Existence of a least subcover of an open cover. If $\mathcal{O}$ is an open cover of $X$, does there exist an open cover $\mathcal{U}\subseteq \mathcal{O}$, such that, if $\mathcal{U}'\subsetneq \mathcal{U}$, $\mathcal{U}'$ does not cover  $X$?
For example, it is true for compact spaces. (Any open cover has a finite subcover. The finite subcover clearly has a minimal subcover.) It is not immediately clear whether in non-compact spaces we have minimal subcovers. (Or whether this is true at least in some spaces.)
 A: No, take the reals and the cover $\mathcal{O} = \{(-n,n): n \in \mathbb{N}\}$. Show that this has no minimal subcover: suppose $\mathcal{U} \subseteq \mathcal{O}$ is a subcover of $\mathcal{O}$. This means exactly that $N_1 = \{n: (-n,n) \in \mathcal{U}\}$ is unbounded above. Let $n_0 = \min(N_1)$ then $\mathcal{U'} = \{(-n,n): n \in N_1 \land n > n_0\}$ is still a cover and a proper subset of $\mathcal{U}$. 
It is true for point-finite covers, so covers where every $x \in X$ is in at most finitely many members of the cover. This is due to Arens and Dugundji (implicit in their paper "Remark on the concept of compactness"), where they show this as follows:
Suppose $\mathcal{U}$ is an open cover of $X$, which is point-finite.
Call a subset $\mathcal{R}$ of $\mathcal{U}$ "redundant" if $\mathcal{U}\setminus \mathcal{R}$ is still a cover of $X$. Order the redundant subsets of $\mathcal{U}$ by inclusion, then we want a maximal redundant subset (i.e., a redundant set such that no larger set is redundant), because then the other sets of $\mathcal{U}$ form a minimal subcover.
We're going to apply Zorn's lemma, so suppose that $\mathcal{R}_i, i \in I$ is some chain of redundant subsets. We claim that $\mathcal{R} = \cup_i \mathcal{R}_i$ is also redundant (and clearly an upperbound). Suppose not. Then there is some $x \in X$ that is not covered by $\mathcal{U}\setminus \mathcal{R}$. As we have a point-finite cover the set $\mathcal{U}_x = \{U \in \mathcal{U}: x \in U\}$ is finite. So $\mathcal{U}_x \subset \mathcal{R} = \cup_i \mathcal{R}_i$ . As the $\mathcal{R}_i$ form a chain (every 2 of them are $\subseteq$-comparable, so every finite subset has a maximal one) there is one $j \in I$ such that $\mathcal{U}_x \subseteq \mathcal{R}_j$. But then the latter set is not redundant, which is a contradiction. So the union $\mathcal{R}$ is redundant and Zorn applies. So $\mathcal{U}$ has minimal subcover in that case.
