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\begin{document}

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\section*{Heine-Borel theorem}

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

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

[[!include analysis - contents]]

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

[[!include topology - contents]]

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

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{versions}{Versions}\dotfill \pageref*{versions} \linebreak
\noindent\hyperlink{logical_status}{Logical status}\dotfill \pageref*{logical_status} \linebreak
\noindent\hyperlink{Proofs}{Proofs}\dotfill \pageref*{Proofs} \linebreak
\noindent\hyperlink{related_theorem}{Related theorem}\dotfill \pageref*{related_theorem} \linebreak
\noindent\hyperlink{references}{References}\dotfill \pageref*{references} \linebreak


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

The classical \emph{Heine--Borel theorem} identifies those [[topological subspaces]] of [[Cartesian spaces]] ($\mathbb{R}^n$s) that are [[compact space|compact]] in terms of simpler properties. A generalisation applies to all [[metric spaces]] and even to [[uniform spaces]].

\hypertarget{versions}{}\subsection*{{Versions}}\label{versions}

This is the classical theorem:

\begin{theorem}
\label{classical}\hypertarget{classical}{}
Let $S$ be a [[topological subspace]] of a [[Cartesian space]] ($S \subset \mathbb{R}^n$). Then $S$ is a [[compact topological space]] (with the [[induced topology]]) precisely if it is [[closed subset|closed]] and [[bounded subset|bounded]] in $\mathbb{R}^n$.

\end{theorem}
It's easy to prove that $S$ is closed precisely if it is a [[complete metric space]] as with the induced [[metric]], and similarly $S$ is bounded precisely if it is [[totally bounded metric space|totally bounded]]. This gives the next version:

\begin{theorem}
\label{cartesian}\hypertarget{cartesian}{}
Let $S$ be a [[topological subspace]] of a [[Cartesian space]] ($S \subset \mathbb{R}^n$). Then $S$ is a [[compact topological space]] (with the [[induced topology]]) precisely if it is [[complete space|complete]] and [[totally bounded space|totally bounded]] (with the [[induced metric]]).

\end{theorem}
This refers entirely to $S$ as a metric space in its own right. In fact it holds much more generally than for subspaces of a cartesian space:

\begin{theorem}
\label{metric}\hypertarget{metric}{}
Let $S$ be a [[metric space]]. Then $S$ is [[compact space|compact]] precisely if it is [[complete space|complete]] and [[totally bounded space|totally bounded]].

\end{theorem}
This theorem refers only to uniform properties of $S$, and in fact a further generalistion is true:

\begin{theorem}
\label{uniform}\hypertarget{uniform}{}
Let $S$ be a [[uniform space]]. Then $S$ is [[compact space|compact]] precisely if it is [[complete space|complete]] and [[totally bounded space|totally bounded]].

\end{theorem}
The hard part is proving that a complete totally bounded space is compact; the converse is easy.

We could also try to generalise Theorem \ref{classical} to subspaces of other metric spaces, but this fails: every compact subspace of a metric space is closed and bounded (which is the easy direction), but not conversely.

Another variant of these statements is:

\begin{itemize}%
\item \emph{[[closed subspaces of compact Hausdorff spaces are equivalently compact subspaces]]}

\end{itemize}
\hypertarget{logical_status}{}\subsection*{{Logical status}}\label{logical_status}

In the old days, one called a closed and bounded [[interval]] in the [[real line]] `compact'; once closedness and boundedness were generalised from intervals to arbitrary [[subsets]] (of the real line), the definition of `compact' also generalised. The content of Theorem \ref{classical}, then, is that this condition is equivalent to the modern definition of `compact' using [[open covers]], and indeed the modern definition was only derived afterwards as a name for the conclusion of the theorem.

In [[constructive mathematics]], one sees several definitions of `compact', which may make the theorem provable, refutable, or undecidable in various constructive systems. Using the open-cover definition of `compact', Theorems \ref{classical} and \ref{cartesian} are equivalent to the [[fan theorem]] (and so hold in [[intuitionism]]), but Theorems \ref{metric} and \ref{uniform} are stronger and [[Jan Brouwer|Brouwer]] could not prove them, leading him to \emph{define} `compact' (for a metric space) to mean complete and totally bounded. In other literature, one sometimes sees the abbreviation `CTB' used instead. In [[Russian constructivism]], already Theorems \ref{classical} and \ref{cartesian} can be refuted using the open-cover definition, but CTB spaces are still important.

