Alexander's Subbasis Theorem

# Alexander's Subbasis Theorem

We are now prepared to look at an extremely important theorem known as Alexander's subbasis that was proven by James Waddell Alexander II.

We first state an important lemma known as Zorn's Lemma which we will use in proving Alexander's subbasis theorem.

Lemma 1 (Zorn's Lemma): If $P$ is a partially ordered set such that every totally ordered subset of $P$ has an upper bound then $P$ contains a maximal element. |

We will now present Alexander's subbasis theorem.

Theorem 1 (Alexander's Subbasis Theorem): Let $(X, \tau)$ be a topological space and let $\mathcal S$ be a subbasis for $\tau$. If every open cover of $X$ consisting only of elements from the subbasis $\mathcal S$ has a finite subcover, then $X$ is a compact space. |

*In other words, consider a subbasis $\mathcal S$ of the topology $\tau$ on a set $X$. Suppose that we take an open cover that is made up of elements from the subbasis $\mathcal S$. If every such cover of this form has a finite subcover, then Alexander's subbasis theorem says that $X$ is then a compact space.*

**Proof:**Let $(X, \tau)$ be a topological space and let $\mathcal S$ be a subbasis for the topology $\tau$ on $X$. Suppose that every open cover of $X$ consisting only of elements from the subbasis $\mathcal S$ has a finite subcover and assume instead that $X$ is NOT compact.

- Consider the collection of all possible open covers of $X$. Since $X$ is NOT compact there exists a subset of covers which do not have a finite subcover. Let:

\begin{align} \quad \mathcal F = \{ \mathcal F^* : \mathcal F^* \: \mathrm{does \: not \: have \: a \: finite \: subcover.} \} \end{align}

- Since $X$ is not compact, this means that $\mathcal F \neq \emptyset$. Furthermore, the set $\mathcal F$ is a partially ordered set (by inclusion). This means that some of the elements in $\mathcal F$ can be compared in the sence that if $\mathcal F_1^*, \mathcal F_2^* \in \mathcal F$ are two distinct sets then we say $\mathcal F_1^* < \mathcal F_2^*$ if $\mathcal F_1^* \subset \mathcal F_2^*$.

- Now let $\{ \mathcal F_i^* \}_{i \in I}$ (where $I$ is an indexing set) be a totally ordered subset of $\mathcal F$. We will show that then $\{ \mathcal F_i^* \}_{i \in I}$ contains an upper bound so that we can apply Zorn's lemma.

- Consider the set $\displaystyle{\tilde{\mathcal F} = \bigcup_{i \in I} \mathcal F_i^*}$. Then clearly, for all $i \in I$ we have that:

\begin{align} \quad \mathcal F_i^* \subset \bigcup_{i \in I}\mathcal F_i^* = \tilde{\mathcal F} \end{align}

- We must show that $\tilde{\mathcal F}$ is contained in $\{ \mathcal F_i \}_{i \in I}$. Now, since $\mathcal F_i^*$ covers $X$ for all $i \in I$ we clearly see that $\tilde{\mathcal F}$ covers $X$.

- Assume that $\tilde{\mathcal F} \not \in \{ \mathcal F_i \}_{i \in I}$, i.e., assume that $\tilde{\mathcal F}$ has a finite subcover, say there exists $F_1, F_2, ..., F_n \in \tilde{\mathcal F}$ such that $\{ F_1, F_2, ..., F_n \}$ covers $X$. Then for each $j \in \{ 1, 2, ..., n \}$ we have that $F_j \in \mathcal F_i^*$ for some $i \in I$, and so there exists a subcollection $\{ \mathcal F_{i_1}^*, \mathcal F_{i_2}^*, ..., \mathcal F_{i_n}^* \}$ of $\tilde{\mathcal F^*}$ where $F_j \in \mathcal F_{i_j}^*$ for each $j \in \{ 1, 2, ..., n \}$.

- Note that $\{ \mathcal F_{i_1}^*, \mathcal F_{i_2}^*, ..., \mathcal F_{i_n}^* \}$ is a finite set that is totally ordered by inclusion. So there exists a maximal set, say without loss of generality that $\mathcal F_{i_n}^*$ is such that $\mathcal F_{i_j}^* \subset \mathcal F_{i_n}^*$ for all $j \in \{1, 2, ..., n - 1\}$. Then $\{ F_1, F_2, ..., F_n\} \subset \mathcal F_{i_n}^*$. So the cover $\mathcal F_{i_n}^*$ has a finite subcover, $\{ F_1, F_2, ..., F_n \}$ of $X$ which is a contradiction since $\mathcal F_{i_n}^* \in \mathcal F$.

