact because they are infinite discrete subspaces of $(Z,\tau(d))$. Hence, $A_n\cap V_n\neq\emptyset$ for each $n\in\omega$. For $n\in\omega$, let $a_n=\min\{m\in\omega: (n, m)\in A_n\cap V_n\}$. We define points $x_n\in A_n\cap V_n$ by putting $x_n=(n, a_n)$ for $n\in\omega$. Notice that the set $K=\{ x_n: n\in\omega\setminus\{0\}\}\cup\{\infty\}$ is $\tau(d)$-compact, while $(0,0)\notin K$. Then $V=Z\setminus K$ is an open neighbourhood of $(0, 0)$ in $(Z, \tau(d)^{\star})$. There must exist  $n\in\omega$ such that $V_n\subseteq V$. This is impossible because $V_n\subseteq V_0$ and $x_n\in V_n$ for each $n\in\omega\setminus\{0\}$. The contradiction obtained proves that  $(Z, \tau(d)^{\star})$ is not first countable. Obviously, $(Z, \tau(d))$ is $\sigma$-compact because $Z$ is countable. Of course, $(Z, \tau(d))$ is second countable as a separable metrizable space.  The point $\infty$ is not a point of local compactness of $(Z, \tau(d))$. This example shows that, in Theorem 2.12, the assumption of local compactness of $(X, \tau)$ cannot be replaced by the assumption that the set of points of non-local compactness of $(X,\tau)$ is finite.
\end{example}

An arbitrary example of a metrizable second countable not $\sigma$-compact space also shows that the assumption of local compactness is essential in Theorem 2.12.

\begin{example} Let $X=\mathbb{R}\setminus\mathbb{Q}$ and let $\tau=\tau_{nat}\vert X$. Then the space of irrationals $(X, \tau)$ is second countable. That $(X, \tau)$ is not $\sigma$-compact in $\mathbf{ZF}$ can be shown by using the facts that the Baire category theorem holds in $\mathbf{ZF}$ in the class of separable completely metrizable spaces (see Theorem 4.102 of \cite{Her}) and that every compact subspace of $(X, \tau)$ is nowhere dense in $(X, \tau)$.  This is why the compact complement topology $(\tau_{nat}\vert X)^{\star}$ is not first countable, so it is not second countable. Of course, the space of irractionals is not locally compact at each one of its points.
\end{example}

\begin{remark} It was shown in Theorem 2.7 of \cite{W} that if $\mathcal{T}$ is the co-finite topology on a set $Z$, then the space $(Z, \mathcal{T})$ is quasi-metrizable if and only if $Z$ is a countable union of finite sets. Now, suppose that $\tau$ is the discrete topology on $X$, i.e. $\tau$  is the power set $\mathcal{P}(X)$ of $X$. Then $\tau^{\star}$ is the co-finite topology on $X$. Hence, for $\tau=\mathcal{P}(X)$, the space $(X, \tau^{\star})$ is quasi-metrizable if and only if $X$ is a countable union of finite sets. In some models of $\mathbf{ZF}$ in which a countable union of finite sets can fail to be countable, even when $X$ is uncountable and $\tau=\mathcal{P}(X)$, then $(X, \tau^{\star})$ can be quasi-metrizable (see \cite{W}).
\end{remark}

The following question does not seem to be trivial:

\begin{question}  What are, expressed in terms of $\tau$,  simultaneously necessary and sufficient conditions for $(X, \tau^{\star})$ to be quasi-metrizable when $(X, \tau)$ is a $\sigma$-compact quasi-metrizable space?
\end{question}

\begin{remark} Let us consider the case when $(X, \tau)$ is not compact. We notice that if $p$ and $\hat{p}$ are properties such that a topological space $Z$ has $p$ if and only if $Z$ is Hausdorff and has $\hat{p}$, then, in view of Proposition 2.9, the space $(X, \tau^{\star})$ does not have $p$. In particular, $(X, \tau^{\star})$ is not a $T_i$-space for $i\in\{2, 3, 3\frac{1}{2}, 4, 5, 6\}$. It is easily seen that $(X, \tau^{\star})$ is neither regular, nor completely regular, nor normal. Every continuous mapping from $(X, \tau^{\star})$ to a Hausdorff space is constant.
\end{remark}

\begin{theorem} Let $A\subseteq X$. Then $A$ is $\tau^{\star}$-compact if and only if $A\cap K$ is $\tau$-closed for each $\tau$-compact set $K$.
\end{theorem}

\begin{proof} \emph{Necessity}. Suppose  that $A$ is $\tau^{\star}$-compact. Let $K$ be a $\tau$-compact set. Since $(X, \tau)$ is Hausdorff, to show that $A\cap K$ is $\tau$-closed, it suffices to check that $A\cap K$ is $\tau$-compact. Let $\mathcal{F}$ be a collection of $\tau$-closed sets such that the collection $\mathcal{H}=\{F\cap A\cap K: F\in\mathcal{F}\}$ is centred.  The sets $A\cap K\cap F$ for $F\in\mathcal{F}$ are all $\tau^{\star}\vert A$-closed. Since $A$ is $\tau^{\star}$-compact and $\mathcal{H}$ is a centred collection of $\tau^{\star}\vert A$-closed sets, we have  that $\bigcap\mathcal{H}\neq\emptyset$. This proves that $A\cap K$ is $\tau$-compact.

\emph{Sufficiency}. Now, suppose that $A\cap K$ is $\tau$-compact for each $\tau$-compact set $K$. We may assume that $A\neq\emptyset$.  Let $\mathcal{U}$ be a non-empty collection of non-empty sets such that $\mathcal{U}\subseteq \tau^{\star}$, while $A\subseteq\bigcup\mathcal{U}$. Fix any  non-empty set $U_0\in\mathcal{U}$. The set $C_0=X\setminus U_0$ is $\tau$-compact, so $A\cap C_0$ is $\tau$-compact as a $\tau$-closed subset of a $\tau$-compact set. Notice that $A\cap C_0\subseteq \bigcup\mathcal{U}$ and, by Theorem 2.2, $\mathcal{U}\subseteq\tau$. By the $\tau$-compactness of $A\cap C_0$, there exists a finite collection $\mathcal{V}\subseteq\mathcal{U}$ such that $A\cap C_0\subseteq\bigcup\mathcal{V}$. Then $A\subseteq U_0\cup\bigcup\mathcal{V}$. This proves that $A$ is $\tau^{\star}$-compact.
\end{proof}

\begin{corollary} For every Hausdorff space $(X, \tau)$, the space $(X, \tau^{\star})$ is compact.
\end{corollary}

A topological space $(Z, \mathcal{T})$ is called \emph{jointly partially metrizable on compact subspaces}, if there is a metric $d$ on $Z$ such that, for every compact subspace $A$ of $(Z, \mathcal{T})$, the restriction of $d$ to $A\times A$ generates the subspace topology $\mathcal{T}\vert A$ on $A$ (see  \cite{ArhShu}).

