Inverse image of compact set under continuous function compact?

Question

I though that the inverse image of a compact set under a continuous function (in $\mathbb{R}$) would also be compact, but I have been shown the following counterexample: consider $f(x)=0$ for all $x\in\mathbb{R}$. Then $\text{Im}(f)=\{0\}$ which is finite, but $f^{-1}(\{0\})=\mathbb{R}$ which is not compact. Is there an issue with this counterexample, or is the statement false?

Answer 1

The statement is false, a necessary and sufficient condition to ensure the compactness of the inverse image of any compact set is that $\lim\limits_{|x|\to+\infty}|f(x)|=+\infty$, these maps are called proper. The key point is that you want the inverse image of a bounded set to remain bounded.

The result holds more generally between real normed vector spaces of finite dimension.