Does every cover have an irredundant subcover?

Question

While composing an answer for this question, I got troubled by a technical point. I wanted to assert the existence of an irredundant subcover of a given open cover, but realized I'm not sure how to guarantee that this exists. I got around this technicality in that particular situation by using other aspects of the setup, but now I want to understand the general situation. I'll phrase the question in the language of topology, but it is really purely a set-theoretic question.

Let us call an open cover $\{U_\lambda\}_{\lambda\in\Lambda}$ irredundant if no $U_\lambda$ can be dropped, i.e. if $\{U_\lambda\}_{\lambda\in\Lambda'}$ is never a cover if $\Lambda'$ is a proper subset of $\Lambda$; equivalently, if for each $\lambda\in\Lambda$, $\bigcap_{\mu\neq \lambda} (U_\mu)^c$ is nonempty.

Given a topological space $X$ with an arbitrary open cover $\{U_\lambda\}_{\lambda\in\Lambda}$, does there always exist an irredundant subcover? Why?

Clearly there is no problem if $\Lambda$ is finite (or if $X$ is compact so that we can first pass to a finite subcover): just drop redundant $U_\lambda$'s one at a time until the cover is irredundant.

But what if $\Lambda$ has much bigger cardinality? At first I thought we ought to be able to construct an irredundant subcover using some appropriate form of the axiom of choice. After all, I thought, all we have to do is drop the redundant $U_\lambda$'s. (I.e. those such that $\bigcap_{\mu\neq \lambda}(U_\mu)^c$ is empty.) We can't drop them all at once, unfortunately, since we might not be left with a cover. But if we drop them one at a time, and reevaluate which are redundant after each drop, then, well, AC is designed for organizing an infinite sequence of choices of this kind.

For example we could try: by the well-ordering theorem, impose a well-order on $\Lambda$. Then the set of $\lambda$'s corresponding to redundant $U_\lambda$'s has a least element, and this is what we drop first. Rinse and repeat.

Or (this was my first idea), the setup seems custom made for Zorn's lemma: the subcovers of a given cover form a nonempty poset under reverse inclusion, and the irredundant covers are precisely the maximal elements of this poset.

The problem is that in the end, I don't actually think one can generally arrive at an irredundant cover, even when it exists, by dropping redundant $U_\lambda$'s one at a time. The example I have in mind is, let $X$ be any nonempty space at all, let $\Lambda = \mathbb{N}$, and let $U_1,U_2,\dots = X$. This example shows that both of the above approaches will fail. If we try to construct an irredundant subcover by repeatedly dropping the first redundant $U_n$, we get the sequence of subcovers

$$ \{U_n\}_{n\geq 1}\supset \{U_n\}_{n\geq 2}\supset \dots$$

and the procedure never lands on an irredundant one since the limit is empty, which isn't a cover. This very sequence also shows the hypotheses of Zorn's lemma aren't met (in the attempt above to use Zorn's lemma): here is an increasing chain in the poset of subcovers that does not have an upper bound.

This example seems to me to show that the AC-based approaches are barking up the wrong tree. On the other hand, evidently there is an irredundant cover here: any individual $U_n$ will do the job. So maybe there is still hope that the question's answer is yes?

Answer 1

As often happens when I write questions here, I realized something while writing. Here I actually realized the answer.

It is no.

Here is an open cover of $\mathbb{R}$ with no irredundant subcover:

$\Lambda=\mathbb{N}$.

$U_n = (-n,n)$.

Because $U_1\subset U_2\subset U_3\subset\dots$, given any pair of $U_n$'s, one is contained in the other. Therefore any subcover is redundant unless it consists of a single $U_n$. But no single $U_n$ forms a cover. Thus no subcover is irredundant.

Answer 2

As you’ve already discovered, the answer is no. In fact, you can’t even guarantee an irreducible open refinement. For $n\in\Bbb N$ let $U_n=\{k\in\Bbb N:k<n\}$, and let $\tau=\{U_n:n\in\Bbb N\}\cup\{\Bbb N\}$; then $\tau$ is a $T_0$ topology on $\Bbb N$, and $\tau\setminus\{\Bbb N\}$ is an open cover of $\Bbb N$ with no irreducible open refinement.

A well-known positive result is that every point-finite open cover of a space has an irreducible subcover. Thus, every open cover of a metacompact space has an irreducible open refinement. (A space $X$ is metacompact if every open cover of $X$ has a point-finite open refinement. A family of sets is point-finite if each point of the space lies in only finitely many of the sets.)

Answer 3

I'm not sure if it helps you with your original question, but let me try proposing an alternative concept capturing the intuitive notion of irredundancy: for each point, take the “smallest” open set in the cover containing that point.

Formally, let $\{U_{\lambda}\}_{\lambda\in\Lambda}$ be an open cover of the topological space $X$, where $\Lambda$ is a non-empty index set. Now endow $\Lambda$ with a well-ordering $\succsim$. For each point $x\in X$, let $\lambda_x$ be the least element of $\{\lambda\in\Lambda\,|\,x\in U_{\lambda}\}$ according to the well-ordering. Then, $\{U_{\lambda_x}\}_{x\in X}$ is a refined subcover of $\{U_{\lambda}\}_{\lambda\in\Lambda}$.