\begin{example}
The space $(\mathbb{R},\tau_{nat})$ is metrizable, hence jointly partially metrizable on compact subspaces. But $(\mathbb{R}, \tau_{nat}^{\star})$ is not jointly partially metrizable on compact subspaces since it is compact and not metrizable for it is not Hausdorff.
\end{example}

A topological space $Z$ is called C-\emph{normal} if there exists a normal space $Y$ and a bijective function $f : Z \rightarrow Y$ such that the restriction $f|A : A \rightarrow f(A)$ is a homeomorphism for each compact subspace $A$ of $Z$ (see \cite{KaZah}).

\begin{example}
The space $(\mathbb{R},\tau_{nat})$ is C-normal. But $(\mathbb{R}, \tau_{nat}^{\star})$ is not C-normal since it is compact and not normal.
\end{example}

\section{$k$-spaces}

Let us recall that a Hausdorff space $Z$ is called a \emph{k-space} if, for every set $A\subseteq Z$, it holds true that  $A$ is closed in $Z$ if and only if $A\cap K$ is closed in $Z$ for each compact set $K$ in $Z$ (see Section 3.3 of \cite{E}).

We deduce directly from Theorem 2.28  the following characterization of $k$-spaces:

\begin{theorem} For every Hausdorff space $(X, \tau)$, it holds true in $\mathbf{ZF}$ that $(X, \tau)$ is a $k$-space if and only if every $\tau^{\star}$-compact subset of $X$ is $\tau$-closed.
\end{theorem}

We recall definitions of sequential and Fr\'echet-Urysohn spaces for completeness.

\begin{definition} Let $Z$ be a topological space and $A\subseteq Z$. Then:
\begin{enumerate}
\item[(i)] $A^{s}$  denotes the set of all points $z\in Z$ such that there exists a
sequence $(z_{n})_{n\in\omega}$ of points of $A\setminus \{z\}$ which converges in $Z$ to the point $z$;
\item[(ii)] $A$ is called \emph{sequentially
closed} if $A^{s}\subseteq A$;
\item[(iii)]  the \emph{sequential closure} of $A$ in $Z$ is the set $\text{scl}_Z(A)=A^{s}\cup A$;
\item[(iv)] $Z$ is called \emph{sequential} (resp. \emph{Fr\'echet-Urysohn}) if every
sequentially closed subset of $Z$ is closed in $Z$ (resp. for every $F\in
\mathcal{P}(Z)$ the equality $\text{scl}_Z(F)=\text{cl}_Z(F)$ holds).
\end{enumerate}
\end{definition}

In some texts,  Fr\'echet-Urysohn spaces are called Fr\'echet spaces (see, for instance, \cite{E} and \cite{Her}). It is well known that the following series of implications hold true in $\mathbf{ZFC}$ and, in general, none of them
is reversible in $\mathbf{ZFC}$ (see, e.g. Sections 1.6 and Theorem 3.3.20 of \cite{E}):

\begin{itemize}
\item $Z$ is Hausdorff and first countable $\rightarrow $ $Z$ is Hausdorff and Fr\'echet-Urysohn $\rightarrow $
$Z$ is Hausdorff and sequential $\rightarrow $ $Z$ is a $k$-space.
\end{itemize}

Of course, the proof to Theorem 3.3.20 of \cite{E} shows that it is true in $\mathbf{ZF}$ that every Hausdorff sequential space is a $k$-space. That even $\mathbb{R}$ can fail to be sequential in a model of $\mathbf{ZF}$ is shown in Theorem 4.55 of \cite{Her}. The second part of Theorem 3.3.20 of \cite{E}, which states that every first countable Hausdorff space is a $k$-space, does not have a proof in $\mathbf{ZF}$. Therefore, since we work in $\mathbf{ZF}$, it is natural to ask about set-theoretical status of the following sentences:
\begin{enumerate}
\item[(a)] Every first countable Hausdorff space is a $k$-space.
\item[(b)] $\mathbb{R}$ is a $k$-space.
\item[(c)] Every subspace of $\mathbb{R}$ is a $k$-space.
\end{enumerate}

In this section,  we are going to prove that (a) is equivalent with the axiom of countable multiple choice (i.e. Form 126 in \cite{HR}), while (b) holds in $\mathbf{ZF}$ and (c) is independent of $\mathbf{ZF}$. We shall also show that even the Sorgenfrey line can fail to be a $k$-space in a model of $\mathbf{ZF}$.

We recall that the axiom of countable multiple choice, denoted by $\mathbf{CMC}$ in \cite{Her}, states that, for every  collection $\{A_n: n\in\omega\}$ of non-empty sets there exists a collection $\{F_n: n\in\omega\}$ of non-empty finite sets such that $F_n\subseteq A_n$ for each $n\in\omega$. It was shown in \cite{Ker1} that $\mathbf{CMC}$ is equivalent with Form 126D of \cite{HR}, i.e with the following sentence denoted by $\mathbf{WCMC}$:

$\textbf{WCMC}$: For every denumerable family $\mathcal{A}$ of disjoint non-empty sets there is an infinite set $C\subseteq \bigcup\mathcal{A}$ such that, for each $A\in\mathcal{A}$ the intersection $A\cap C$ is finite.

More information about $\mathbf{WCMC}$ can be found in \cite{Ker1} and in  Note 132 of \cite{HR}.

If $\mathcal{A}$ is a denumerable collection of pairwise disjoint non-empty sets, then every infinite set $C\subseteq\bigcup\mathcal{A}$ such that $C\cap A$ is finite for each $A\in\mathcal{A}$ is called a \emph{partial multiple choice} set of $\mathcal{A}$.

\begin{theorem} The following conditions are all equivalent in $\mathbf{ZF}$:
\begin{enumerate}
\item[(i)] $\mathbf{CMC}$;
\item[(ii)] every Hausdorff first countable space is a $k$-space;
\item[(iii)] every metrizable space is a $k$-space.
\end{enumerate}
\end{theorem}

\begin{proof}  Let $Y$ be a first countable Hausdorff space and let $D$ be a subset of $Y$ which is not closed in $Y$. Fix in $Y$ an accumulation point $y$ of $D$ such that $y\notin D$. Let $\mathcal{B}(y)=\{ U_n: n\in\omega\}$ be a countable base of open neighbourhoods of $y$ in $Y$ such that $U_{n+1}\subset U_n$ for each $n\in\omega$. Since $Y$ is Hausdorff, we can find a strictly increasing sequence $(k_n)_{n\in\omega}$ of positive integers such that the set $D_n= D\cap(U_{k_n}\setminus U_{k_{n+1}})$ is non-empty for each $n\in\omega$. Suppose  that $\mathbf{CMC}$ holds. By $\mathbf{CMC}$,  there exists a sequence $(C_n)_{n\in\omega}$ of non-empty finite sets such that $C_n\subseteq D_n$ for each $n\in\omega$. Then the set $C=\{y\}\cup\bigcup_{n\in\omega} C_n$ is compact in $Y$, while $y$ is an accumulation point of $D\cap C$ and $y\notin D\cap C$. Thus $D\cap C$ is not closed in $Y$. Therefore,  $Y$ is a $k$-space if $\mathbf{CMC}$ holds. Hence, (i) implies (ii). It is obvious that (ii) implies (iii). To complete the proof, it suffices to show that (iii) implies $\mathbf{WCMC}$.

