Separable Metacompact Space is Lindelöf/Proof 1

Theorem

Let $T = \struct {S, \tau}$ be a separable topological space which is also metacompact.

Then $T$ is a Lindelöf space.


Proof

$T$ is separable if and only if there exists a countable subset of $S$ which is everywhere dense.

$T$ is metacompact if and only if every open cover of $S$ has an open refinement which is point finite.

$T$ is a Lindelöf space if every open cover of $S$ has a countable subcover.


Having established the definitions, we proceed.

Let $T = \left({S, \tau}\right)$ be separable and metacompact.

Aiming for a contradiction, suppose there exists an open cover $\UU$ of $S$ which has no countable subcover.


As $T$ is metacompact, $\UU$ has an open refinement $\VV$ which is point finite.

By nature of $\UU$, which has no countable subcover, $\VV$ is uncountable.


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By hypothesis, $T$ is separable.

Let $\SS$ be a countable subset of $S$ which is everywhere dense.

Then each $V \in \VV$ contains some $s \in \SS$.


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So some $s \in \SS$ is contained in an uncountable number of elements of $\VV$.

Thus, by definition, $\VV$ is not point finite.


Thus no uncountable open refinement $\VV$ of $\UU$ exists which is point finite.

It follows that $\VV$ must be countable.

Thus $\UU$ has a countable subcover.

That is, by definition, $T$ is a Lindelöf space.

$\blacksquare$


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: Paracompactness
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