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.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $3$: Compactness: Global Compactness Properties