Now, let us assume that $\mathbf{WCMC}$ is false. Suppose that $\mathcal{A}=\{A_n : n\in\omega\}$ is a collection of pairwise
disjoint non-empty sets without a partial multiple choice set. Put $A=\bigcup_{n\in\omega}A_n$. Take a point $\infty\notin A$ and put $Z=A\cup\{\infty\}$. Consider the metric $d$ on $Z$ defined in Lemma 2.22, as well as the topology $\tau(d)$ on $Z$ induced by $d$. Let $K$ be a compact subspace of $(Z, \tau(d))$. Since each $A_n$ is a discrete clopen subspace of $(Z, \tau(d))$, the sets $K\cap A_n$ are all finite. If $K$ were infinite, then $K$ would be a multiple choice set of $\mathcal{A}$. Hence, $K$ is finite, so $A\cap K$ is compact in $(Z, \tau(d))$. By Lemma 2.22,  $A$ is not closed in $(Z, \tau(d))$. This shows that $(Z, \tau(d))$ is not a $k$-space. Hence, (iii) implies (i).
\end{proof}

\begin{corollary} It is consistent with $\mathbf{ZF}$ that not every metrizable space is a $k$-space.
\end{corollary}

\begin{theorem} If $\mathcal{M}$ is a model of $\mathbf{ZF}$ in which every metrizable space is sequential, then $\mathbf{CMC}$ holds in $\mathcal{M}$.
\end{theorem}
\begin{proof} Suppose $(Z, \tau(d))$ is the space from Lemma 2.22 and the proof to Theorem 3.3 where $\mathcal{A}=\{A_n : n\in\omega\}$ is a collection of pairwise disjoint non-empty sets without a partial multiple choice set. Then the set $A$ is sequentially closed but not closed in $(Z, \tau(d))$.
\end{proof}

\begin{remark} Let us notice that since $\mathbf{CMC}$ implies $\mathbf{CC}(\mathbb{R})$, it follows directly  from  Exercise E.3 to Section 4.6  of \cite{Her} that in every  model of $\mathbf{ZF}$ in which $\mathbf{CMC}$ holds, every second countable $T_ 0$-space (in particular, every second countable metrizable space) is Fr\'echet-Urysohn, so sequential.
\end{remark}

\begin{theorem} $\mathbb{R}$ is a $k$-space in every model of $\mathbf{ZF}$.
\end{theorem}
\begin{proof} Let $A$ be a  subset  of $\mathbb{R}$ such that $A\cap K$ is closed in $\mathbb{R}$ for each compact set $K$ in $\mathbb{R}$. Suppose that  $x\in (\text{cl}_{\mathbb{R}}A)\setminus A$. Let $K_n=A\cap [x-\frac{1}{2^n}, x+\frac{1}{2^n}]$ for each $n\in\omega$. The sets $K_n$ are all non-empty and compact in $\mathbb{R}$. We put $x_n=\inf( K_n)$ for each $n\in\omega$. It follows from the compactness of $K_n$ that $x_n\in K_n$ for each $n\in\omega$. In this way, we define a sequence $(x_n)_{n\in\omega}$ of points of $A$ which converges in $\mathbb{R}$ to $x$. The set $K=\{x\}\cup\{ x_n: n\in\omega\}$ is compact in $\mathbb{R}$ but $A\cap K$ is not closed in $\mathbb{R}$ which is a contradiction. Hence, $A$ must be closed in $\mathbb{R}$. This implies that $\mathbb{R}$ is a $k$-space in $\mathbf{ZF}$.
\end{proof}

\begin{proposition} (i) It is consistent with $\mathbf{ZF}$ that a subspace of $\mathbb{R}$ can fail to be a $k$-space.\\
(ii) It is consistent with $\mathbf{ZF}$ that all subspaces of $\mathbb{R}$ are $k$-spaces.
\end{proposition}
\begin{proof} (i) Suppose that $X$ is an infinite Dedekind finite subset of $\mathbb{R}$. Since $X$ as a subspace of $\mathbb{R}$  is not discrete, there exists a set $A\subseteq X$ such that $A$ is not closed in $X$. Let $K$ be a compact subset of $X$. Then $K$ is compact in $\mathbb{R}$, so, if $K$ were infinite, then $K$ would be Dedekind infinite. Since $K$ is Dedekind finite, we deduce that $K$ is finite. This implies  $A\cap K$ is closed in $X$ because $A\cap K$ is finite. To complete the proof to (i), it  suffices to notice that in the model $\mathcal{M}1$ of \cite{HR} there is an infinite Dedekind finite subset of $\mathbb{R}$.

(ii) Let $\mathcal{M}$ be a model of $\mathbf{ZF}$ in which $\mathbf{CC}(\mathbb{R})$ holds. For instance, the model $\mathcal{M}2$ of \cite{HR} can be taken as $\mathcal{M}$. Since, by Theorem 4.54 of \cite{Her}, it is true in $\mathcal{M}$ that every subspace of $\mathbb{R}$  is sequential, we infer that, in $\mathcal{M}$, every subspace of $\mathbb{R}$ is a $k$-space.
\end{proof}

\begin{corollary} It is independent of $\mathbf{ZF}$ that all subspaces of $\mathbb{R}$ are $k$-spaces.
\end{corollary}

In what follows, as a metric space, $\mathbb{R}$ is considered with the metric $\rho$ defined by $\rho(x, y)=\vert x-y\vert$ for all $x, y\in\mathbb{R}$.

