Evaluation map of closed subspace compact? I have the following bit of information:
Let $X$ be a topological space.  If $f \in C^*(X)$, the set of closed, bounded continuous functions on the interval $I_f \subseteq \mathbb{R}$ with $f(X) \subseteq I_f$.  Define the evaluation map
$$e: X \longrightarrow \prod_{f \in C^*(X)}I_f \quad \text{where } e(x)(f)=f(x).$$
I have managed to show that this map is well defined and that $e$ is continuous.  How can I show that $\overline{e(X)}$ is a compact subspace of $\prod_{f \in C^*(X)}I_f$?
 A: More commonly for this construction of the maximal compactification of $X$ (that is what you’re doing, right?) we use all continuous $X \to [0,1]$ (so we need $X$ to be $T_{3\frac12}$ to have enough of them), a proper subset of $C^*(X)$ which is the set of all bounded continuous (not necessarily closed) functions on $X$ with values in $\Bbb R$.
But here the text (I think) wants you to fix a closed interval $I_f \subseteq \Bbb R$ (so of the form $[a,b]$ for some $a < b$ in $\Bbb R$) such that $f[X]\subseteq I_f$; this is exactly possible because $f$ is bounded; for definiteness we could define $I_f= \left[ \inf f[X], \sup f[X] \right]$ using the completeness of $\Bbb R$ (so we need not make a choice at all).
Then $\prod_{f \in C^*(X)} I_f$ is compact by Tychonoff's theorem (as all $I_f$ are) and $\overline{e[X]}$ is a closed subset of that product so compact by standard facts too.
It's also a standard fact that for Tychonoff (i.e. $T_{3\frac12}$) spaces $e:X \to e[X]$ is a homeomorphism. For a general space $X$ we can only prove it is continuous.
