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,¹²³

given a topological space ${\displaystyle X}$ and a subbase ${\displaystyle S}$ for it, a subset ${\displaystyle K\subset X}$ is compact if and only if for each cover of ${\displaystyle K}$ consisting of sets in ${\displaystyle S}$, 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.

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

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

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

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

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

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

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

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

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.²

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 ${\displaystyle \prod _{i}X_{i}}$ 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 ${\displaystyle C}$ of the product that does not have a finite subset which covers ${\displaystyle X}$ (we do not require ${\displaystyle C}$ is a cover of ${\displaystyle X}$), we can partition ${\displaystyle C=\cup _{i}C_{i}}$ into subfamilies that consist of exactly those cylinder sets corresponding to a given factor space. By assumption, ${\displaystyle C_{i}}$ does not have a finite cover of ${\displaystyle X}$. Being cylinder sets, this means their projections onto ${\displaystyle X_{i}}$ have no finite cover of ${\displaystyle X_{i}}$. Since each ${\displaystyle X_{i}}$ is compact, these projections do not cover ${\displaystyle X_{i}}$. Find a point ${\displaystyle x_{i}\in X_{i}}$ not contained in any of the projections of ${\displaystyle C_{i}}$ onto ${\displaystyle X_{i}}$. Repeating this for all ${\displaystyle C_{i}}$ yields a point ${\displaystyle \left(x_{i}\right)_{i}\in \prod _{i}X_{i}}$ which is not covered by ${\displaystyle C}$. ${\displaystyle \blacksquare }$

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 ${\displaystyle \left(x_{i}\right)_{i}.}$