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?


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.)

  • $\begingroup$ +1 "Does every open cover have an irredundant open refinement?" was going to be my next question! (Aside: I see the argument for the positive result because the Zorn's lemma approach goes through: point-finiteness shows that the poset of subcovers satisfies the ascending chain condition.) $\endgroup$ – Ben Blum-Smith Apr 26 '15 at 16:55
  • $\begingroup$ Actually I think I'm waving my hands my aside here. Let $X=\mathbb{N}$ with the discrete topology, let $\Lambda = \mathbb{N}\times \{0,1\}$, and let $U_{(n,i)} = \{n\}$ for both $i=0,1$. Let $\mathcal{F}_k = \{U_{n,1}\}_{n\geq k} \cup \{U_{n,0}\}_{n\geq 1}$. Then $\mathcal{F}_1\prec \mathcal{F}_2\prec\dots$ is a nonterminating ascending chain in the poset of subcovers. $\endgroup$ – Ben Blum-Smith May 2 '15 at 15:49
  • $\begingroup$ But the Zorn's lemma approach still goes through! What point-finiteness does is to force the poset of subcovers to be closed under taking intersections of chains, thus ascending chains have upper bounds. If $\mathcal{F}_1\prec\mathcal{F}_2\prec\dots$ is an ascending chain of subcovers (i.e. descending under inclusion), and the intersection is not a cover, there is an uncovered $x\in X$. Then the set of $k$ for which some set of $\mathcal{F}_K\setminus\mathcal{F}_{k+1}$ contains $x$ must be cofinal in $\mathbb{N}$. But this implies infinitely many sets of the cover contain $x$. $\endgroup$ – Ben Blum-Smith May 2 '15 at 16:01
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    $\begingroup$ @Ben: Yes, that’s the argument that I gave here and essentially the argument in Engelking. $\endgroup$ – Brian M. Scott May 2 '15 at 18:39

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:


$U_n = (-n,n)$.

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.


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 $\{U_{\lambda}\}_{\lambda\in\Lambda}$.


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