In [[locale theory]] and other approaches to [[pointless topology]], the open-cover definition of `compact' is clearly correct, and the failure of CTB spaces to be compact (constructively) may be seen as a consequence of working with points. Already in Bishop's weak system of constructivism, every CTB metric space $X$ gives rise to a compact locale, which classically (assuming [[excluded middle]] and [[dependent choice]]) is the [[locale of open subsets]] of $X$ but constructively requires a more nuanced construction; see \hyperlink{CTBlocale}{Vickers}.

\hypertarget{Proofs}{}\subsection*{{Proofs}}\label{Proofs}

For definiteness, we restate the versions which we prove:

\begin{lemma}
\label{CompactClosedInterval}\hypertarget{CompactClosedInterval}{}
\textbf{(closed interval is compact)}

In [[classical mathematics]]:

For any $a \lt b \in \mathbb{R}$ the [[closed interval]]

\begin{displaymath}
[a,b] \subset \mathbb{R}
\end{displaymath}
regarded with its [[topological subspace|subspace topology]] is a [[compact topological space]].

\end{lemma}
\begin{proof}
Since all the closed intervals are [[homeomorphism|homeomorphic]] it is sufficient to show the statement for $[0,1]$. Hence let $\{U_i \subset [0,1]\}_{i \in I}$ be an open cover. We need to show that it has an open subcover.

Say that an element $x \in [0,1]$ is \emph{admissible} if the closed sub-interval $[0,x]$ is covered by finitely many of the $U_i$. In this terminlogy, what we need to show is that $1$ is admissible.

Observe from the definition that

\begin{enumerate}%
\item 0 is admissible,


\item if $y \lt x \in [0,1]$ and $x$ is admissible, then also $y$ is admissible.



\end{enumerate}
This means that the set of admissible $x$ forms either an [[open interval]] $[0,g)$ or a [[closed interval]] $[0,g]$, for some $g \in [0,1]$. We need to show that the latter is true, and for $g = 1$. We do so by observing that the alternatives lead to contradictions:

\begin{enumerate}%
\item Assume that the set of admissible values were an open interval $[0,g)$. By assumption there would be a finite subset $J \subset I$ such that $\{U_i \subset [0,1] \}_{i \in J \subset I}$ were a finite open cover of $[0,g)$. Accordingly, since there is some $i_g \in I$ such that $g \in U_{i_g}$, the union $\{U_i\}_{i \in J } \sqcup \{U_{i_g}\}$ were a finite cover of the closed interval $[0,g]$, contradicting the assumption that $g$ itself is not admissible (since it is not contained in $[0,g)$).


\item Assume that the set of admissible values were a closed interval $[0,g]$ for $g \lt 1$. By assumption there would then be a finite set $J \subset I$ such that $\{U_i \subset [0,1]\}_{i \in J \subset I}$ were a finite cover of $[0,g]$. Hence there would be an index $i_g \in J$ such that $g \in U_{i_g}$. But then by the nature of open subsets in the Euclidean space $\mathbb{R}$, this $U_{i_g}$ would also contain an open ball $B^\circ_g(\epsilon) = (g-\epsilon, g + \epsilon)$. This would mean that the set of admissible values includes the open interval $[0,g+ \epsilon)$, contradicting the assumption.



\end{enumerate}
This gives a [[proof by contradiction]].

\end{proof}
\begin{prop}
\label{BorelHeine}\hypertarget{BorelHeine}{}
\textbf{([[Heine-Borel theorem]] classically)}

For $n \in \mathbb{N}$, regard $\mathbb{R}^n$ as the $n$-dimensional [[Euclidean space]], regarded as a [[topological space]] via its [[metric topology]].

Then in [[classical mathematics]] for a [[topological subspace]] $S \subset \mathbb{R}^n$ the following are equivalent:

\begin{enumerate}%
\item $S$ is [[compact topological space|compact]],


\item $S$ is [[closed subset|closed]] and [[bounded subset|bounded]].



\end{enumerate}
\end{prop}
\begin{proof}
First consider a [[subset]] $S \subset \mathbb{R}^n$ which is closed and bounded. We need to show that regarded as a [[topological subspace]] it is [[compact topological space|compact]].

