Pseudocompact Normal Space is Countably Compact

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Theorem

Let $T = \struct {S, \tau}$ be a normal space.

Then $T$ is pseudocompact if and only if $T$ is countably compact.


Proof

Let $T = \struct {S, \tau}$ 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.


Aiming for a contradiction, suppose $T$ is not countably compact.

By definition of countably compact space, $S$ contains an infinite subset $H = \set {x_n}$ with no $\omega$-accumulation point in $T$.

Since, by definition of normal, $S$ is a $T_1$ space, $H$ is closed and discrete in the subspace topology.


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By definition, $T$ is pseudocompact if and only if every continuous real-valued function on $S$ is bounded.

By definition of normal, we have that $S$ is a $T_4$ space.

Hence the Tietze Extension Theorem guarantees a continuous extension to $S$ of the unbounded continuous mapping $f: H \to \R$ defined by $\map f {x_n} = n$.

This contradicts our hypothesis that $S$ is pseudocompact.

It follows by Proof by Contradiction that $T$ is countably compact.

$\blacksquare$


Sources

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