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.


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    $\begingroup$ 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. $\endgroup$ – Simon Jun 2 '18 at 16:45
  • $\begingroup$ Related to but not the same as compact spaces: math.stackexchange.com/questions/1833824/… $\endgroup$ – Davislor Jun 2 '18 at 19:45
  • $\begingroup$ @Simon: That's a kind of convoluted way of saying it. It is just that every finite partial order has a minimal element. $\endgroup$ – tomasz Jun 2 '18 at 21:01
  • $\begingroup$ 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. $\endgroup$ – Simon Jun 2 '18 at 21:24
  • $\begingroup$ 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... $\endgroup$ – Itai Jun 4 '18 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.

  • $\begingroup$ Or $S=\Bbb N$ and $C= \{[[n]] : n\in \Bbb N\}$ where $[[n]]=\{j\in \Bbb N: j\leq n\}.$... Or, in Set Theory, $S=C=\omega$. $\endgroup$ – DanielWainfleet Jun 24 '18 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.

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    $\begingroup$ $S=C=\omega$ also works. $\endgroup$ – chi Jun 3 '18 at 10:17

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