\documentclass[12pt,titlepage]{article}

\usepackage{amsmath}
\usepackage{mathrsfs}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{mathtools}
\usepackage{graphicx}
\usepackage{color}
\usepackage{ucs}
\usepackage[utf8x]{inputenc}
\usepackage{xparse}
\usepackage{hyperref}

%----Macros----------
%
% Unresolved issues:
%
%  \righttoleftarrow
%  \lefttorightarrow
%
%  \color{} with HTML colorspec
%  \bgcolor
%  \array with options (without options, it's equivalent to the matrix environment)

% Of the standard HTML named colors, white, black, red, green, blue and yellow
% are predefined in the color package. Here are the rest.
\definecolor{aqua}{rgb}{0, 1.0, 1.0}
\definecolor{fuschia}{rgb}{1.0, 0, 1.0}
\definecolor{gray}{rgb}{0.502, 0.502, 0.502}
\definecolor{lime}{rgb}{0, 1.0, 0}
\definecolor{maroon}{rgb}{0.502, 0, 0}
\definecolor{navy}{rgb}{0, 0, 0.502}
\definecolor{olive}{rgb}{0.502, 0.502, 0}
\definecolor{purple}{rgb}{0.502, 0, 0.502}
\definecolor{silver}{rgb}{0.753, 0.753, 0.753}
\definecolor{teal}{rgb}{0, 0.502, 0.502}

% Because of conflicts, \space and \mathop are converted to
% \itexspace and \operatorname during preprocessing.

% itex: \space{ht}{dp}{wd}
%
% Height and baseline depth measurements are in units of tenths of an ex while
% the width is measured in tenths of an em.
\makeatletter
\newdimen\itex@wd%
\newdimen\itex@dp%
\newdimen\itex@thd%
\def\itexspace#1#2#3{\itex@wd=#3em%
\itex@wd=0.1\itex@wd%
\itex@dp=#2ex%
\itex@dp=0.1\itex@dp%
\itex@thd=#1ex%
\itex@thd=0.1\itex@thd%
\advance\itex@thd\the\itex@dp%
\makebox[\the\itex@wd]{\rule[-\the\itex@dp]{0cm}{\the\itex@thd}}}
\makeatother

% \tensor and \multiscript
\makeatletter
\newif\if@sup
\newtoks\@sups
\def\append@sup#1{\edef\act{\noexpand\@sups={\the\@sups #1}}\act}%
\def\reset@sup{\@supfalse\@sups={}}%
\def\mk@scripts#1#2{\if #2/ \if@sup ^{\the\@sups}\fi \else%
  \ifx #1_ \if@sup ^{\the\@sups}\reset@sup \fi {}_{#2}%
  \else \append@sup#2 \@suptrue \fi%
  \expandafter\mk@scripts\fi}
\def\tensor#1#2{\reset@sup#1\mk@scripts#2_/}
\def\multiscripts#1#2#3{\reset@sup{}\mk@scripts#1_/#2%
  \reset@sup\mk@scripts#3_/}
\makeatother

% \slash
\makeatletter
\newbox\slashbox \setbox\slashbox=\hbox{$/$}
\def\itex@pslash#1{\setbox\@tempboxa=\hbox{$#1$}
  \@tempdima=0.5\wd\slashbox \advance\@tempdima 0.5\wd\@tempboxa
  \copy\slashbox \kern-\@tempdima \box\@tempboxa}
\def\slash{\protect\itex@pslash}
\makeatother

% math-mode versions of \rlap, etc
% from Alexander Perlis, "A complement to \smash, \llap, and lap"
%   http://math.arizona.edu/~aprl/publications/mathclap/
\def\clap#1{\hbox to 0pt{\hss#1\hss}}
\def\mathllap{\mathpalette\mathllapinternal}
\def\mathrlap{\mathpalette\mathrlapinternal}
\def\mathclap{\mathpalette\mathclapinternal}
\def\mathllapinternal#1#2{\llap{$\mathsurround=0pt#1{#2}$}}
\def\mathrlapinternal#1#2{\rlap{$\mathsurround=0pt#1{#2}$}}
\def\mathclapinternal#1#2{\clap{$\mathsurround=0pt#1{#2}$}}

