Alexander's subbase lemma

In mathematics, specifically topology, Alexander's subbase lemma states that for a topological space to be a compact, it is necessary and sufficient that each open cover of the space consisting of sets in a (fixed) subbase has a finite subcover. Precisely,

given a topological space

X

{\displaystyle X}

and a subbase

S

{\displaystyle S}

for it, a subset

K

⊂

X

{\displaystyle K\subset X}

is compact if and only if for each cover of

K

{\displaystyle K}

consisting of sets in

S

{\displaystyle S}

, there exists a finite subcover of it.

The lemma, also often called Alexander's subbase theorem, is due to James Waddell Alexander II. The lemma is typically used to prove Tychonoff's theorem.

Proof

If

K

{\displaystyle K}

is compact, the existence of a finite cover holds by definition. So, we shall show the converse and without loss of generality, replace

X

{\displaystyle X}

by

K

{\displaystyle K}

.

Suppose for the sake of contradiction that the space

X

{\displaystyle X}

is not compact (so

X

{\displaystyle X}

is an infinite set), yet every subbasic cover from

S

{\displaystyle {\mathcal {S}}}

has a finite subcover.

Let

S

{\displaystyle \mathbb {S} }

denote the set of all open covers of

X

{\displaystyle X}

that do not have any finite subcover of

X

.

{\displaystyle X.}

Partially order

S

{\displaystyle \mathbb {S} }

by subset inclusion and use Zorn's Lemma to find an element

C

∈

S

{\displaystyle {\mathcal {C}}\in \mathbb {S} }

that is a maximal element of

S

.

{\displaystyle \mathbb {S} .}

Observe that:

Since

C

∈

S

,

{\displaystyle {\mathcal {C}}\in \mathbb {S} ,}

by definition of

S

,

{\displaystyle \mathbb {S} ,}

C

{\displaystyle {\mathcal {C}}}

is an open cover of

X

{\displaystyle X}

and there does not exist any finite subset of

C

{\displaystyle {\mathcal {C}}}

that covers

X

{\displaystyle X}

(so in particular,

C

{\displaystyle {\mathcal {C}}}

is infinite).

The maximality of

C

{\displaystyle {\mathcal {C}}}

in

S

{\displaystyle \mathbb {S} }

implies that if

V

{\displaystyle V}

is an open set of

X

{\displaystyle X}

such that

V

∉

C

{\displaystyle V\not \in {\mathcal {C}}}

then

C

∪

{

V

}

{\displaystyle {\mathcal {C}}\cup \{V\}}

has a finite subcover, which must necessarily be of the form

{

V

}

∪

C

V

{\displaystyle \{V\}\cup {\mathcal {C}}_{V}}

for some finite subset

C

V

{\displaystyle {\mathcal {C}}_{V}}

of

C

{\displaystyle {\mathcal {C}}}

(this finite subset depends on the choice of

V

{\displaystyle V}

).