The assumption that $S$ is bounded by (hence contained in) some [[open ball]] $B^\circ_x(\epsilon)$ in $\mathbb{R}^n$ implies that it is contained in $\{ (x_i)_{i = 1}^n \in \mathbb{R}^n \,\vert\, -\epsilon \leq x_i \leq \epsilon \}$. This topological subspace is homeomorphic to the $n$-cube $[-\epsilon, \epsilon]^n$. Since the closed interval $[-\epsilon, \epsilon]$ is compact by lemma \ref{CompactClosedInterval}, the [[Tychonoff theorem]] implies that this $n$-cube is compact.

Since [[subsets are closed in a closed subspace precisely if they are closed in the ambient space]] the closed subset $S \subset \mathbb{R}^n$ is still closed as a subset $S \subset [-\epsilon, \epsilon]^n$. Since [[closed subspaces of compact spaces are compact]] this implies that $S$ is compact.

Conversely, assume that $S \subset \mathbb{R}^n$ is a compact subspace. We need to show that it is closed and bounded.

The first statement follows since the [[Euclidean space]] $\mathbb{R}^n$ is [[Hausdorff topological space|Hausdorff]] and since [[compact subspaces of Hausdorff spaces are closed]].

Hence what remains is to show that $S$ is bounded.

To that end, choose any [[positive number|positive]] [[real number]] $\epsilon \in \mathbb{R}_{\gt 0}$ and consider the [[open cover]] of all of $\mathbb{R}^n$ by the open [[n-cubes]]

\begin{displaymath}
(k_1-\epsilon, k_1+1+\epsilon) 
    \times 
  (k_2-\epsilon, k_2+1+\epsilon)
    \times  
      \cdots
    \times
  (k_n-\epsilon, k_n+1+\epsilon)
\end{displaymath}
for [[n-tuples]] of [[integers]] $(k_1, k_2 , \cdots, k_n ) \in \mathbb{Z}^n$. The restrictions of these to $S$ hence form an open cover of the subspace $S$. By the assumption that $S$ is compact, there is then a finite subset of $n$-tuples of integers such that the corresponding $n$-cubes still cover $S$. But the union of any finite number of bounded closed $n$-cubes in $\mathbb{R}^n$ is clearly a bounded subset, and hence so is $S$.

\end{proof}
\hypertarget{related_theorem}{}\subsection*{{Related theorem}}\label{related_theorem}

\begin{itemize}%
\item [[closed subspaces of compact Hausdorff spaces are equivalently compact subspaces]]


\item [[Tietze extension theorem]]


\item [[Tychonoff theorem]]


\item [[topological invariance of dimension]]


\item [[Brouwer's fixed point theorem]]


\item [[Jordan curve theorem]]



\end{itemize}
\hypertarget{references}{}\subsection*{{References}}\label{references}

\href{https://en.wikipedia.org/wiki/Heine%E2%80%93Borel_theorem#History_and_motivation}{According to Wikipedia}, the theorem was first proved by [[Pierre Cousin]] in 1895. It is named after [[Eduard Heine]] (who used it but did not prove it) and [[Émile Borel]] (who proved a limited version of it), an instance of [[Baez's law]].

A proof is spelled out for instance at

\begin{itemize}%
\item \href{http://www.proofwiki.org/wiki/Main_Page}{ProofWiki}, \emph{Heine-Borel Theorem \href{http://www.proofwiki.org/wiki/Heine-Borel_Theorem_%28General_Case%29}{(General case)}, \href{http://www.proofwiki.org/wiki/Heine-Borel_Theorem_%28Special_Case%29}{(Special case)}}

\end{itemize}
On constructing a compact locale from a CTB metric space:

\begin{itemize}%
\item [[Steve Vickers]], \href{http://www.tac.mta.ca/tac/volumes/14/15/14-15abs.html}{Localic completion of generalized metric spaces}

\end{itemize}
[[!redirects Heine-Borel theorem]] [[!redirects Heine-Borel theorems]] [[!redirects Heine--Borel theorem]] [[!redirects Heine--Borel theorems]] [[!redirects Heine–Borel theorem]] [[!redirects Heine–Borel theorems]]



\end{document}