% Renames \sqrt as \oldsqrt and redefine root to result in \sqrt[#1]{#2}
\let\oldroot\root
\def\root#1#2{\oldroot #1 \of{#2}}
\renewcommand{\sqrt}[2][]{\oldroot #1 \of{#2}}

% Manually declare the txfonts symbolsC font
\DeclareSymbolFont{symbolsC}{U}{txsyc}{m}{n}
\SetSymbolFont{symbolsC}{bold}{U}{txsyc}{bx}{n}
\DeclareFontSubstitution{U}{txsyc}{m}{n}

% Manually declare the stmaryrd font
\DeclareSymbolFont{stmry}{U}{stmry}{m}{n}
\SetSymbolFont{stmry}{bold}{U}{stmry}{b}{n}

% Manually declare the MnSymbolE font
\DeclareFontFamily{OMX}{MnSymbolE}{}
\DeclareSymbolFont{mnomx}{OMX}{MnSymbolE}{m}{n}
\SetSymbolFont{mnomx}{bold}{OMX}{MnSymbolE}{b}{n}
\DeclareFontShape{OMX}{MnSymbolE}{m}{n}{
    <-6>  MnSymbolE5
   <6-7>  MnSymbolE6
   <7-8>  MnSymbolE7
   <8-9>  MnSymbolE8
   <9-10> MnSymbolE9
  <10-12> MnSymbolE10
  <12->   MnSymbolE12}{}

% Declare specific arrows from txfonts without loading the full package
\makeatletter
\def\re@DeclareMathSymbol#1#2#3#4{%
    \let#1=\undefined
    \DeclareMathSymbol{#1}{#2}{#3}{#4}}
\re@DeclareMathSymbol{\neArrow}{\mathrel}{symbolsC}{116}
\re@DeclareMathSymbol{\neArr}{\mathrel}{symbolsC}{116}
\re@DeclareMathSymbol{\seArrow}{\mathrel}{symbolsC}{117}
\re@DeclareMathSymbol{\seArr}{\mathrel}{symbolsC}{117}
\re@DeclareMathSymbol{\nwArrow}{\mathrel}{symbolsC}{118}
\re@DeclareMathSymbol{\nwArr}{\mathrel}{symbolsC}{118}
\re@DeclareMathSymbol{\swArrow}{\mathrel}{symbolsC}{119}
\re@DeclareMathSymbol{\swArr}{\mathrel}{symbolsC}{119}
\re@DeclareMathSymbol{\nequiv}{\mathrel}{symbolsC}{46}
\re@DeclareMathSymbol{\Perp}{\mathrel}{symbolsC}{121}
\re@DeclareMathSymbol{\Vbar}{\mathrel}{symbolsC}{121}
\re@DeclareMathSymbol{\sslash}{\mathrel}{stmry}{12}
\re@DeclareMathSymbol{\bigsqcap}{\mathop}{stmry}{"64}
\re@DeclareMathSymbol{\biginterleave}{\mathop}{stmry}{"6}
\re@DeclareMathSymbol{\invamp}{\mathrel}{symbolsC}{77}
\re@DeclareMathSymbol{\parr}{\mathrel}{symbolsC}{77}
\makeatother

% \llangle, \rrangle, \lmoustache and \rmoustache from MnSymbolE
\makeatletter
\def\Decl@Mn@Delim#1#2#3#4{%
  \if\relax\noexpand#1%
    \let#1\undefined
  \fi
  \DeclareMathDelimiter{#1}{#2}{#3}{#4}{#3}{#4}}
\def\Decl@Mn@Open#1#2#3{\Decl@Mn@Delim{#1}{\mathopen}{#2}{#3}}
\def\Decl@Mn@Close#1#2#3{\Decl@Mn@Delim{#1}{\mathclose}{#2}{#3}}
\Decl@Mn@Open{\llangle}{mnomx}{'164}
\Decl@Mn@Close{\rrangle}{mnomx}{'171}
\Decl@Mn@Open{\lmoustache}{mnomx}{'245}
\Decl@Mn@Close{\rmoustache}{mnomx}{'244}
\makeatother

