Proof. Let \(U\) be a collection of open sets in \(\mathbb {R}\) with the property that each point of the interval \([a,b]\) belongs to at least one of these sets.
We must show that \([a,b]\) is covered by finitely many of these open sets.
Let \(S\) be the set of all \(\tau \in [a,b]\) with the property that \([a,\tau ]\) is covered by some finite collection of open sets belonging to \(U\), and let \(s=\sup S\). Now, \(s\in W\) for some open set \(W\) belonging to \(U\). Moreover, \(W\) is open in \(\mathbb {R}\) and therefore there exists some \(\delta >0\) such that \((s-\delta ,s+\delta )\subseteq W\). Moreover, \(s-\delta \) is not an upper bound for the set \(S\), hence there exists some \(\tau \in S\) satisfying \(\tau >s-\delta \). It follows from the definition of \(S\) that \([a,\tau ]\) is covered by some collection \(V_1,V_2,.........,V_n\) of open sets belonging to \(U\). Let \(t\in [a,b]\) satisfy \(\tau \leq t<s+\delta \). Then \([a,t]\subseteq [a,\tau ]\cup (s-\delta ,s+\delta )\subseteq V_1\cup V_2\cup ...............\cup V_n\cup W\), and thus \(t\in S\). In particular \(s\in S\) and moreover, \(s=b\), since otherwise \(s\) would not be an upper bound of the set \(S\). Thus \(b\in S\), and therefore \([a,b]\) is covered by a finite collection of open sets belonging to \(U\), as required.

The Heine-Borel theorem still holds true in any Euclidean space \(\mathbb {R}^n\), that is; a closed and bound subset of \(\mathbb {R}^n\) is compact.

 

We prove that a closed subset of a compact topological space is compact. The following lemma will be required.

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Proof. Let \(V=\{B_{\lambda }:\lambda \in \Omega \}\) be an open covering of \(A\). Then each \(B_{\lambda }\) takes the form \(B_{\lambda }=V_{\lambda }\cap A\), for \(V_{\lambda }\) in the collection U. Since \(A\subseteq V_1\cup V_2\cup ...........\cup V_n\). It follows that \begin {align*} A &=(V_1\cup V_2\cup .............\cup V_n)\cap A\\ &=(V_1\cap A)\cup (V_2\cap A)\cup ...............\cup (V_n\cap A)\\ &=B_1\cup B_2\cup ...............\cup B_n, \end {align*}

where \(B_1,B_2,...........,B_n\) are in the collection \(V\).

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Proof. Let \(U\) be any collection of open sets in \(X\) covering \(A\). On adjoining the open set \(X-A\) to \(U\), we obtain an open covering for \(X\). Since \(X\) is compact, this open covering contains a finite sub-covering. Moreover, A is covered by the open sets in the collection \(U\) that belongs to this finite sub-covering. It follows from the lemma that \(A\) is compact.

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Proof. Let \(\nu \) be a collection of open sets in \(Y\) which covers \(f(A)\). Then \(A\) is covered by a collection of open sets of the form \(f^{-1}(V)\) for some \(V\in \nu \). It follows from compactness of \(A\) that there exists a finite collection \(V_1,V_2,.........,V_k\) of open sets belonging to \(\nu \) such that \(A\subseteq f^{-1}(V_1)\cup f^{-1}(V_2)\cup ............\cup f^{-1}(V_k)\). But then \(f(A)\subseteq V_1\cup V_2\cup .............\cup V_k\). This shows that \(f(A)\) is compact.

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