2.4 Continuity and Compactness
Theorem 2.4.1. Suppose \(f\) is a continuous function from a compact set \(X\subseteq \mathbb {R}\) into \(Y\subseteq \mathbb {R}\). Then \(f(X)\) is compact; that is, the continuous image of a compact set is compact.
Proof. Let \(\{G_i : i\in I\}\) be an open covering of \(f(X)\). For each \(i\), the set \(f^{-1}(G_i)\) is open in \(X\), since \(G_i\cap Y\) is open in \(Y\) and \(f\) is continuous. So \(f^{-1}(G_i) = X\cap O_i\) for some open set \(O_i\subseteq \mathbb {R}\). Every \(x\in X\) has \(f(x)\) in some \(G_i\), so the sets \(O_i\) form an open covering of \(X\). Since \(X\) is compact, finitely many of them, say \(O_1,\dots ,O_n\), cover \(X\). Then every \(x\in X\) lies in some \(f^{-1}(G_i)\) with \(1\leq i\leq n\), so \[f(X) \subseteq G_1\cup G_2\cup \dots \cup G_n .\] Thus every open covering of \(f(X)\) has a finite subcovering, and \(f(X)\) is compact. □
Corollary 2.4.2. Let \(f\) be a continuous function from a compact set \(X\subseteq \mathbb {R}\) into \(Y\subseteq \mathbb {R}\). Then \(f(X)\) is a bounded and closed subset of \(\mathbb {R}\).
Proof. By the theorem \(f(X)\) is compact, and by the Heine–Borel Theorem a compact subset of \(\mathbb {R}\) is bounded and closed. □
Theorem 2.4.3. Suppose \(f\) is a continuous real valued function on a non-empty compact set \(X\). Then there exist points \(p, q\in X\) such that \[f(p) = \sup \{ f(x): x \in X\} \qquad \text {and}\qquad f(q) = \inf \{f(x): x\in X\}.\]
Proof. By the theorem above, \(f(X)\) is a non-empty compact set of real numbers. By the corollary to the Heine–Borel Theorem it has a largest element \(M\) and a smallest element \(m\). Since \(M,m\in f(X)\), we have \(M = f(p)\) and \(m = f(q)\) for some \(p, q\in X\). □
Remark. The proof that the continuous image of a compact set is compact used no special property of \(\mathbb {R}\), only open coverings. That is why the statement, and with it the extreme value theorem, carries over unchanged to general topological spaces.
Theorem 2.4.4. Suppose \(f\) is a continuous one-one function from a compact set \(X\subseteq \mathbb {R}\) onto \(Y\subseteq \mathbb {R}\). Then the inverse function \(g = f^{-1}\), defined by \(g(f(x)) = x\) for \(x\in X\), is continuous on \(Y\).
Proof. By the corollary on closed sets, it suffices to show that \(g^{-1}(C) = f(C)\) is closed in \(Y\) for every set \(C\) closed in \(X\). Such a set is \(C = X\cap F\) with \(F\) closed in \(\mathbb {R}\); since \(X\) is closed and bounded, so is \(C\), and \(C\) is therefore compact by the Heine–Borel Theorem. By the theorem above, \(f(C)\) is compact, hence closed in \(\mathbb {R}\), and so \(f(C) = Y\cap f(C)\) is closed in \(Y\). Hence \(g\) is continuous. □
Example 2.4.5. Find the image of \(f(x) = \frac {1}{1+x^2}\) on \([-3,3]\) and on \(\mathbb {R}\), and explain the difference.
Solution. The function is even, decreasing on \([0,\infty )\), with \(f(0) = 1\) and \(f(3) = \frac {1}{10}\). So \(f([-3,3]) = \left [\frac {1}{10},1\right ]\), a compact interval, as the theorems of this section predict.
On \(\mathbb {R}\), \(f(x)\) takes every value in \((0,1]\) but never \(0\), since \(f(x)>0\) and \(f(x)\longrightarrow 0\) as \(|x|\longrightarrow \infty \). So \(f(\mathbb {R}) = (0,1]\), which is not compact: the infimum \(0\) is not attained. Nothing is contradicted, because \(\mathbb {R}\) is not compact.
Example 2.4.6. Show that compactness of the domain is needed in the theorem on continuous inverses: give a continuous bijection whose inverse is not continuous.
Solution. Let \(X = [0,1)\cup [2,3]\) and define \(f: X\longrightarrow [0,2]\) by \(f(x) = x\) on \([0,1)\) and \(f(x) = x-1\) on \([2,3]\). Then \(f\) is continuous on \(X\), one-one, and onto \([0,2]\). Its inverse is \(g(y) = y\) for \(0\leq y<1\) and \(g(y) = y+1\) for \(1\leq y\leq 2\). As \(y\longrightarrow 1^-\), \(g(y)\longrightarrow 1\), but \(g(1) = 2\). So \(g\) is not continuous at \(1\). The domain \(X\) is not compact, because it is not closed.
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