% Widecheck
\makeatletter
\DeclareRobustCommand\widecheck[1]{{\mathpalette\@widecheck{#1}}}
\def\@widecheck#1#2{%
    \setbox\z@\hbox{\m@th$#1#2$}%
    \setbox\tw@\hbox{\m@th$#1%
       \widehat{%
          \vrule\@width\z@\@height\ht\z@
          \vrule\@height\z@\@width\wd\z@}$}%
    \dp\tw@-\ht\z@
    \@tempdima\ht\z@ \advance\@tempdima2\ht\tw@ \divide\@tempdima\thr@@
    \setbox\tw@\hbox{%
       \raise\@tempdima\hbox{\scalebox{1}[-1]{\lower\@tempdima\box
\tw@}}}%
    {\ooalign{\box\tw@ \cr \box\z@}}}
\makeatother

% \mathraisebox{voffset}[height][depth]{something}
\makeatletter
\NewDocumentCommand\mathraisebox{moom}{%
\IfNoValueTF{#2}{\def\@temp##1##2{\raisebox{#1}{$\m@th##1##2$}}}{%
\IfNoValueTF{#3}{\def\@temp##1##2{\raisebox{#1}[#2]{$\m@th##1##2$}}%
}{\def\@temp##1##2{\raisebox{#1}[#2][#3]{$\m@th##1##2$}}}}%
\mathpalette\@temp{#4}}
\makeatletter

% udots (taken from yhmath)
\makeatletter
\def\udots{\mathinner{\mkern2mu\raise\p@\hbox{.}
\mkern2mu\raise4\p@\hbox{.}\mkern1mu
\raise7\p@\vbox{\kern7\p@\hbox{.}}\mkern1mu}}
\makeatother

%% Fix array
\newcommand{\itexarray}[1]{\begin{matrix}#1\end{matrix}}
%% \itexnum is a noop
\newcommand{\itexnum}[1]{#1}

%% Renaming existing commands
\newcommand{\underoverset}[3]{\underset{#1}{\overset{#2}{#3}}}
\newcommand{\widevec}{\overrightarrow}
\newcommand{\darr}{\downarrow}
\newcommand{\nearr}{\nearrow}
\newcommand{\nwarr}{\nwarrow}
\newcommand{\searr}{\searrow}
\newcommand{\swarr}{\swarrow}
\newcommand{\curvearrowbotright}{\curvearrowright}
\newcommand{\uparr}{\uparrow}
\newcommand{\downuparrow}{\updownarrow}
\newcommand{\duparr}{\updownarrow}
\newcommand{\updarr}{\updownarrow}
\newcommand{\gt}{>}
\newcommand{\lt}{<}
\newcommand{\map}{\mapsto}
\newcommand{\embedsin}{\hookrightarrow}
\newcommand{\Alpha}{A}
\newcommand{\Beta}{B}
\newcommand{\Zeta}{Z}
\newcommand{\Eta}{H}
\newcommand{\Iota}{I}
\newcommand{\Kappa}{K}
\newcommand{\Mu}{M}
\newcommand{\Nu}{N}
\newcommand{\Rho}{P}
\newcommand{\Tau}{T}
\newcommand{\Upsi}{\Upsilon}
\newcommand{\omicron}{o}
\newcommand{\lang}{\langle}
\newcommand{\rang}{\rangle}
\newcommand{\Union}{\bigcup}
\newcommand{\Intersection}{\bigcap}
\newcommand{\Oplus}{\bigoplus}
\newcommand{\Otimes}{\bigotimes}
\newcommand{\Wedge}{\bigwedge}
\newcommand{\Vee}{\bigvee}
\newcommand{\coproduct}{\coprod}
\newcommand{\product}{\prod}
\newcommand{\closure}{\overline}
\newcommand{\integral}{\int}
\newcommand{\doubleintegral}{\iint}
\newcommand{\tripleintegral}{\iiint}
\newcommand{\quadrupleintegral}{\iiiint}
\newcommand{\conint}{\oint}
\newcommand{\contourintegral}{\oint}
\newcommand{\infinity}{\infty}
\newcommand{\bottom}{\bot}
\newcommand{\minusb}{\boxminus}
\newcommand{\plusb}{\boxplus}
\newcommand{\timesb}{\boxtimes}
\newcommand{\intersection}{\cap}
\newcommand{\union}{\cup}
\newcommand{\Del}{\nabla}
\newcommand{\odash}{\circleddash}
\newcommand{\negspace}{\!}
\newcommand{\widebar}{\overline}
\newcommand{\textsize}{\normalsize}
\renewcommand{\scriptsize}{\scriptstyle}
\newcommand{\scriptscriptsize}{\scriptscriptstyle}
\newcommand{\mathfr}{\mathfrak}
\newcommand{\statusline}[2]{#2}
\newcommand{\tooltip}[2]{#2}
\newcommand{\toggle}[2]{#2}

