Countably Compact Space is Pseudocompact

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Theorem

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


Then $T$ is a pseudocompact space.


Proof

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

Then every countable open cover of $S$ has a finite subcover.

By definition:

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

Let $f: S \to \R$ be such that $f$ is continuous.

Consider the sets:

$S_n = \set {x \in S: \size {\map f x} < n}$

$S_n$ forms a countable cover of $S$ whose finite subcover yields a bound for the absolute value of $f$.

Hence the result.

$\blacksquare$


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