Using the notation from Theorem 4.55 of \cite{Her}, we denote by $\mathbf{CC}(c\mathbb{R})$ the following sentence: Every non-empty countable collection of non-empty complete subspaces of $\mathbb{R}$ has a choice function.

\begin{theorem}  (i) If the Sorgenfrey line is a $k$-space, then $\mathbf{CC}(c\mathbb{R})$ holds.\\
(ii) If $\mathbf{CC}(\mathbb{R})$ holds, then the Sorgenfrey line is a $k$-space.
\end{theorem}
\begin{proof} (i) Suppose that $\mathbf{CC}(c\mathbb{R})$ does not hold. Then, by Theorem 4.55 of \cite{Her}, $\mathbb{R}$ is not sequential. Let $A$ be a sequentially closed subset  of $\mathbb{R}$ which is not closed in $\mathbb{R}$. Let $a\in(\text{cl}_{\mathbb{R}}A)\setminus A$. The set $B=[(A-a)\cup (-A+a)]\cap (0, +\infty)$ is sequentially closed in $\mathbb{R}$ and not closed in $\mathbb{R}$. Since $0\in(\text{cl}_{\mathbb{S}}B)\setminus B$, the set $B$ is not closed in $\mathbb{S}$. Let $K$ be a compact set in $\mathbb{S}$. Then $K$ is countable and compact in $\mathbb{R}$. The set $K\cap B$ is sequentially closed in $\mathbb{R}$ and since, in addition, $K\cap B$ is countable, we deduce that $K\cap B$ is closed in $\mathbb{R}$. This implies $K\cap B$ is closed in $\mathbb{S}$. Therefore, $\mathbb{S}$ is not a $k$-space.

(ii) Now, suppose that $\mathbf{CC}(\mathbb{R})$ holds. Let $F\subseteq \mathbb{R}$ be not closed in $\mathbb{S}$ and let $x\in\text{cl}_{\mathbb{S}}F\setminus F$. Then $G=F\cap (x, +\infty)$ is not closed in  $\mathbb{R}$ and $x\in(\text{cl}_{\mathbb{R}}G)\setminus G$. In the light of Theorem 4.54 of \cite{Her}, $\mathbb{R}$ is Fr\'echet. This implies that there exists a sequence $(x_n)_{n\in\omega}$ of points of $G$ which converges in $\mathbb{R}$ to $x$. The set $K=\{x\}\cup\{x_n: n\in\omega\}$ is compact in $\mathbb{S}$ but $K\cap F$ is not closed in $\mathbb{S}$. This proves that $\mathbb{S}$ is a $k$-space.
\end{proof}

\begin{corollary} It  is consistent with $\mathbf{ZF}$ that the Sorgenfrey line is not a $k$-space.
\end{corollary}

From Theorems 2.28  and 3.10, we immediately obtain the following:

\begin{corollary} Let $\tau$ be the topology of the Sorgenfrey line. If every compact in $(\mathbb{R}, \tau^{\star})$ set is closed in $(\mathbb{R}, \tau)$, then $\mathbf{CC}(c\mathbb{R})$ holds.
\end{corollary}

\section{Compact complement partial topology}

Let us slightly reformulate Definition 2.1 of \cite{OPSW}:

\begin{definition} A \emph{partial topology} of a set $X$ is a collection $Cov_X\subseteq\mathcal{P}(\mathcal{P}(X))$ which satisfies the following conditions:
\begin{enumerate}
\item[(i)] $\tau_X=\bigcup Cov_X$ is a topology on $X$;
\item[(ii)] if $\mathcal{U}\subseteq \tau_X$ and $\mathcal{U}$ is finite, then $\mathcal{U}\in Cov_X$;
\item[(iii)] if $\mathcal{U}\in Cov_X$ and $V\in\tau_X$, then $\{U\cap V: U\in\mathcal{U}\}\in Cov_X$;
\item[(iv)] if $\mathcal{U}\in Cov_X$ and, for each $U\in\mathcal{U}$, we have $\mathcal{V}(U)\in Cov_X$ such that $U=\bigcup\mathcal{V}(U)$, then $\bigcup_{U\in\mathcal{U}}\mathcal{V}(U)\in Cov_X$;
\item[(v)] if $\mathcal{U}\subseteq \tau_X$ and $\mathcal{V}\in Cov_X$ are such that $\bigcup\mathcal{U}=\bigcup\mathcal{V}$ and, for each $V\in\mathcal{V}$, there exists $U\in\mathcal{U}$ such that $V\subseteq U$, then $\mathcal{U}\in Cov_X$.
\end{enumerate}
\end{definition}

\begin{definition} If  $Cov_X$ is a partial topology on a set $X$, then the ordered pair $(X, Cov_X)$ is called a \emph{partially topological space}, while $\tau_X=\bigcup Cov_X$ is called the \emph{topology corresponding to} $Cov_X$.
\end{definition}

 \begin{remark} Let us notice that if $(X, Cov_X)$ is a partially topological space, then the triple $(X,\bigcup Cov_X, Cov_X)$ is a Delfs-Knebusch generalized topological space (in abbreviation a D-K gts) in the sense of Definition 2.2.1 of \cite{Pie1} and, moreover, this D-K gts is partially topological in the sense of Definition 2.2.67 of \cite{Pie1}.  Delfs-Knebusch gtses were studied, for instance, in \cite{DK,K,OPSW,Pie1,Pie2,Pie3,PW, PW1}. We recall that, according to Remark 2.2.3 of \cite{Pie1}, a D-K gts is an ordered pair $(X, Cov_X)$ such that, for $Op_X=\bigcup Cov_X$, the triple $(X,Op_X, Cov_X)$ satisfies the conditions of Definition 2.2.2 of \cite{Pie1}. In general, $Op_X$ need not be a topology on $X$. If $(X, Cov_X)$ is a D-K gts, then $Cov_X$ is called a D-K (Delfs-Knebusch) generalized topology on $X$.
\end{remark}

If $\psi$ is a topological property, then we say that a partially topological space $(X, Cov_X)$ has $\psi$ if the topological space $(X, \bigcup Cov_X)$ has $\psi$. In particular:

\begin{definition} We say that a partially topological space $(X. Cov_X)$ is:
\begin{enumerate}
\item[(i)] \emph{Hausdorff} if $(X, \bigcup Cov_X)$ is Hausdorff;
\item[(ii)]\emph{compact} if $(X, \bigcup Cov_X)$  is compact.
\end{enumerate}
\end{definition}

\begin{definition}
Let  $(X, Cov_X)$ be a Hausdorff partially topological space,  $\tau_X$  the topology corresponding to $Cov_X$ and $\tau_X^{\star}$ the compact complement  topology of $(X, \tau_X)$. Then the collection
$$ Cov_X^{\star}=Cov_X \cap \mathcal{P}(\tau_X^{\star})$$
will be called the \emph{compact complement partial topology} of $(X, Cov_X)$.
\end{definition}

\begin{remark}
Let  $(X, Cov_X)$ be a Hausdorff partially topological space.  That $Cov_X^{\star}$ is a D-K generalized topology follows from Fact 2.2.31 in \cite{Pie1} which says that  the intersection of any non-empty family of D-K generalized topologies on $X$ is a D-K generalized topology on $X$. Since $\bigcup (Cov_X \cap \mathcal{P}(\tau_X^{\star}))=\tau_X^{\star}$, the D-K generalized topology  $Cov_X^{\star}$ is a partial topology on $X$.
\end{remark}