% Theorem Environments
\theoremstyle{plain}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
\newtheorem{prop}{Proposition}
\newtheorem{cor}{Corollary}
\newtheorem*{utheorem}{Theorem}
\newtheorem*{ulemma}{Lemma}
\newtheorem*{uprop}{Proposition}
\newtheorem*{ucor}{Corollary}
\theoremstyle{definition}
\newtheorem{defn}{Definition}
\newtheorem{example}{Example}
\newtheorem*{udefn}{Definition}
\newtheorem*{uexample}{Example}
\theoremstyle{remark}
\newtheorem{remark}{Remark}
\newtheorem{note}{Note}
\newtheorem*{uremark}{Remark}
\newtheorem*{unote}{Note}

%-------------------------------------------------------------------

\begin{document}

%-------------------------------------------------------------------


\section*{compact spaces equivalently have converging subnet of every net}

\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{topology}{}\paragraph*{{Topology}}\label{topology}

[[!include topology - contents]]

\hypertarget{analysis}{}\paragraph*{{Analysis}}\label{analysis}

[[!include analysis - contents]]

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{statement}{Statement}\dotfill \pageref*{statement} \linebreak
\noindent\hyperlink{related_concepts}{Related concepts}\dotfill \pageref*{related_concepts} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

For [[metric spaces]] with their [[metric topology]] it is true that [[sequentially compact metric spaces are equivalently compact metric spaces]]. The analogous statement fails for more general [[topological spaces]]: for them, being [[sequentially compact topological space|sequentially compact]] in general neither implies nor is implied by being [[compact topological space|compact]] (see the counter-examples \href{sequentially+compact+topological+space#Examples}{here}).

But the failure of this equivalence is not due to a deficit in the concept of [[convergence]] but in the concept of [[sequences]] and sub-sequences. If the latter are generalized to [[nets]] and [[sub-nets]] (beware that the definition of sub-nets is slightly non-obvious), then the analogue of the statement remains true in generality: A topological space is [[compact topological space|compact]] precisely if every [[net]] in it has a [[sub-net]] that [[convergence]].

\hypertarget{statement}{}\subsection*{{Statement}}\label{statement}

\begin{prop}
\label{CompactSpacesEquivalentlyHaveConvergetSubnets}\hypertarget{CompactSpacesEquivalentlyHaveConvergetSubnets}{}
\textbf{(compact spaces are equivalently those for which every net has a converging subnet)}

Assuming [[excluded middle]] and the [[axiom of choice]], then:

A [[topological space]] $(X,\tau)$ is [[compact topological space|compact]] precisely if every [[net]] in $X$ has a [[sub-net]] that [[convergence|converges]].

