Under what conditions can we establish a converse to the extreme value theorem? That is, for what topological spaces $(X, \tau)$ can we say that if $(\forall f \in C(X))(\exists c \in E) \left( f(c) = \sup_{x \in E}f(x)\right)$, then $E$ is compact, where $C(E)$ denotes the family of all continuous maps $f: E \to \mathbb{R}$ and $E$ has the subspace topology?
I believe this claim holds in any metric space $X$ where a subset is compact iff it is closed and bounded. Clearly if $E$ is unbounded we can consider $x \mapsto d(x, p)$ restricted to $E$ for appropriate $p$. So let $E$ be bounded and not closed, thus not compact. Then there exists $p \in \operatorname{cl}(E) \setminus E$. Let $f : x \mapsto - d( x , p )$. Then since $p$ is a limit point of $E$, we have that for every neighborhood $B(p, 1 / k)$ of $p$ exists $x_{k} \in B(P, 1 / k)$, and thus $- 1 / k < f(x_{k}) < 0$, so $\sup_{x \in E} f(x) = 0$. But $f$ won't attain the supremum on $E$, as $f(x) = 0 \iff x = p$.
Is this correct? How might this be extended to a broader class of spaces $X$? Thanks.
It’s true in any metric space. If $X$ is metric, then $E$ is compact if and only if $E$ is countably compact. If $E$ is not countably compact, then $E$ has an infinite discrete subset $D$ that is closed in $E$. Let $D=\{x_n:n\in\Bbb N\}$, and define $f:D\to\Bbb R:x_n\mapsto n$; $f$ is continuous on $D$,so by the Tietze extension theorem there is a $g\in C(E)$ such that $g\upharpoonright D=f$, and clearly $g$ is unbounded.
However, it fails for Mrówka’s space $\Psi$. This is a pseudocompact Tikhonov space, i.e., a completely regular Hausdorff space on which every continuous real-valued function is bounded. In fact, each $f\in C(\Psi)$ attains its supremum. I’ll use the notation at the link, save that I’ll write $\Psi$ for $\Psi(\mathscr{A})$.
Suppose that $f\in C(\Psi)$, let $r=\sup_{x\in\Psi}f(x)$, and suppose that $r\notin f[\Psi]$. For each $k\in\omega$ there is an $n_k\in\omega$ such that $r-2^{-k}<f(n_k)<r$, and without loss of generality we may assume that the points $n_k$ are distinct. Let $B=\{n_k:k\in\omega\}$; $\mathscr{A}$ is a maximal family of almost disjoint subsets of $\omega$, so there is an $A\in\mathscr{A}$ such that $B\cap A$ is infinite. Let $N=\{k\in\omega:n_k\in A\}$; then $\langle n_k:k\in N\rangle$ converges to $A$ in $\Psi$, so $\langle f(n_k):k\in N\rangle$ converges to $f(A)$ in $\Bbb R$, and hence $f(A)=r$. This contradiction shows that $f$ does in fact attain its supremum.
However, $\Psi$ is not even countably compact, let alone compact: $\Psi\setminus\omega$ is an infinite, closed, discrete subset of $\Psi$.