Show that if \(f:X\to Y\) is continuous and \(X\) is compact then \(f(X)\) is compact.
Solution
Goal: Show that if \(f: X \to Y\) is continuous and \(X\) is compact, then \(f(X)\) is compact.
1. Let \(\theset{U_\alpha}\) be an open cover of \(f(X)\).
Arbitrary open cover in \(Y\) (intersect with \(f(X)\) if needed).
2. \(\theset{f^{-1}(U_\alpha)}\) is an open cover of \(X\).
Each \(f^{-1}(U_\alpha)\) is open (continuity), and for every \(x \in X\), \(f(x) \in U_\alpha\) for some \(\alpha\), so \(x \in f^{-1}(U_\alpha)\).
3. There is a finite subcover \(\theset{f^{-1}(U_{\alpha_1}), \ldots, f^{-1}(U_{\alpha_k})}\) of \(X\).
\(X\) is compact and 2 gives an open cover.
4. \(\theset{U_{\alpha_1}, \ldots, U_{\alpha_k}}\) covers \(f(X)\).
For \(y \in f(X)\), \(y = f(x)\) with \(x \in f^{-1}(U_{\alpha_j})\) for some \(j\) (<1>3); then \(y = f(x) \in U_{\alpha_j}\).
5. Q.E.D.
<1>1–<1>4 show every open cover of \(f(X)\) has a finite subcover, i.e. \(f(X)\) is compact.