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The problem states: Let $X$ be a completely regular space such that $|X|<c=|\mathbb{R}|$. Prove that there exists a basis $\mathcal{B}$ for the topology on $X$ such that for all $B\in\mathcal{B}$, $B$ is both open and closed.

I let $\mathcal{B}$ be the collection of all clopen sets since any other choice for $\mathcal{B}$ would yield a coarser topology. It is easy to verify that $\mathcal{B}$ forms a valid basis. Let $\tau$ be the topology on $X$ and let $\tau_\mathcal{B}$ be the topology generated by $\mathcal{B}$. Then clearly $\tau_\mathcal{B}\subset\tau$. I couldn't figure out how to show that $\tau_\mathcal{B}=\tau$, so I assumed it wasn't and tried coming to a contradiction. If $\tau_\mathcal{B}\neq\tau$, then that means there must exist an open set $U\in\tau$ and a point $x\in U$ such that every clopen set containing $x$ must intersect $X\setminus U$. I thought that maybe using this along with the completely regular condition, I could prove the existence of some surjective function $f:X\to[0,1]$, thus contradicting the cardinality restriction, but I couldn't find any way of doing that. Any ideas?

Anonymous
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    This question isn't a duplicate because in the other question the space was countable, this version is a stronger version. – G. Ottaviano Jul 16 '18 at 19:14
  • That's absolutely right and that's what first came to my mind as well. However, the answer/hint given to the other question gives a solution to this one as well, so I allowed it to be marked as a duplicate. – Anonymous Jul 17 '18 at 01:58

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As you say, let us take all Clopen sets in our topology. To show that it is a base, we need to show that for every open set $O$ and $p \in O$ there is a member of the base $M$ with $p \in M \subseteq O$. Now how can we find $M$? Well because the space is completely regular, we have a function $f$ with the image of $p$ being $0$ and the image of $O^{c}$ (a closed set that does not contain $p$) being $1$. Now what is the inverse image of $1$? It will be a superset of $O^{c}$. It will also be closed (as the inverse image of a closed set via a continuous function) and so its complement will be open, will contain $p$ (because $p$ maps to $0$) and will be a subset of $O$. Now can we see that this set if also closed? We must use the cardinality condition for this. OK so let $I$ be the image of $X$ under $f$. This cannot contain any open intervals (because if it did then $X$ would have cardinality greater than or equal to $|R|$. Hence every point of $I$ is open and closed in the subspace topology (of $I$ derived from $R$) and so the inverse image of $1$ must be open and closed, its complement must be open and closed and the set of Clopen sets is a basis for $X$.

Simon Terrington
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    OK this wouldn't quite work as it would contain $p$. But you are really just asking why not take the inverse image of a small open interval around $1$. I think the issue is that it is not immediately obvious that this would be closed (it would clearly be open!) and we need both. I think it would be closed but we would need to prove that. – Simon Terrington Jul 16 '18 at 07:24
  • Yea, it's not necessarily the case that every point in $I$ will be both open and closed since $I$ could be, for example, $\mathbb{Q}\cap[0,1]$. I think the key part is that there must exist some $a\in(0,1)\setminus I$ so that $f^{-1}([0,a))=f^{-1}([0,a])$ is a closed and open set containing $p$ inside of $M$. Thanks for the help! – Anonymous Jul 16 '18 at 07:49
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    Yes, that's a good point about taking all the rationals in the range $[0,1]$. You are right to say that none of those points are open and closed. They are all close right? :) but not open. I'll have a think about your other points; I'm too tired at the moment. – Simon Terrington Jul 16 '18 at 21:02