Prove that if $f$ is a bounded continuous function then $G$ is compact.

Question

Let $X$ be a compact topological space. Let $f:X\to\Bbb R$. Define $G(f)=\{(x,f(x)):\,x\in X\}$.

Prove that if $f$ is a bounded continuous function then $G$ is compact.

Since $\Bbb R$ is Hausdorff then $G(f)$ is closed. Now $G(f)\subset X\times f(X)$ which is compact since $X$ is so. Since every closed subset of a compact space is compact so $G(f)$ is compact.

Is this proof correct? Where is the hypothesis $f$ is bounded needed?

Answer 1

Your proof is fine. The hypothesis "$f$ is bounded" is not needed, but that is fine. It is automatically true, as $f[X]$, being a compact subsets of $\mathbf R$ is bounded for any continuous $f$.

Answer 2

We must assume $X$ to be Hausdorff, otherwise I believe that your statement is false.

Since $X$ is compact and $\Bbb R$ is Hausdorff, it follows that $f$ is closed, therefore $f(X)$ is closed (not $G(f)$ as you write). Since $f$ is continuous on a bounded space, it is bounded (no need to assume it!), therefore $f(X)$ (being closed) is compact. Therefore, $X \times f(X)$ is compact. The map $X \ni x \mapsto (x, f(x)) \in X \times f(X) \subseteq X \times \Bbb R$ is closed, so its image will be a closed subset of the compact space $X \times f(X)$, therefore compact itself.

Note that you have used the closed map lemma twice: if $f : X \to Y$ is continuous, $X$ is compact and $Y$ Hausdorff, then $f$ is a closed map (i.e. the image of any closed subset is closed).