Countably Compact Lindelöf Space is Compact
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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