In what follows, we use the  symbols $\cap_1, \setminus_1$ introduced on page 219 of \cite{Pie1}. We recall that, for  collections $\mathcal{U}, \mathcal{V}$ of subsets of $X$, we have $\mathcal{U}\cap_1\mathcal{V}=\{ U\cap V: U\in\mathcal{U}, V\in\mathcal{V}\}$ and, analogously,  $\mathcal{U}\setminus_1\mathcal{V}=\{U\setminus V: U\in\mathcal{U}, V\in\mathcal{V}\}$. Moreover, for a collection $\mathcal{A}\subseteq\mathcal{P}(\mathcal{P}(X))$, we denote by $\langle \mathcal{A}\rangle_X$ the intersection of all D-K generalized topologies on $X$ that contain $\mathcal{A}$ (see page 242 of \cite{PW}).

\begin{definition}
 For a subset $Y$ of $X$ and a partial topology $Cov$ on $X$, let
$$Cov\vert Y=\langle \{\mathcal{V}\cap_1 \{ Y\} : \mathcal{V}\in  Cov\} \rangle_Y.$$
Then $Cov\vert Y$ is called the \emph{partial topology on $Y$ induced by} $Cov$ and
$(Y, Cov\vert Y)$ is called  a partially topological \emph{subspace} of
$(X, Cov)$.
\end{definition}

The following  theorem is an adaptation of Theorem 2.2 to partial topologies:

\begin{theorem} Let $(X, Cov_X)$ be a Hausdorff partially topological space, $\tau_X=\bigcup Cov_X$ and
 $Y\subseteq X$. The following conditions are fuilfilled:
\begin{enumerate}
\item[(i)] $Cov_X^{\star}\vert Y\subseteq Cov_X\vert Y$;
\item[(ii)] if $Y$ is compact in $(X, \tau_X)$, then $Cov_X^{\star}\vert Y=Cov_X\vert Y$;
\item[(iii)] $Cov_X^{\star}\vert Y=Cov_X\vert Y$ if and only if there exists a $\tau_X$-compact set $K$ such that $Y\subseteq K$.
\end{enumerate}
\end{theorem}

\begin{proof}  Since  $Cov_X^{\star}\subseteq Cov_X$, it is obvious that (i) is satisfied.

(ii)  Suppose that $Y$ is $\tau_X$-compact and $\mathcal{V}\in Cov_X$. Since $(X, \tau_X)$ is Hausdorff, the set $Y$ is $\tau_X$-closed, so $\mathcal{A}=\{Y\}\cap_1 (\{X\}\setminus_1 \mathcal{V})$ is a collection of  $\tau_X$-compact subsets of $Y$.
Notice that $\mathcal{V}\cap_1 \{Y\}=\{Y\}\cap_1 (\{X\}\setminus_1 \mathcal{A})$. This implies that $\mathcal{V}\cap_1 \{Y\}\in Cov_X^{\star}\vert Y$ and, in consequence,  $Cov_X\vert Y\subseteq  Cov_X^{\star}\vert Y$.

(iii) Now, assume that $K$ is a $\tau_X$-compact set such that $Y\subseteq K$. It follows from (ii) that $Cov_X\vert K=Cov_X^{\star}\vert K$. Hence, in view of Fact 10.3 of \cite{PW}, we have $Cov_X\vert Y= (Cov_X\vert K)\vert Y=(Cov_X^{\star}\vert K)\vert Y= Cov_X^{\star}\vert Y$.

Finally,  suppose that $Y$ is a subset of $X$ such that $Cov_X^{\star}\vert Y=Cov_X\vert Y$. Let $V\in \tau_X$ be such that $\emptyset\neq V \cap Y\neq Y$. Then $\{V\}\in Cov_X$.  Since $\{V\cap Y\}\in Cov_X^{\star}\vert Y$, there exists a $\tau_X$-compact  set $K_0$ such that $V \cap Y=Y\setminus {K}_0$. Reasoning as in the proof to Theorem \ref{theorem1} (iii), we get that there exists a $\tau_X$-compact set $K$ such that  $Y\subseteq K$.
\end{proof}

\begin{corollary} A Hausdorff partially topological space $(X, Cov_X)$ is compact if and only if
$Cov_X=Cov_X^{\star}$.
\end{corollary}

In view of Proposition 2.8 and Corollary 2.29,  the following proposition holds:

\begin{proposition} If $(X, Cov_X)$ is a Hausdorff partially topological space,  then the partially topological space  $(X, Cov_X^{\star})$ is compact and $T_1$.
\end{proposition}

\begin{remark}
Similarly to the situation in  Remark 2.4,  we have that, in general,  $Cov_X^{\star}\vert Y$ need not be equal to $(Cov_X \vert Y)^{\star}$.
\end{remark}

Although it can be said more about compact complement partial topologies, let us finish with the following example:

\begin{example}\label{example2-1}
Consider the partially topological real lines considered in
Definition 1.2 of  \cite{PW}: $\mathbb{R}_{st}=(\mathbb{R},Cov_{st}), \mathbb{R}_{lst}=(\mathbb{R},Cov_{lst}), \mathbb{R}_{l^+st}=(\mathbb{R},Cov_{l^+st})$. Let $I$ be a  bounded interval of $\mathbb{R}$.
Then we get the following  equalities of the induced partial topologies:
 $Cov_{st}\vert I=Cov_{lst}\vert I=Cov_{l^+st}\vert I=
Cov_{st}^{\star}\vert I=Cov_{lst}^{\star}\vert I=
Cov_{l^+st}^{\star}\vert I$.
\end{example}

config: arxiv


TITLE: How to construct orthogonal complement subspace of any subspace?
QUESTION [1 upvotes]: If I have one subspace $V$ belong to $\mathbb{R}^{n}$, the subspace has basis $v_{1},v_{2},\cdot\cdot\cdot,v_{k}$,where $k<n$. I want to find the orthogonal complement subspace $V^{\perp}$ of $V$.
$$V^{\perp}=span(u_{1},u_2,\cdot\cdot\cdot,u_{n-k})$$
Could anyone tell me how to construct the $V^{\perp}$?

REPLY [1 votes]: What you want is the kernel (null-space) of the matrix
$$
\pmatrix{
-&v_1^T&-\\
-&v_2^T&-\\
&\vdots\\
-&v_k^T&-
}
$$
this may be found by row-reduction.  If each of the $v_i$ are mutually orthogonal, however, using the Gram Schmidt process is faster.

Gram-Schmidt process: let $e_1,\dots,e_n$ denote the standard basis vectors of $\Bbb R^n$.  Begin by row-reducing the matrix
$$
\pmatrix{
v_1 & \cdots & v_k & e_1 & \cdots & e_n
}
$$
there will be $n$ pivot columns once this matrix is row-reduced. $k$ of them will be in the first $k$ columns, and the rest fall in the last $n$ positions.
Let $x_1,\dots,x_{n-k}$ be the vectors $e_i$ such that $e_i$ became a pivot.
Apply the Gram Schmidt process to $\{v_1,\dots,v_k,x_1,\dots,x_{n-k}\}$.


