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

and a subbase

S

for it, a subset

K \subset X

is compact if and only if for each cover of

K

consisting of sets in

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

Suppose for the sake of contradiction that the space $X$ is not compact (so $X$ is an infinite set), yet every subbasic cover from $𝒮$ has a finite subcover. Let $𝕊$ denote the set of all open covers of $X$ that do not have any finite subcover of $X .$ Partially order $𝕊$ by subset inclusion and use Zorn's Lemma to find an element $𝒞 \in 𝕊$ that is a maximal element of $𝕊 .$ Observe that:

Since $𝒞 \in 𝕊,$ by definition of $𝕊,$ $𝒞$ is an open cover of $X$ and there does not exist any finite subset of $𝒞$ that covers $X$ (so in particular, $𝒞$ is infinite).
The maximality of $𝒞$ in $𝕊$ implies that if $V$ is an open set of $X$ such that $V \in̸ 𝒞$ then $𝒞 \cup {V}$ has a finite subcover, which must necessarily be of the form ${V} \cup 𝒞_{V}$ for some finite subset $𝒞_{V}$ of $𝒞$ (this finite subset depends on the choice of $V$ ).

We will begin by showing that $𝒞 \cap 𝒮$ is not a cover of $X .$ Suppose that $𝒞 \cap 𝒮$ was a cover of $X,$ which in particular implies that $𝒞 \cap 𝒮$ is a cover of $X$ by elements of $𝒮 .$ The theorem's hypothesis on $𝒮$ implies that there exists a finite subset of $𝒞 \cap 𝒮$ that covers $X,$ which would simultaneously also be a finite subcover of $X$ by elements of $𝒞$ (since $𝒞 \cap 𝒮 \subseteq 𝒞$ ). But this contradicts $𝒞 \in 𝕊,$ which proves that $𝒞 \cap 𝒮$ does not cover $X .$

Since $𝒞 \cap 𝒮$ does not cover $X,$ there exists some $x \in X$ that is not covered by $𝒞 \cap 𝒮$ (that is, $x$ is not contained in any element of $𝒞 \cap 𝒮$ ). But since $𝒞$ does cover $X,$ there also exists some $U \in 𝒞$ such that $x \in U .$ It follows that $U \neq X$ , because otherwise it would imply $𝒞$ has a finite subcover of $X$ , namely the subcover ${U} = {X},$ contradicting $𝒞 \in 𝕊 .$ Since $U \neq X,$ and $𝒮$ is a subbasis generating $X$ 's topology (together with $X$ ), from the definition of the topology generated by $𝒮,$ there must exist a finite collection of subbasic open sets $S_{1}, \dots, S_{n} \in 𝒮$ with $n \geq 1$ such that $x \in S_{1} \cap \dots \cap S_{n} \subseteq U .$

We will now show by contradiction that $S_{i} \in̸ 𝒞$ for every $i = 1, \dots, n .$ If $i$ was such that $S_{i} \in 𝒞,$ then also $S_{i} \in 𝒞 \cap 𝒮$ so the fact that $x \in S_{i}$ would then imply that $x$ is covered by $𝒞 \cap 𝒮,$ which contradicts how $x$ was chosen (recall that $x$ was chosen specifically so that it was not covered by $𝒞 \cap 𝒮$ ).

As mentioned earlier, the maximality of $𝒞$ in $𝕊$ implies that for every $i = 1, \dots, n,$ there exists a finite subset $𝒞_{S_{i}}$ of $𝒞$ such that ${S_{i}} \cup 𝒞_{S_{i}}$ forms a finite cover of $X .$ Define $𝒞_{F} := 𝒞_{S_{1}} \cup \dots \cup 𝒞_{S_{n}},$ which is a finite subset of $𝒞 .$ Observe that for every $i = 1, \dots, n,$ ${S_{i}} \cup 𝒞_{F}$ is a finite cover of $X$ so let us replace every $𝒞_{S_{i}}$ with $𝒞_{F} .$

Let $\cup 𝒞_{F}$ denote the union of all sets in $𝒞_{F}$ (which is an open subset of $X$ ) and let $Z$ denote the complement of $\cup 𝒞_{F}$ in $X .$ Observe that for any subset $A \subseteq X,$ ${A} \cup 𝒞_{F}$ covers $X$ if and only if $Z \subseteq A .$ In particular, for every $i = 1, \dots, n,$ the fact that ${S_{i}} \cup 𝒞_{F}$ covers $X$ implies that $Z \subseteq S_{i} .$ Since $i$ was arbitrary, we have $Z \subseteq S_{1} \cap \dots \cap S_{n} .$ Recalling that $S_{1} \cap \dots \cap S_{n} \subseteq U,$ we thus have $Z \subseteq U,$ which is equivalent to ${U} \cup 𝒞_{F}$ being a cover of $X .$ Moreover, ${U} \cup 𝒞_{F}$ is a finite cover of $X$ with ${U} \cup 𝒞_{F} \subseteq 𝒞 .$ Thus $𝒞$ has a finite subcover of $X,$ which contradicts the fact that $𝒞 \in 𝕊 .$ Therefore, the original assumption that $X$ is not compact must be wrong, which proves that $X$ is compact. $◼$

Although this proof makes use of Zorn's Lemma, the proof does not need the full strength of choice. Instead, it relies on the intermediate Ultrafilter principle.

Application

Tychonoff's theorem, which states that the product of non-empty compact spaces is compact, has a short proof using the lemma.

The product topology on $\prod_{i} X_{i}$ has, by definition, a subbase consisting of cylinder sets that are the inverse projections of an open set in one factor. Given a subbasic family $C$ of the product that does not have a finite subset which covers $X$ (we do not require $C$ is a cover of $X$ ), we can partition $C = \cup_{i} C_{i}$ into subfamilies that consist of exactly those cylinder sets corresponding to a given factor space. By assumption, $C_{i}$ does not have a finite cover of $X$ . Being cylinder sets, this means their projections onto $X_{i}$ have no finite cover of $X_{i}$ . Since each $X_{i}$ is compact, these projections do not cover $X_{i}$ . Find a point $x_{i} \in X_{i}$ not contained in any of the projections of $C_{i}$ onto $X_{i}$ . Repeating this for all $C_{i}$ yields a point ${(x_{i})}_{i} \in \prod_{i} X_{i}$ which is not covered by $C$ . $◼$

Note, that in the last step we implicitly used the axiom of choice (which is actually equivalent to Zorn's lemma) to ensure the existence of ${(x_{i})}_{i} .$