Space is Compact iff exists Basis such that Every Cover has Finite Subcover

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Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Then $T = \struct {S, \tau}$ is compact if and only if $\tau$ has a basis $\BB$ such that:

from every cover of $S$ by elements of $\BB$, a finite subcover of $S$ can be selected.

Proof

Sufficient Condition

Let every open cover for $S$ have a finite subcover.

Let $\BB$ be a basis $\BB$.

Then every cover of $S$ by elements of $\BB$ is an open cover for $S$.

So:

from every cover of $S$ by elements of $\BB$, a finite subcover of $S$ can be selected.

$\Box$

Necessary Condition

Let $\tau$ have a basis $\BB$ such that:

from every cover of $S$ by elements of $\BB$, a finite subcover of $S$ can be selected.

Let $\AA$ be an open cover of $S$.

Let $f: \AA \to \powerset B$ be the mapping defined by:

$\forall A \in \AA : \map f A = \set{B \in \BB : B \subseteq A}$

Since each element of $\AA$ is open:

$A = \bigcup \map f A$ for each $A \in \AA$.

Let $\AA' = \bigcup f \sqbrk {\AA}$.

Then $\AA'$ is an cover of $S$ by elements of $\BB$

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

Let $g: \FF' \to \AA$ be the map each element of $\FF'$ to an element of $\AA$ that contains it.

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

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

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

It follows that every open cover of $S$ has a finite subcover of $S$.

$\blacksquare$