TITLE: Fourier transform of a Gaussian over a square window
QUESTION [1 upvotes]: I am trying to evaluate an integral of the form
$$
G_T(\omega)
=
\int_{-T/2}^{T/2} \mathrm{d}t ~ e^{-\sigma^2 t^2} e^{i\omega t}
$$
and I am looking for a closed-form expression. I know this integral can be rearranged into a convolution between a Gaussian and a sinc function, and an "incomplete" expression for that integral can be found here. I am however rather confused by the complex erf function in that answer, and I really need some clarifications.

REPLY [0 votes]: If, after @Sal's  comment, you still need to be conviced that
$$G_T(\omega)=\int_{-\frac T2}^{+\frac T2} e^{-\sigma^2 t^2} e^{i\omega t}\,dt$$
$$G_T(\omega)=i\,\frac{ \sqrt{\pi } }{2 \sigma }\,e^{-\frac{\omega ^2}{4 \sigma ^2}}\left(\text{erfi}\left(\frac{\omega -i \sigma ^2 T}{2 \sigma
   }\right)-\text{erfi}\left(\frac{\omega +i \sigma ^2 T}{2 \sigma
   }\right)\right)$$ is not "intimidating", expand it as a Taylor series around $T=0$ and let $k=\frac{\omega ^2}{ \sigma ^2}$
$$G_T(\omega)=T\Bigg[1+\sum_{n=1}^\infty(-1)^n\, \frac{\sigma^{2n}}{4^n\,(2 n+1)!} a_n\,T^{2n}\Bigg]$$ The first $a_n$ are
$$\left(
\begin{array}{cc}
n & a_n \\
 1 & k+2 \\
 2 & k^2+12 k+12 \\
 3 & k^3+30 k^2+180 k+120 \\
 4 & k^4+56 k^3+840 k^2+3360 k+1680 \\
 5 & k^5+90 k^4+2520 k^3+25200 k^2+75600 k+30240 \\
\end{array}
\right)$$


TITLE: Prove limit by Cauchy definition $\lim_{x \to -4} \frac{x-10}{x+6} = -7$
QUESTION [0 upvotes]: I need to prove:  $ \lim_{x \to -4} \frac{x-10}{x+6} = -7 $ so I said let ${\epsilon}$ > 0. the instructor told us to find a $δ$ such that for any x that $0<|x-a|<δ$ the following happens: $|{f(x) - L| = |\frac{x-10}{x+6} + 7| <\epsilon }$ and I got to this:
$ {8* |\frac{x+4}{x+6}| < \epsilon } $
Now how do I choose a correct $δ$ value to say that this expression is correct for any ${\epsilon}$? We just finished sequences and this new definition is really confusing to me. Thanks

REPLY [0 votes]: $f(x):=8\dfrac{|x-(-4)|}{|x+6|}$.
Consider
$|x-(-4)|<1$, then
$-1 < x-(-4)<1$; $-1-4<x< 1-4$;
$-5<x<-3$; finally $-5+6<x+6<-3+6$;
$f(x):=|\dfrac{x-10}{x+4}+7| =8\dfrac{|x+4|}{|x+6|} <$
$8\dfrac{|x-(-4)|}{1}$.
Choose $\delta = \min (1,\epsilon/8)$.
Then $|x-(-4)|<\delta$ implies
$f(x)<8|x-(-4)| <8\delta <\epsilon$.


TITLE: Cohomology of Symmetric Group 3 using Lyndon-Hochschild-Serre spectral sequence
QUESTION [4 upvotes]: For the symmetric group $S_{3}$ we have the short exact sequence $$0\rightarrow C_{3}\rightarrow S_{3}\rightarrow C_{2}\rightarrow 0,$$ where $C_{n}$ is the cyclic group of order $n$. Using the Lyndon-Hochschild-Serre spectral sequence we obtain $$E_{2}^{p,q}=H^{p}(C_{2},H^{q}(C_{3},\mathbb{Z})),$$ where we would have $0$ for $q$ odd (right?).
So my doubt is that I'm not sure of how to obtain the non-trivial action of $C_{2}$ on $\mathbb{Z}$ or $C_{3}$ when $q$ is even.
A trivial action doesn't lead to the correct result, since after adding the diagonals in $E_{\infty}$ it should be the cohomology of $S_{3}$, which is
$$H^{n}(S_{3},\mathbb{Z})=\begin{cases}
\mathbb{Z} & n=0 \\
0 & n \hbox{ odd} \\
C_{2} & n\equiv2 \hbox{ mod 4} \\
C_{6} & n\equiv0 \hbox{ mod 4.}  \end{cases}$$

