# Pseudocompactness in the $m$-topology in $C(X)$

The sets $\{\{f\in C(X):|g−f|\leq u\} : g\in C(X) \text{ and } u \text{ a positive unit of } C(X)\}$ form a base for some topology on $C(X)$ which is called the $m$-topology on $C(X)$. A norm on $C^{*}(X)$ is given by $\|f\|=\sup|f(x)|$ and the resulting metric topology is called the uniform norm topology on $C^{*}(X)$.

How to show the above two topologies on $C^{*}(X)$ (consider relative top. for the m-topology) coincide iff $X$ is pseudocompact.

(Hints: When $X$ is not pseudo-compact, the set of constant functions in $C^{*}(X)$ is discrete, in the $m$-topology, so that $C^{*}(X)$ is not even a topological vector space whereas $C^{*}(X)$ forms a Banach algebra w.r.t. uniform norm topology).

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What is $C^*(X)$? – Rasmus Jul 21 '12 at 12:43
@Rasmus: it's a somewhat old-fashioned notation for the ring of bounded continuous (real valued) functions on a topological space. It is still used by some people. This question is (essentially verbatim) part 2. of problem 2N on page 35 of Gillman-Jerison, Rings of continuous functions, at least in my 1960 edition of that book. – t.b. Jul 21 '12 at 23:33
@Gerry: What's done is done. I'll merge the old copy into this one. – Willie Wong Jul 22 '12 at 14:33
Deleted some comments about which question to close as the duplicate of which; note that the other question has been closed and merged into this one. Please try not to create a self-referential exact-duplicate closure! – Willie Wong Jul 22 '12 at 14:37

Suppose that $X$ is not pseudocompact; then there is a continuous unbounded $f:X\to\Bbb R$. Without loss of generality we may assume that $f$ is strictly positive: if not, replace it with the function $x\mapsto\max\{|f(x)|,1\}$. Let
$$u:X\to\Bbb R:x\mapsto\frac1{f(x)}\;,$$
and let $U=\{g\in C(X):|f(x)|\le u(x)\text{ for all }x\in X\}$. Then $U$ is a nbhd of $0$ in the $m$-topology, but clearly there is no $\epsilon>0$ such that $\{g\in C(X):\|g\|<\epsilon\}\subseteq U$, so $U$ is not a nbhd of $0$ in the uniform norm topology.
If $X$ is pseudocompact, let $u$ be a positive unit of $C(X)$, and let $f(x)=\frac1{u(x)}$ for $x\in X$; clearly $f\in C(X)$. Suppose that for every $\epsilon>0$ there is an $x_\epsilon\in X$ such that $u(x_\epsilon)<\epsilon$. Then $f(x_\epsilon)>\frac1\epsilon$, so $f$ is unbounded, contradicting the pseudocompactness of $X$.