Separable Metacompact Space is Lindelöf

Theorem

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

Then $T$ is a Lindelöf space.

Proof 1

$T$ 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.

$\blacksquare$

Proof 2

$T$ 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 $\UU$ be an open cover of $S$.

Let $\VV$ be a point finite open refinement of $\UU$.

By Point Finite Set of Open Sets in Separable Space is Countable, $\VV$ is countable.

Define a mapping $H$ on $\VV$ thus:

$\forall V \in \VV: \map H V = \set {U \in \UU: V \subseteq U}$

By Image of Countable Set under Mapping is Countable, the image of $H$ is countable.

Call this image $I$.

Since $\VV$ is a refinement of $\UU$, $\O \notin I$.

By the Axiom of Countable Choice, $I$ has a choice function $c$.

Then $G = c \circ H: \VV \to \UU$ is a mapping such that:

$\forall V \in \VV: V \subseteq \map G V$

Then $\QQ = \map G \VV$ is countable by Image of Countable Set under Mapping is Countable.

Each element of $\QQ$ is an element of $\UU$ by the definition of $G$.

Let $x \in S$.

Then since $\VV$ is a cover for $S$:

$\exists V \in \VV: x \in V$

Then $x \in V \subseteq \map G V \in \QQ$.

Thus $\QQ$ is a countable subcover of $\UU$.

Thus each open cover of $S$ has a countable subcover, so $T$ is a Lindelöf space.

$\blacksquare$