Countably Compact Lindelöf Space is Compact

Theorem

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

Then $T$ is compact.

Proof

By the definitions:

If $T = \struct {S, \tau}$ is a Lindelöf space then every open cover of $S$ has a countable subcover.

If $T = \struct {S, \tau}$ is a countably compact space then every countable open cover of $S$ has a finite subcover.

It follows trivially that every open cover of $S$ has a finite subcover.

Hence the result by definition of compact.

$\blacksquare$

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