# Maximal open cover in the context of Alexander subbase theorem

I'm trying to understand a proof of Alexander subbase theorem. One thing that is not clear to me is an open cover $C$ without finite subcover which is maximal amongst such covers. Quote:

Use Zorn's Lemma to find an open cover $C$ without finite subcover that is maximal amongst such covers. That means that if $V$ is an open set of $X$ which is not in $C$, then $C \cup \{V\}$ has a finite subcover, necessarily of the form $\{V\} \cup C_V$ , where the choice of the finite subset $C_V$ of the cover $C$ depends on the picked additional set $V$.

Since the proof is made by contradiction, such maximal cover $C$ does not exist, and it is hard to imagine.

But let's consider a subspace $([0, 1), \tau)$ of $R$ with induced topology, which is not compact. Then how would $C$ be constructed in $[0, 1)$? As I understand $C = \tau \setminus \{U : (x, 1) \subseteq U \in \tau,\ \ x \in [0, 1)\}$ - all possible open sets that don't contain intervals $(x, 1)$. Or $C = \{U : (x, 1) \not\subseteq U \in \tau,\ \ x \in [0, 1)\}$ . Am I correct?

Could you please provide more details how to apply Zorn's Lemma here. It states that "if every chain is bounded ...". In case of open covers how can we show that every chain is bounded?

• As is usual in context of Zorn's lemma, the upper bound is just the union. Union of covers is a cover, and having a finite subcover is a finitary condition. – tomasz May 3 '18 at 22:01

It's a completely standard Zorn argument, but being pedantic:

First a general lemma/fact:

Claim: if any poset $(P, \le)$ if we have $n$ many elements in a chain $C \subseteq P$ then one of them is their maximum.

Proof of claim by induction on $n$: $n=1$ is trivial, $x$ is a maximum for $\{x\}$. Suppose the claim holds for $n \ge 1$ elements, then let $x_1,\ldots, x_n, x_{n+1}$ be $n+1$ many points from $C$. By the induction hypothesis there is some $i \in \{1,\ldots,n\}$ with $x_i = \max(x_1, \ldots, x_n)$. Then as $x_i$ and $x_{n+1}$ are in a chain, either $x_i \le x_{n+1}$ and then $x_{n+1} = \max(x_1, \ldots, x_n, x_{n+1})$ or $x_{n+1} \le x_i$ and then $x_i = \max(x_1, \ldots, x_n, x_{n+1})$. In either case we have found the maximum. This proves the induction step and thus the claim.

In the proof, the partial order is the set of all open covers of $X$ that have no finite subcover (call it an "bad" cover for short), ordered by inclusion (so that a maximal cover is one that we can add no open set to). The assumption is that this poset is non-empty (as in the proof we assume that $X$ is not compact, so there is at least one such bad cover), which is necessary for Zorn to apply (there can be no maximal element if there are no elements at all!).

If $\mathfrak{U}$ is a chain of bad covers, so if $\mathcal{U}_1, \mathcal{U}_2 \in \mathfrak{U}$ then $\mathcal{U}_1 \subseteq \mathcal{U}_2$ or $\mathcal{U}_2 \subseteq \mathcal{U}_1$, let $\mathcal{O} = \bigcup \mathfrak{U} = \{O \subseteq X: \exists \mathcal{U} \in \mathfrak{U}: O \in \mathcal{U}\}$.

Then clearly $\forall \mathcal{U} \in \mathfrak{U}: \mathcal{U} \subseteq \mathcal{O}$ so that $\mathcal{O}$ is an upperbound for $\mathfrak{U}$, if is indeed in the poset: is $\mathcal{O}$ a bad cover? Suppose not, then there are $O_1, \ldots, O_n \in \mathcal{O}$ such that $\cup_{i=1}^n O_i = X$. By the definition of $\mathcal{O}$ we thus have $\mathcal{U}_i \in \mathfrak{U}$ such that $O_i \in \mathcal{U}_i$, for $i=1,\ldots,n$.

The lemma then tells us there is some $m \in \{1, \ldots,n\}$ such that $\mathcal{U_i} \subseteq \mathcal{U}_m$. But then all $O_i$ are already in $\mathcal{U}_m$ which contradicts that $\mathcal{U}_m$ is a bad cover. So indeed $\mathcal{O}$ must also be bad and thus a member of the poset and a valid upperbound.

Now Zorn applies: we have a non-empty poset where every chain has an upperbound. So it has a maximal element: i.e. a bad cover such that no larger open cover can be bad (or it would contradict the maximality).

The last immediately implies that if $\mathcal{U}$ is such a maximal bad cover, and $O \notin \mathcal{U}$, $\{O\} \cup \mathcal{U}$ is no longer bad, so must have a finite subcover (which must include $O$, as otherwise the $\mathcal{U}$ wouldn't have been bad itself), so there are $U_1, \ldots, U_n \in \mathcal{U}$ such that $O \cup U_1 \cup \ldots U_n = X$.

The proof now goes on to reason about this weird maximal cover $\mathcal{U}$, to derive a final contradiction with the starting assumption that all subbasic open covers have a finite subcover. This last contradiction shows that the starting assumption (there are bad covers) must have been false.

Even in concrete spaces such a maximal bad cover will be hard to construct; these are non-constructive objects, closely related to ultrafilters on $X$.

• Thanks for the answer! You proved Zorn's Lemma for finite $n$. As I understand, the proof for the infinite sets requires the Axiom Of Choice. That's why the Zorn's Lemma is equivalent to the Axiom Of Choice? – Andreo May 6 '18 at 20:45
• im my concrete example about interval $[0, 1)$. The "bad" open cover can be constructed only "around" point 1 - as I understand? – Andreo May 6 '18 at 20:48
• @Andreo the Zorn Lemma is trivial for finite sets, indeed. You can prove AC from Zorn and vice versa (in ZF) that’s why they’re equivalent. – Henno Brandsma May 6 '18 at 21:02

If $C_1\subset C_2\subset\dots$ is a chain of covers without finite subcover, then $C=\bigcup C_i$ is a maximal element of this chain. Indeed, if $C$ had a finite subcover, then by finteness this would be contained in some $C_i$ for some sufficiently large $i$.