REPLY [3 votes]: Every short exact sequence of groups induces an action of the quotient on the normal subgroup by conjugation. Here, the action is inversion in $C_3$; to see this explicitly, observe that $$(12)(123)(12)^{-1} = (132).$$ Anyway, you want to know what the map $i^*: H^*(C_3; \Bbb Z) \to H^*(C_3;\Bbb Z)$ induced by inversion is. The first useful fact is that because inversion is a group homomorphism, this map is an algebra homomorphism.
Because $H^*(C_3;\Bbb Z) = \Bbb Z[c]/(3c)$, where $|c| = 2$, we see that we just need to determine what this map does in $H^2(C_3;\Bbb Z)$; it will either be the identity, or multiplication by $-1$; in the former case the induced map on $H^*(C_3;\Bbb Z)$ is the identity. In the latter case, $c^k \mapsto (-c)^k = (-1)^k c^k$, and so $i^*: H^{4n+2}(C_3;\Bbb Z)$ is multiplication by $-1$, and $i^*: H^{4n}(C_3;\Bbb Z)$ is the identity.
To pin this down, recall the universal coefficient theorem gives a natural isomorphism for any finite group $G$ $$H^2(G;\Bbb Z) \cong \text{Ext}(H_1(G;\Bbb Z), \Bbb Z),$$ because group cohomology in positive degrees is torsion, and in particular $$\text{Hom}(H_2(G;\Bbb Z), \Bbb Z) = 0.$$
Lastly, conclude with the fact that $H_1(G;\Bbb Z) \cong G^{\text{ab}}$ is a natural isomorphism, and $\text{Ext}(A, \Bbb Z)$ is naturally equivalent to $\text{Hom}(A, S^1)$ for any finite group $A$; essentially, this is saying that $\text{Ext}(A, \Bbb Z)$ is non-canonically isomorphic to $A$ itself, but that the induced maps are dualized; they go in the opposite direction.
In any case, applying all this to $G = C_3$, we find that the induced map of inversion on $H^2(C_3;\Bbb Z) = \text{Ext}(\Bbb Z/3, \Bbb Z)$ is multiplication by $-1$.
Now you need to know the group cohomology of $H^*(C_2;\Bbb Z/3)$, where $\Bbb Z/3$ is either given the trivial action or the negation action. Using the calculation here, as well as the fact that $H^0(G;M) = M^G$ by definition, we see that $H^*(C_2;\Bbb Z/3) = 0$ in positive degrees, and that $H^0(C_2;\Bbb Z/3) = \Bbb Z/3$ for the trivial action, but $H^0(C_2;\Bbb Z/3) = 0$ with the negation action.
(A high-tech calculation: if $G$ is a finite group and $M$ is a $G$-module in which multiplication by $|G|$ is an isomorphism, the existence of the transfer map implies that $H^*(G;M) = 0$ in positive degrees. In particular, this applies to $G = \Bbb Z/2$ and $M = \Bbb Z/3$ with any action.)
Now run your spectral sequence. The $q = 0$ line is $$\Bbb Z, 0, \Bbb Z/2, 0, \Bbb Z/2, 0, \cdots $$ while the $p = 0$ line is $$\Bbb Z, 0, 0, 0, \Bbb Z/3, 0, 0, 0, \Bbb Z/3, 0, \cdots $$
That is, there is a $\Bbb Z/3$ on every $E^{0,4n}_2$, and it is otherwise zero.
The spectral sequence is completely supported on these axes. Because everything is supported in even bidegree, there can be no nontrivial differentials, and $E_2 = E_\infty$. Along with the observation that there are no nontrivial extension problems (because $\Bbb Z/2$ and $\Bbb Z/3$ only give rise to a single abelian group as extension), we get the desired calculation.
If you also want to know what the product structure is, observe that the map $H^*(C_2;\Bbb Z) \to H^*(\Bbb S_3; \Bbb Z)$ sends the degree $2$ generator to the degree $2$ generator (you can see this at the level of the $E^2$ page of your spectral sequence), and is a ring homomorphism.
Similarly, one sees that the homomorphism $H^*(S_3; \Bbb Z) \to H^*(C_3; \Bbb Z)$ sends a 3-torsion generator in degree $4$ to the generator of $H^4(C_3;\Bbb Z)$.
These two facts, combined with the description of the underlying cohomology groups above, gives an isomorphism $$H^*(S_3;\Bbb Z) \cong \Bbb Z[c_1, c_2]/(2c_1, c_2-3c_1^2),$$ where $|c_i| = 2i$; this means that $c_1$ is $2$-torsion and $c_2 = 3c_1^2$ is $6$-torsion.


TITLE: Can I determine whether it is night at a location by calculating only the sun's elevation?
QUESTION [3 upvotes]: There are many formulas for determining sunrise and sunset times for locations, but for my application I only need to know if it is day or night (given a time, latitude and longitude).
After studying the terms azimuth, altitude/elevation, and zenith, I am curious if I can calculate the sun's elevation, and use that value to determine day/night status alone.
As I understand it, any value below the horizon (taking into consideration the various values for twilight) should tell me that it is night. I should be able to use -6 for civil twilight, -12 for nautical twilight, etc.
Do I need to worry about azimuth also?

REPLY [4 votes]: Yes, knowing the sun's elevation is generally enough to determine how dark it is, at least to a first approximation.
However, if you're using a simple geometric formula for the elevation, keep in mind that you may need to adjust it for atmospheric refraction.  Wikipedia's sunrise article has a pretty nice explanation of this:

"Sunrise occurs before the Sun actually reaches the horizon because the Sun's image is refracted by the Earth's atmosphere. The average amount of refraction is 34 arcminutes, though this amount varies based on atmospheric conditions.
Also, unlike most other solar measurements, sunrise occurs when the Sun's upper limb, rather than its center, appears to cross the horizon. The apparent radius of the Sun at the horizon is 16 arcminutes.
These two angles combine to define sunrise to occur when the Sun's center is 50 arcminutes below the horizon, or 90.83° from the zenith."

In general, if you're just comparing the elevation to some fixed threshold, it's enough to adjust the threshold to account for this.

REPLY [2 votes]: All you need is altitude. Azimuth is strictly what compass heading to look toward to hunt for the Sun.


TITLE: difference between "let" and "for all"
QUESTION [0 upvotes]: What is the difference between "let" and "for all"?
Consider the following example

For all natural numbers n, if n is even, then n squared is even.

Let n be a natural number. If n is even, then n squared is even.

REPLY [3 votes]: "For all" means just what it says. It means that the rest of the statement is true when $n$ is replaced by any natural number.
"Let" is used for introducing a "supplemental hypothesis". That is, a statement that we are assuming is true for the rest of the discussion. That statement can introduce a new constant value, such as it does here for $n$. Or it can used to add additional conditions on existing values (though usually we say "suppose" in that case). This is a common technique in proofs. You introduce the supplemental hypothesis $A$, then prove some other statement $B$ making use of it. The theorem is then "$A \implies B$". So in case (2). you are really stating
"If $n$ is a natural number, then (if $n$ is even, then $n^2$ is even)."
Which is an awkward way of stating what you mean. It also has an implicit quantization. I.e., it really means:
"For all $n$, if $n$ is a natural number, then (if $n$ is even, then $n^2$ is even)."
Fortunately, logic and grammar allow us to reduce this to the form of (1). The important thing to note, though, is that (1) is the proper statement of the "theorem", while (2) is appropriate only in limited discussions, where the supplemental hypothesis is only assumed to hold within that discussion.


TITLE: Express the Induced Curvature Forms on the Pullback Bundle through the connection forms
QUESTION [3 upvotes]: This question came up while reading Differential Geometry by Loring W.Tu.
Suppose $\pi: E \to M$ is a $\mathcal{C}^{\infty}$ vector bundle. And $f:N \to M$ a $\mathcal{C}^{\infty}$ map of manifolds. Let $\nabla$ be a connection on $E$ with connection matrix $\omega_e$ relative to a local frame $e$ for $E$. Then there is a unique connection $\nabla'$ on the pullback bundle $f^{*}E \to N$ whose connection matrix relative to the frame $f^{*}e$ is $f^{*}(\omega_e)$.
That makes sense so far.
The author then writes:
The induced curvature form on $f^{*}E$ is therefore
$f^{*} \Omega_e = f^{*} \omega_e + \frac{1}{2} [f^{*} \omega_e, f^{*} \omega_e]$.
Now I don't understand why the induced curvature forms are $f^{*} \Omega_e$ or why this formula holds.

