While I was studying Functional Analysis, this question arised:
Let $K \subseteq \mathbb{R}$ be a subset with the propertie that, for all $f$ continuous ($f \in \mathcal{C}^0(\mathbb{R},\mathbb{R})$), the image $f(K)$ has a global maximum (notice that if we consider $(-f)$, then it also has a global minimum). Is $K$ a compact subset?
We can, of course, try to generalize and conjecture and ask ourselves if $f$ being continuous only at the topological space induced by $K$, or if $K$ is any compact topological space, but for a while I am trying that "simple" question.
(If $K$ has "nice" properties, e.g. $K$ is a connected space, the answer is clear, but the general case is a sweet challenge to me.)
If $K$ is not bounded then the function $f(t)=|t|$ has no maximum on $K$. If $K$ is not closed, say $x$ is in the closure but not in $K$; then $f(t)=-|t-x|$ has no maximum on $K$.
(If the notation means you want bounded $f$ then $|t|/(1+|t|)$ and $\max(-|t-x|,-1)$.)