# Does every cover of a set have a minimal subcover?

I imagine this problem is a common one, however without having any source to refer to I don't know its usual name, and am having trouble finding an answer.

I am taking the definition: A cover $C$ of a set $S$ is a set such that $\cup C = S$

I want to know if every cover has a minimal subcover.

Thanks.

• If C is finite then the answer is yes, by considering the power set of C with its usual partial order (by inclusion) and partitioning that power set into those collections which cover S and those which don't. But then there may be more than one minimal subcover. – Simon Jun 2 at 16:45
• Related to but not the same as compact spaces: math.stackexchange.com/questions/1833824/… – Davislor Jun 2 at 19:45
• @Simon: That's a kind of convoluted way of saying it. It is just that every finite partial order has a minimal element. – tomasz Jun 2 at 21:01
• You are right - we could transform my explanation into yours by deleting mention of those collections which fail to cover S, and pointing out that inclusion is a partial order on those collections which do cover S. – Simon Jun 2 at 21:24
• It probably depends on your definition of "minimal", but assuming the axiom of choice, I suspect there always is a minimal subcover w.r.t. inclusion (i.e. using Zorn's lemma). Edit: on second thought, not so sure, as probably not every chain has an upper (lower) bound... – Itai Jun 4 at 7:22

How about $S=\Bbb R$ and the cover composed of the intervals $(-n,n)$? Any subcover of this cover remains a subcover if you omit one of its elements.
• Or $S=\Bbb N$ and $C= \{[[n]] : n\in \Bbb N\}$ where $[[n]]=\{j\in \Bbb N: j\leq n\}.$... Or, in Set Theory, $S=C=\omega$. – DanielWainfleet Jun 24 at 8:57
Consider $S=\aleph_\omega$ and $C=\{\aleph_n:n <\omega\}$. Any infinite subset of $C$ is a cover, and therefore there is no minimal cover.
(As I was writing this, another answer was posted. Note that although the spaces $S$ in both cases are different, the idea is the same: $S$ is covered by a countable increasing family of sets, so that any infinite subfamily is again a cover.)
For a different example, any limit ordinal $\alpha$ (of any cofinality) works as $S$, with $C$ any cofinal subset. A subset of $C$ is a cover if and only if it is itself cofinal. Thus, there is no minimal subcover, and any subfamily of $C$ that works has size at least $\operatorname{cf}(\alpha)$. More general still, we can take as $S$ any linearly ordered set without a maximum, pick a cofinal sequence $a_\tau$ of members of $S$, and let $C$ consist of the intervals $(-\infty,a_\tau)$. The example in the other answer is essentially this one, taking advantage that both the coinitiality and cofinality of $\mathbb R$ as an ordered set coincide. It would be interesting to see an essentially different source of examples.
• $S=C=\omega$ also works. – chi Jun 3 at 10:17