- So the assumption that $\tilde{\mathcal F} \not \in \{ \mathcal F_i \}_{i \in I}$ was false. So the totally ordered set $\{ \mathcal F_i \}_{i \in I}$ contains an upper bound (namely $\displaystyle{\tilde{\mathcal F} = \bigcup_{i \in I} \mathcal F_i}$. So every totally ordered subset of $\mathcal F$ has an upper bound. By Zorn's lemma, this implies that $\mathcal F$ has a maximal element, call it $\mathcal M \in \mathcal F$.

- Define a new set $\mathcal S^{\dagger}$ as follows:

\begin{align} \quad \mathcal S' =\mathcal S \cap \mathcal M \end{align}

- We claim that $\mathcal S'$ covers $X$. Suppose not, i.e., suppose that $\displaystyle{X \not \subseteq \bigcup_{F \in \mathcal S'} F}$. Then there exists an $\displaystyle{x \in \bigcup_{F \in \mathcal S'} F}$. Since $\mathcal M$ is a (maximal) cover of $X$ (from $\mathcal F$) there exists an open set $U \in \mathcal M$ such that $x \in U$.

- Furthermore, $\mathcal S$ is a subbasis of the topology $\tau$ on $X$. Since $U$ is an open set, there exists a finite subcollection of subbasis elements, $S_1, S_2, ..., S_n \in \mathcal S$ whose intersection is contained in $U$, i.e.:

\begin{align} \quad x \in \bigcap_{i=1}^{n} S_i \subseteq U \end{align}

- Furthermore, since $S_i \in \mathcal S$ and $x \in S_i$ we must have that $S_i \not \in \mathcal M$ for all $i \in \{ 1, 2, ..., n \}$. If so, then $S_i \in \mathcal S \cap \mathcal M = \mathcal S'$ and so $x$ can be covered by $S_i$ - a contradiction.

- Now since $\mathcal M$ is the maximal element of $\mathcal F$ we must have that for each $i \in \{ 1, 2, ..., n \}$ that $\mathcal M \cup \{ S_i \}$ has a finite subcover of $X$ . For each $i \in \{ 1, 2, ..., n \}$ let $\mathcal M_i$ be union of sets contained in the finite subcover of $\mathcal M \cup \{ S_i \}$ except for the set $S_i$. Then $\mathcal M_i \cup S_i$ covers $X$, so $\mathcal M_i \cup S_i = X$.

\begin{align} \quad U \cup \left ( \bigcup_{i=1}^{n} \mathcal M_i \right ) \supseteq \left ( \bigcap_{i=1}^{n} S_i \right ) \cup \left ( \bigcup_{i=1}^{n} \mathcal M_i \right ) \end{align}

- Since the intersection of a set with itself is itself, we see that:

\begin{align} \quad \left ( \bigcap_{i=1}^{n} S_i \right ) \cup \left ( \bigcup_{i=1}^{n} \mathcal M_i \right ) = \left ( \bigcap_{i=1}^{n} S_i \right ) \cup \bigcap_{i=1}^{n} \left ( \bigcup_{i=1}^{n} \mathcal M_i \right ) = \bigcap_{i=1}^{n} \left ( S_i \cup \bigcup_{i=1}^{n} \mathcal M_i \right ) \supseteq \bigcap_{i=1}^{n} (S_i \cup \mathcal M_i) = \bigcap_{i=1}^{n} X = X \end{align}

- However, $U$ is contained in $\mathcal M$, and each $\mathcal M_i$ is a finite union of sets from $\mathcal M$ such that with $S_i$ covers $X$. So $\displaystyle{U \cup \bigcup_{i=1}^{n} \mathcal M_i}$ is a finite collection of sets from $\mathcal M$ that covers $X$.

- But this is a contradiction since $\mathcal M$ is the maximal element from $\mathcal F$ and every open cover in $\mathcal F$ does not have a finite subcover of $X$. So the assumption that $\displaystyle{\mathcal S' = \mathcal S \cap \mathcal M}$ does not cover $X$ was false. So $\displaystyle{\mathcal S'}$ covers $X$.

- Note that since $\displaystyle{\mathcal S' = \mathcal S \cap \mathcal M}$ that then:

\begin{align} \quad \mathcal S' \subset \mathcal S \: (*) \quad \mathrm{and} \quad \mathcal S' \subset \mathcal M \: (**) \end{align}

- From $(*)$ we see that $\mathcal S'$ is an open cover of $X$ consisting only of elements from the subbasis $\mathcal S$. So we're given that then $\displaystyle{\mathcal S'}$ has a finite subcover.

- But $(**)$ implies that $\mathcal M$ also has a finite subcover of $X$ which contradicts that $\mathcal M \in \mathcal F$.

- So the original assumption that $X$ is not compact was false. So, if every open cover of $X$ consisting only of elements from the subbasis $\mathcal S$ has a finite subcover, then $X$ is a compact space.