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.
 A: 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.
A: 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.
