Continuous Images of Compact Sets Are Compact

Proof. Let \((y_n)_{n\in\mathbb{N}}\) be a sequence in \(f(I)\). Then for each \(n\in\mathbb{N}\) there is an \(x_n\in I\) such that \(y_n=f(x_n)\). Since \(I\) is compact, there exists a subsequence \((x_{n_k})_{k\in\mathbb{N}}\) that converges to some \(x\in I\). Now, since \(f\) is continuous, we have \[\lim_{k\rightarrow \infty} y_{n_k} = \lim_{k\rightarrow \infty} f(x_{n_k}) = f(x)=:y \in f(I).\] Hence we found a subsequence \((y_{n_k})_{k\in\mathbb{N}}\) of \((y_n)_{n\in\mathbb{N}}\) that converges in \(f(I)\). Therefore \(f(I)\) is compact. ◻