Heine-Borel Theorem

For fun and to improve my proof-writing, I am working through the book Measure, Integration & Real Analysis by Sheldon Axler. Each week (as of 7/27/24), I plan to write a blog post re-writing a proof from Axler’s book.

Open Cover: Let \(A\subset \mathbb{R}\). A collection \(\mathcal{C}\) of open subsets of \(\mathbb{R}\) is called an open cover of \(A\) if \(A\) is contained in the union of all the sets in \(\mathcal{C}\).

Heine-Borel Theorem: every open cover \(\mathcal{C}\) of a closed bounded subset \(F\) of \(\mathbb{R}\) has a finite subcover.

Proof: To begin, we consider the case \(F=[a,b]\) such that \(a,b\in\mathbb{R}\) and \(a<b\). Let \(D=\{d\in [a,b] \mid [a,d]\; \text{has a finite subcover from }\mathcal{C} \}\). To prove there exists a finite subcover from \(\mathcal{C}\) over \(F\), we will show that \(b\in D\). Outlining the proof, we first show that \(D\) exists and that \(\sup D\in [a,b]\). Hence, by the definition of \(D\), a finite subcover over \([a,\sup D]\) exists. Finally, we show that \(\sup D = b\), completing the proof.

\((D\neq \emptyset )\): Since \(\mathcal{C}\) is an open cover on \(F\), there exists an open set \(G\in\mathcal{C}\) such that \(a\in G\) and thus \(a\in D\). Therefore \(D\neq \emptyset\).

\((\sup D\in [a,b] )\): Let \(s=\sup D\). Since \(D\subseteq [a,b]\), a closed set, \(s\in [a,b]\).

\((b=\sup D \; \& \; b\in D )\): By \(\mathcal{C}\) being an open cover on \(F\), there exists \(G'\in \mathcal{C}\) such that \(s\in G'\). By \(G'\) being an open set, there exists a \(\delta >0\) such that \((s-\delta ,s+\delta )\subset G'\). Since \(s\in [a,b]\), there exists \(d'\in (s-\delta ,s)\) and a finite open cover \(G_1\dots G_n\) such that \([a,d']\subseteq G_1\cup \dots \cup G_n\). Observer for all \(d''\in [s,s+\delta)\) it hold that \([a,d'']\subseteq G'\cup G_1\cup \dots \cup G_n\). Therefore, for any \(\hat{d}\in [s,s+\delta )\cap [a,b]\) it holds that \(\hat{d}\in D\).
We claim that \(b=s\). If \(b\neq s\) then some element of \([s,s+\delta )\) would be greater than \(\sup D\) forming a contradiction. Hence, \(b\in [b,b+\delta )\cap [a,b]\) meaning \(b\in D\).

We now cover the general case for a closed bounded subset \(F\) of \(\mathbb{R}\) with an open cover \(\mathcal{C}\). Pick \(a',b'\in \mathbb{R}\) such that \(F\subset [a',b']\). Note \(\mathcal{C}\cup \{\mathbb{R}\setminus F \}\) is an open cover of \(\mathbb{R}\) ane hence \([a',b']\). By our first proven case, there exists \(G_1\dots G_n\in\mathcal{C}\) such that \([a',b']\subset G_1\cup\dots \cup G_n\cup (\mathbb{R}\setminus F)\). Since \(F\subset [a',b']\) and \(F\cap (\mathbb{R}\setminus F)=\emptyset\), it implies that \(F\subset G_1\cup\dots \cup G_n\). \(\square\)