Pseudocompact Normal Space is Countably Compact
Theorem
Let $T = \left({X, \vartheta}\right)$ be a normal space.
Then $T$ is pseudocompact iff $T$ is countably compact.
Proof
Let $T = \left({X, \vartheta}\right)$ be a normal space.
By Countably Compact Space is Pseudocompact we already have that if $T$ is countably compact then $T$ is pseudocompact.
It remains to prove that if $T$ is pseudocompact then $T$ is countably compact.
Suppose this were not the case.
Then $X$ would contain an infinite subset $S = \left\{{x_n}\right\}$ with no $\omega$-accumulation point.
Since, by definition of normal, $X$ is a $T_1$ space, $S$ is closed and discrete in the subspace topology.
$S$ is closed and discrete in the subspace topology.
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Since, by definition of normal, $X$ is a $T_4$ space, the Tietze Extension Theorem guarantees a continuous extension to $X$ of the unbounded continuous mapping $f: S \to \R$ defined by $f \left({x}\right) = n$.
So $X$ could not have been pseudocompact after all.
$\blacksquare$
Sources
- Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970)... (previous)... (next): $\text{I}: \ \S 3$: Global Compactness Properties
