Compactness from Basis

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Theorem

Let $\left({X, \tau}\right)$ be a topological space.

Let $B$ be a basis for $\tau$.


Then the following propositions are equivalent:

\((1)\)   $:$   $\left({ X, \tau }\right)$ is compact.             
\((2)\)   $:$   Every open cover of $X$ by elements of $B$ has a finite subcover.             

Proof

$(1)$ implies $(2)$

This follows immediately from the definition of compactness,

$\Box$


$(2)$ implies $(1)$

Suppose that $(2)$ holds.

Let $\mathcal A$ be an open cover of $X$.

Let $f:\mathcal A \to \mathcal P(B)$ map each element of $\mathcal A$ to the set of all elements of $B$ it contains.

Since each element of $\mathcal A$ is open, $A = \bigcup f(A)$ for each $A \in \mathcal A$.

Let $\mathcal A' = \bigcup f(\mathcal A)$.

Then $\mathcal A'$ is an open cover of $X$ consisting solely of elements of $B$.

By the premise, $\mathcal A'$ has a finite subset $\mathcal F'$ that covers $X$.

Let $g:\mathcal F' \to \mathcal A$ map each element of $\mathcal F'$ to an element of $\mathcal A$ containing it.

Note that since $\mathcal F'$ is finite, this does not require the Axiom of Choice.

Let $\mathcal F = g(\mathcal F')$.

Then $\mathcal F$ is a finite subcover of $\mathcal A$.

Since every open cover of $X$ has a finite subcover, $X$ is compact.

$\blacksquare$