Compactness from Basis

From ProofWiki
Jump to navigation Jump to search


Theorem

Let $\struct {X, \tau}$ be a topological space.

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


Then the following propositions are equivalent:

\((1)\)   $:$   $\struct {X, \tau}$ 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 $\AA$ be an open cover of $X$.

Let $f: \AA \to \powerset B$ map each element of $\AA$ to the set of all elements of $B$ it contains.

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

Let $\AA' = \bigcup \map f \AA$.

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

By the premise, $\AA'$ has a finite subset $\FF'$ that covers $X$.

Let $g: \FF' \to \AA$ map each element of $\FF'$ to an element of $\AA$ containing it.

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

Let $\FF = \map g {\FF'}$.

Then $\FF$ is a finite subcover of $\AA$.

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

$\blacksquare$