REPLY [5 votes]: First of all, I think that there is a typo, either in your question or in the book. The right equality should be this
$$f^*\Omega_e = d(f^*\omega_e) + \frac{1}{2}[f^*\omega_e,f^*\omega_e]$$
or this
$$
f^*\Omega_e = f^*d\omega_e + \frac{1}{2}f^*[\omega_e,\omega_e],
$$
which are pretty much the same thing, since the pullback commutes with both the exterior differential $d$ and the bracket $[\cdot,\cdot]$. These equalities are trivial, since by definition, $\Omega_e = d\omega_e + \frac{1}{2}[\omega_e,\omega_e]$ (apply the pullback through $f$ to that equality).
The real question is: Why is this the curvature form of $\nabla'$ on $f^*E$ associated to the frame $f^*e$?
The answer is because, as you said, the connection matrix $\omega_{f^*e}$ associated to the frame $f^*e$ is $f^*\omega_e$. Therefore
$$\Omega_{f^*e} :=d\omega_{f^*e} + \frac{1}{2}[\omega_{f^*e},\omega_{f^*e}]\underset{\omega_{f^*e}=f^*\omega_e}{=} d(f^*\omega_e)+\frac{1}{2}[f^*\omega_e,f^*\omega_e]$$
which has already been shown above to be equal to $f^*\Omega_e$.


TITLE: Is it possible to improve the weight of perfect bipartite matchings faster than with Bellman-Ford?
QUESTION [3 upvotes]: If $G\left(A\cup B,\ E=\lbrace\lbrace a, b\rbrace\,|\, a\in A,\, b\in B\rbrace\right)$ is a weighted bipartite graph and $M_0$ an initial perfect matching, then the optimality of $M_0$ can be verified by the absence of negative cycles in the associated residual network $N\left(V=A\cup B,\,F=\lbrace(a_i,b_j)\,|\,e_{ij}\notin M_0\rbrace\cup\lbrace(b_j,a_i)\,|\,e_{ij}\in M_0\rbrace\right)$ and weights $\omega(a_i,b_j)=w(e_{ij}),\,\omega(b_j,a_i)=-w(e_{ij})$.
If however a negative cycles exists in $N$ then $M_1\ :=\ (M_0\setminus\nu)\cup \pi$, where $\nu$ and $\pi$ denote the edges of $G$ that correspond  to negative, resp. positive arcs of negative cycles in $N$, has lower weight.
The apparently unconditionally recommended method for finding a the arcs of a negative cycle in $N$ is the Bellman-Ford shortest path algorithm with $O(mn)$ runtime complexity.
However it seems to me that something taylored for the special properties of the residual network $N$ could bring about a performance boost.
The underlying idea is simple enough:
Construct from the edges in $M_0$ an alternative network $\mathcal{N}$, that contains arc set $\lbrace (b_j, a_i)\,|\,e_{ij}\in M_0\rbrace\cup\lbrace(a_i,b_{j_0})\,|\,e_{ij}\notin M_0\,\land\,\omega(a_i,b_{j_0})\le\omega(a_i,b_j)\rbrace$ where the weights of arcs are as defined for $N$. Put differently, connect the end of negative arcs with the nearest start of a different negative arc.
As the outdegree of every arc in $\mathcal{N}$ is exactly $1$, the underlying undirected graph has the topology of a $1$-forest, a graph, in which every connected component has exactly one cycle.
Repeatedly deleting from $\mathcal{N}$ all arcs with indegree 0 (the sources) leaves a collection of directed cycles in which the negatives ones can be trivially identified as the (strongly) connected components with negative arc-weight sum.
The complexity of the suggested alternative method can be easily determined as $O(m-n/2)$ for finding the shortest positive outgoing arc of a negative one plus $O(n)$ for repeatedly removing the source nodes in $\mathcal{N}$ when executed sequentially.

Question:
when will the proposed alternative way of detecting negative cycles in the residual network of minimum weight perfect bipartite matching fail, resp. will it always detect one if it exists?
Could it ever report false positives, i.e. yield a negative cycle if there actually isn't one? If not, the proposed method could be used to do the "bulk work" before resorting to Bellman-Ford.

Addendum:
In view of John Machacek's answer I see the need for clarification and making things explicit.

the task is to determine the perfect matching of minimum weight in a symmetric bipartite graph, however as $e_{ij}:=\lbrace a_i\in A,\,b_j\in B\rbrace$ is a shorthand notation for $e_{a_i b_j}=e_{b_j a_i}$, the order of indices is significant and $e_{ii}$ doesn't denote a self loop in $G$.
Consequently $w(e_{ij})$ can be different from $w(e_{ji})$

the end node $a_i$ and start node $b_{j_0}$ of negative arcs that are connected by the positive arc $\lbrace a_i, b_{j_0}\rbrace$ must not belong to the same negative arc, i.e. $\mathcal{N}$ must not contain pairs of antiparallel arcs.

REPLY [3 votes]: If I understand notation correctly $e_{ij}$ is the edge $\{a_i, b_j\}$ in $G$. I'll let $w_{ij}$ be the weight $e_{ij}$. I'll give an example showing the alternative method can fail to detect a negative cycle in $N$. Consider
$$w_{11} = \epsilon$$
$$w_{12} = B$$
$$w_{13} = B$$
$$w_{21} = B$$
$$w_{22} = A$$
$$w_{23}= B - \epsilon$$
$$w_{31} = B + 3\epsilon$$
$$w_{32} = B + 2\epsilon$$
$$w_{33} = \epsilon$$
Where $\epsilon > 0$ is small while $A$ and $B$ are large with  $B < A < 2B - \epsilon$. Let $M_0 = \{e_{12}, e_{21}, e_{33}\}$. Then $N$ contains the negative cycle
$$(a_1, b_1), (b_1, a_2), (a_2, b_2), (b_2,a_1)$$
which has weight $\epsilon - B + A -B = A + \epsilon - 2B < 0$. Indeed, $M_0$ is not minimal weight. The minimal weight matching is $M^* = \{e_{11}, e_{22}, e_{33}\}$.
In this example $\mathcal{N}$ consists of arcs
$$\{(b_2, a_1), (b_1, a_2), (b_3, a_3), (a_1, b_1), (a_2, b_3), (a_3, b_2)\}$$
or
$$\{(b_2, a_1), (b_1, a_2), (b_3, a_3), (a_1, b_1), (a_2, b_3), (a_3, b_3)\}$$
(it depends on my understanding of the definition of $\mathcal{N}$ and if $e_{i,j_0}$ is allowed to be in $M_0$ or not).
Either way we fail to find a negative cycle. In the latter case the only cycle is $(a_3, b_3), (b_3, a_3)$ which has weight $0$. In the first case we have a cycle will all the arcs
$$(a_1,b_1), (b_1,a_2), (a_2, b_3), (b_3, a_3), (a_3,b_2), (b_2, a_1)$$
which has weight $\epsilon - B + (B - \epsilon) - \epsilon + (B + 2\epsilon) - B = \epsilon > 0$.