\end{prop}
We break this up into lemmas \ref{InACompactSpaceEveryNetHasAConvergentSubnet} and \ref{IfEveryNetHasConvergentSubnetThenSpaceIsCompact}:

\begin{example}
\label{InACompactSpaceEveryNetHasAConvergentSubnet}\hypertarget{InACompactSpaceEveryNetHasAConvergentSubnet}{}
\textbf{(in a compact space, every net has a convergent subnet)}

Let $(X,\tau)$ be a [[compact topological space]]. Then every net in $X$ has a convergent subnet.

\end{example}
\begin{proof}
Let $\nu \colon A \to X$ be a net. We need to show that there is a subnet which converges.

For $a \in A$ consider the [[topological closures]] $Cl(S_a)$ of the sets $S_a$ of elements of the net beyond some fixed index:

\begin{displaymath}
S_a 
    \;\coloneqq\;
  \left\{
    \nu_b  \in X \;\vert\; b \geq a
  \right\}
  \subset X
  \,.
\end{displaymath}
Observe that the set $\{S_a \subset X\}_{a \in A}$ and hence also the set $\{Cl(S_a) \subset X\}_{a \in A}$ has the [[finite intersection property]], by the fact that $A$ is a [[directed set]]. Therefore \href{finite+intersection+property#CompactnessInTermsOfFiniteIntersectionProperty}{this prop.} implies from the assumption of $X$ being compact that the intersection of \emph{all} the $Cl(S_a)$ is non-empty, hence that there is an element

\begin{displaymath}
x \in \underset{a \in A}{\cap}  Cl(S_a)
  \,.
\end{displaymath}
In particular every [[neighbourhood]] $U_x$ of $x$ intersects each of the $Cl(S_a)$, and hence also each of the $S_a$. By definition of the $S_a$, this means that for every $a \in A$ there exists $b \geq a$ such that $\nu_b \in U_x$, hence that $x$ is a [[cluster point]] of the net.

We will not produce a [[sub-net]]

\begin{displaymath}
\itexarray{
     B && \overset{f}{\longrightarrow} && A
     \\
     & \searrow && \swarrow_{\nu}
     \\
     && X
  }
\end{displaymath}
that converges to this cluster point. To this end, we first need to build the domain directed set $B$. Take it to be the sub-directed set of the Cartesian product directed set of $A$ with the directed neighbourhood set $Nbhd_X(x)$ of $x$

\begin{displaymath}
B \subset A_{\leq} \times Nbhd_X(x)_{\supset}
\end{displaymath}
on those pairs such that the element of the net indexed by the first component is is contained in the second component:

\begin{displaymath}
B 
    \;\coloneqq\;
  \left\{
     (a,U_x)  \,\vert \, \nu_a \in U_X
  \right\}
  \,.
\end{displaymath}
It is clear $B$ is a [[preordered set]]. We need to check that it is indeed directed, in that every pair of elements $(a_1, U_1)$, $(a_2, U_2)$ has a common upper bound $(a_{bd}, U_{bd})$. Now since $A$ itself is directed, there is an upper bound $a_3 \geq a_1, a_2$, and since $x$ is a cluster point of the net there is moreover an $a_{bd} \geq a_3 \geq a_1, a_3$ such that $\nu_{a_{bd}} \in U_1 \cap U_2$. Hence with $U_{bd} \coloneqq U_1 \cap U_2$ we have obtained the required pair.

Next take the function $f$ to be given by

\begin{displaymath}
\itexarray{
     B &\overset{f}{\longrightarrow}& A
     \\
     (a, U) &\overset{\phantom{AAA}}{\mapsto}& a
  }
  \,.
\end{displaymath}
This is clearly order preserving, and it is cofinal since it is even a surjection. Hence we have defined a subnet $\nu \circ f$.

It now remains to see that $\nu \circ f$ converges to $x$, hence that for every open neighbourhood $U_x$ of $x$ we may find $(a,U)$ such that for all $(b,V)$ with $a \leq b$ and $U \supset V$ then $\nu(f(b,V)) = \nu(b) \in U_x$. Now by the nature of $x$ there exists some $a$ with $\nu_a \in U_x$, and hence if we take $U \coloneqq U_x$ then nature of $B$ implies that with $(b, V) \geq (a,U_x)$ then $b \in V \subset U_x$.

