Show that if \(f:X\to Y\) is continuous and \(X\) is compact then \(f(X)\) is compact.
Exercise
Solution
1. Let \(\mathcal U\) be an open cover of \(f(X)\). Then \(\{f^{-1}(U):U\in\mathcal U\}\) is an open cover of \(X\).
Continuity makes each preimage open, and every \(x\in X\) maps into some member of \(\mathcal U\).
2. Compactness of \(X\) gives finitely many \(U_1,\dots,U_r\in\mathcal U\) whose preimages cover \(X\).
Apply the definition of compactness to the cover in 1.
3. Then \(U_1,\dots,U_r\) cover \(f(X)\), so \(f(X)\) is compact.
For \(y=f(x)\), some \(f^{-1}(U_i)\) contains \(x\), hence \(y\in U_i\).