Prove that if \(f:X\to Y\) is continuous and \(C\) is a compact subset of \(X\), then \(f(C)\) is a compact subset of \(Y\).
Solution
Let \(\mathcal U\) be a cover of \(f(C)\) by open subsets of \(Y\).
1. \(\{f^{-1}(U):U\in\mathcal U\}\) is an open cover of \(C\).
Each \(f^{-1}(U)\) is open in \(X\) because \(f\) is continuous. If \(x\in C\), then \(f(x)\in f(C)\), so some \(U\in\mathcal U\) contains \(f(x)\); hence \(x\in f^{-1}(U)\).
2. There exist \(U_1,\ldots,U_n\in\mathcal U\) such that \(C\subseteq f^{-1}(U_1)\cup\cdots\cup f^{-1}(U_n)\).
This follows from step 1 and compactness of \(C\).
3. \(f(C)\subseteq U_1\cup\cdots\cup U_n\).
If \(y\in f(C)\), write \(y=f(x)\) with \(x\in C\). By step 2, \(x\in f^{-1}(U_i)\) for some \(i\), so \(y\in U_i\).
4. Q.E.D.
By step 3, every open cover \(\mathcal U\) of \(f(C)\) has a finite subcover, so \(f(C)\) is compact.