\end{proof}
\begin{example}
\label{IfEveryNetHasConvergentSubnetThenSpaceIsCompact}\hypertarget{IfEveryNetHasConvergentSubnetThenSpaceIsCompact}{}
Assuming [[excluded middle]], then:

Let $(X,\tau)$ be a [[topological space]]. If every [[net]] in $X$ has a [[subnet]] that [[convergence|converges]], then $(X,\tau)$ is a [[compact topological space]].

\end{example}
\begin{proof}
By [[excluded middle]] we may equivalently prove the [[contrapositive]]: If $(X,\tau)$ is not compact, then not every net in $X$ has a convergent subnet.

Hence assume that $(X,\tau)$ is not compact. We need to produce a net without a convergent subnet.

Again by [[excluded middle]], then by \href{finite+intersection+property#CompactnessInTermsOfFiniteIntersectionProperty}{this prop.} $(X,\tau)$ not being compact means equivalently that there exists a set $\{C_i \subset X\}_{i \in I}$ of [[closed subsets]] satisfying the [[finite intersection property]], but such that their intersection is empty: $\underset{i \in I}{\cap} C_i = \emptyset$.

Consider then $P_{fin}(I)$, the set of [[finite set|finite]] [[subsets]] of $I$. By the assumption that $\{C_i \subset X\}_{i \in I}$ satisfies the [[finite intersection property]], we may [[axiom of choice|choose]] for each $J \in P_{fin}(I)$ an element

\begin{displaymath}
x_J \in \underset{i \in J \subset I}{\cap} C_i
  \,.
\end{displaymath}
Now $P_{fin}(X)$ regarded as a [[preordered set]] under inclusion of subsets is clearly a [[directed set]], with an upper bound of two finite subsets given by their [[union]]. Therefore we have defined a net

\begin{displaymath}
\itexarray{
     P_{fin}(X)_{\subset} &\overset{\nu}{\longrightarrow}& X
     \\
     J &\overset{\phantom{AAA}}{\mapsto}& x_J
  }
  \,.
\end{displaymath}
We will show that this net has no converging subnet.

Assume on the contrary that there were a subnet

\begin{displaymath}
\itexarray{
     B && \overset{f}{\longrightarrow} && P_{fin}(X)
     \\
     & \searrow && \swarrow_{\nu}
     \\
     && X
  }
\end{displaymath}
which converges to some $x \in X$.

By the assumption that $\underset{i \in I}{\cap} C_i = \emptyset$, there would exist an $i_x \in I$ such that $x \neq C_{i_x}$, and because $C_i$ is a [[closed subset]], there would exist even an [[open neighbourhood]] $U_x$ of $x$ such that $U_x \cap C_{i_x} = \emptyset$. This would imply that $x_J \neq U_x$ for all $J \supset \{i_x\}$.

Now since the function $f$ defining the subset is cofinal, there would exist $b_1 \in B$ such that $\{i_x\} \subset f(b_1)$. Moreover, by the assumption that the subnet converges, there would also be $b_2 \in B$ such that $\nu_{b_2 \leq \bullet} \in U_x$. Since $B$ is directed, there would then be an upper bound $b \geq b_1, b_2$ of these two elements. This hence satisfies both $\nu_{f(e)} \in U_x$ as well as $\{i_x\} \subset f(b_1) \subset f(b)$. But the latter of these two means that $\nu_{f(b)}$ is not in $U_x$, which is a contradiction to the former. Thus we have a [[proof by contradiction]].

\end{proof}
\hypertarget{related_concepts}{}\subsection*{{Related concepts}}\label{related_concepts}

\begin{itemize}%
\item [[sequentially compact metric spaces are equivalently compact metric spaces]]


\item [[countably compact metric spaces are equivalently compact metric spaces]]


\item [[sequentially compact metric spaces are totally bounded]]


\item [[Lebesgue number lemma]]



\end{itemize}


\end{document}