2.4 Continuity and Compactness
Theorem 2.4.1. Suppose \(f\) is a continuous function of a compact set \(X\big (\subset \mathbb {R}\big )\) into \(Y\big (\subset \mathbb {R}\big )\). Then \(f(x)\) is compact
i.e the continuous image of a compact set is compact.
Proof. Let \(G = \big \{G_i\big \}\) be an open covering of \(f(x)\) i.e \(f(x) \subseteq \bigcup _{i\in I}\, G_i\). Since \(f\) is continuous, \(f^{-1}(G)\) is an open subset of \(X\) for every
open set \(G_i\) in \(G\). Then
\[X\subset f^{-1}\big (f(x)\big )\subset f^{-1}\Bigg (\bigcup _{i\in I}\, G_i\Bigg )= \bigcup _{i\in I}\, f^{-1}(G_i)\]
Hence \(\zeta = \big \{f^{-1}(G_i)\big \}\) is an open covering of \(X\). Since \(X\) is compact a finite number of sets \(f^{-1}(G_i)\, , \, i = 1, \, 2,\, \cdots \cdots , \, n\,\) from \(\zeta \) form a finite
subcovering of \(X\). This gives
\(X\subset \bigcup ^n_{i = 1}f^{-1}(G_i)\). Then \begin {align*} f(x) & \subseteq f\Bigg (\bigcup ^n_{i = 1}\, f^{-1}(G_i)\Bigg )\\\\ & = f\Bigg (f^{-1}\Big (\bigcup ^n_{i = 1} G_i\Big )\Bigg )\\\\ & = \bigcup ^n_{i = 1} G_i \end {align*}
Thus, every open covering \(G\) of \(f(X)\) has a finite subcovering of \(f(x)\). Hence \(f(x)\) is compact.
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Corollary 2.4.2. Let \(f\) be continuous function from a compact set \(X\big (\subset \mathbb {R}\big )\) into \(Y\big (\subset \mathbb {R}\big )\). Then the image \(f(x)\) is a
bounded and closed subset of \(\mathbb {R}\).
Proof. Let \(\{V_i\}\) be an open cover of \(f(X)\). Continuity makes each \(f^{-1}(V_i)\) open, and these sets cover \(X\) since every \(x\in X\) has \(f(x)\) in some \(V_i\). Compactness of \(X\) gives finitely many, \(f^{-1}(V_1),\dots ,f^{-1}(V_n)\), that cover \(X\); applying \(f\), the sets \(V_1,\dots ,V_n\) cover \(f(X)\). Hence \(f(X)\) is compact. □
Remark. With Heine–Borel this delivers the extreme value theorem at once: \(f(X)\) is compact, hence closed and bounded, and a non-empty closed bounded set contains its supremum and infimum. The proof used no property of \(\mathbb {R}\) at all, only open covers, which is why the statement carries over unchanged to a general topological space — as it does in the Topology course.
Theorem 2.4.3. Suppose \(f\) is a continuous real valued function on a compact set \(X\). Then there
exists points \(p\, , q\in X\) such that \(f(p) = \sup \big \{ f(x): \, x \in X\big \}\) and \(f(q) = \big \{f(x):\ x\in X\big \}\).
Proof. Let \(\{V_i\}\) be an open cover of \(f(X)\). Continuity makes each \(f^{-1}(V_i)\) open in \(X\), and these sets cover \(X\), since every \(x\) has \(f(x)\) in some \(V_i\). Compactness of \(X\) yields finitely many, \(f^{-1}(V_1),\dots ,f^{-1}(V_n)\), covering \(X\); applying \(f\), the sets \(V_1,\dots ,V_n\) cover \(f(X)\). Hence \(f(X)\) is compact. □
Remark. Together with Heine–Borel this gives the extreme value theorem in one line: a continuous real function on a compact set has compact, hence closed and bounded, image, and a closed bounded non-empty set contains its supremum and infimum. It is worth noticing that the proof used no property of \(\mathbb {R}\) at all — only open covers — which is why the statement survives unchanged in a general topological space.
Proof. By the above corollary \(f(x)\) is a closed and bounded set of real numbers. Hence \(\, \exists \, M\, , \, m\in f(x) \, \ni \, M= \sup \, f(x) = f(p)\) and \(m = \inf f(x) = f(q)\, ,\, p\, , q\in X\).
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Theorem 2.4.4. Suppose \(f\) is continuous one-one function of a compact set \(X(\subset \mathbb {R})\) onto \(Y(\subset \mathbb {R})\). Then the
inverse function \(f^{-1}\) defined by \(f^{-1}\big (f(x)\big ) = X, \, x\in X\,\) is a continuous function on \(X\).
Proof. Thus, it suffices to prove that \(f(V)\) is an open set in \(Y\) for every open set \(V\) in \(X\).
Taking \(V\big (\subset X\big )\) to be an open set, the complement \(V^C\) is closed in \(X\). Hence, \(f\big (V^C\big )\) is closed in \(Y\). But since \(f\) is
one-one and onto and \(f\big (V^C\big ) = \big [f(V)\big ]^C\, ,\, f(V)\,\) is the complement of \(f\big (V^C\big )\). Hence \(f(V)\) is open.
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