Proving that a continuous function on a compact set gives a compact image

Question

Why is being continuity important when we prove that a compact set is mapped to a compact set? Is it because if the function isn’t continuous at an x belonging to my domain then my limit might be infinity and so my codomain doesn’t remain closed and bounded anymore?

Answer 1

Let $f\colon X \to Y$ be a continuous surjective map between topological spaces, with $X$ compact. Let's show that $Y$ is compact. Let $(V_\alpha)_{\alpha \in A}$ be an open cover for $Y$. Then each $f^{-1}[V_\alpha]$ is open, since $f$ is continuous. Now $(f^{-1}[V_\alpha])_{\alpha \in A}$ is an open cover for $X$, since $f$ is surjective. Now $X$ is compact, and so there is a finite subcover $(f^{-1}[V_{\alpha_i}])_{i=1}^n$ for $X$, where $\alpha_1,\ldots, \alpha_n \in A$ are fixed indices. Then $(V_{\alpha_i})_{i=1}^n$ is the corresponding finite subcover for $Y$.

So to concretely answer your question: if $f$ is not continuous, one cannot bring back the open cover of the image of $f$ back to $X$, to use compactness of $X$.