Let \(f : X \to Y\) be a continuous function, with \(X\) compact. Show that \(f(X)\) is compact.
Solution
Let \(f:X\to Y\) be continuous with \(X\) compact, and \(\theset{U_\alpha} \covers f(X)\) be an open cover. Then \(\theset{\inverseof{f}(U_\alpha)} \covers X\) is an open cover of \(X\), since \(x\in X \implies f(x) \in f(X) \implies f(x) \in U_\alpha\) for some \(\alpha\), so \(x\in \inverseof{f}(U_\alpha)\) by definition. By compactness of \(X\) there is a finite subcover \(\theset{\inverseof{f}(U_j) \suchthat j\leq N} \covers X\). Then the finite subcover \(\theset{U_j\suchthat j\leq N} \covers f(X)\), since if \(y\in f(X)\), \(y\in U_\alpha\) for some \(\alpha\) and thus \(\inverseof{f}(y) \in \inverseof{f}(U_j)\) for some \(j\) since \(\theset{U_j}\) is a cover of \(X\).