Munkres' leaves the following as an exercise:

Let $X$ be some topological space and $\mathcal{B}$ be a basis of $X$. If every cover of basis elements has a countable subcover, then any open cover has a countable subcover.

I have a proof and am hoping someone will critique it. Let $\mathcal{A}$ be some open cover of $X$, whose elements are indexed by the set $I$, and let $x \in X$. Then there exists $i \in I$ such that $x \in A_i$. Since $\mathcal{B}$ is a basis, there exists $B_i$ such that $x \in B_i \subseteq A_i$. Let $\mathcal{B}'$ be the collection of all such basis elements. By definition, $\mathcal{B}'$ covers $X$, and so by hypothesis there exists $J \subseteq I$ countable such that $\{B_i\}_{i \in J} \subseteq \mathcal{B}'$ covers $X$. It's clear that the subcollection $\mathcal{A}' = \{A_i \mid i \in J\}$ is a countable subcover of $X$, since if $x \in X$, there exists $x \in B_i$ and $B_i \subseteq A_i$.

As you may have noticed, I didn't explicitly define $\mathcal{B}'$ but merely described it as "the collection of all such basis elements." I was thinking it could be explicitly defined as $\mathcal{B}' = \{B_i \mid \exists A_i \in \mathcal{A} \mbox{ s.t. } B_i \subseteq A_i \}$, but this doesn't feel right for some reason. Perhaps it doesn't seem right because of set-theoretic subtleties of which I am not fully aware.

  • $\begingroup$ it looks good to me; to define $\mathcal{B}'$ more explicitly, you'd need to define $B_i$ as a function of $x$ (this would require invoking the axiom of choice, if you want to get really technical). What you suggest is not right for what you want it to say, because it doesn't refer to how the $B_i$ was chosen; hence, it loses the correspondence between the $B_i$'s and the $A_i$'s which allows you to define $\mathcal{A}'$. If you want to work with that definition, you have to start with it, show that it is a cover, and then define $A_i$ for each $B_i$ in its subcover. $\endgroup$ – Nick Pavlov Oct 22 '17 at 18:58
  • $\begingroup$ by the way, is this the exact wording in the book? I was a bit confused at first, because I would have said "cover by basis elements." What you wrote, I understood to mean "if every basis element is Lindelof" and only in reading your proof figured out what was meant. $\endgroup$ – Nick Pavlov Oct 22 '17 at 19:02
  • $\begingroup$ @NickPavlov Regarding your first comment, I figured there was some subtlety I was missing; as this is my first time learning topology (in fact, I'm learning it on my own), it would probably be best not to worry about such things at this state and just focus on the topological aspect. Regarding your second comment, no this isn't an exact quotation; I tried my best to capture what he was saying. Sorry about the confusion $\endgroup$ – user193319 Oct 22 '17 at 19:39

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