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This question comes up just from curiosity.

Let $B$ a Banach space and $A\subset B$ a topological subspace such that all the points of $A$ are isolated. Then, it is necessarily $A$ countable?

I assume that the answer is yes but I dont know how to do it (I tried build some proofs based in the radius of open balls but I failed). Some proof (or link to a proof) or counterexample will be appreciated, thank you.

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    If $a \in A$ then also $\epsilon a \in A$ for all $\epsilon > 0$... – 2017-02-08
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    @Alqatrkapa yes, fixed. I mean just a subset, a topological subspace, not a vector subspace. – 2017-02-08
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    I think saying just "subset" is clearest then. – 2017-02-08

2 Answers 2

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No, they are not. In $l^\infty$, the set of bounded sequences, with the standard topology, the set of sequences that just contain 0 and 1 is uncountable; and they are isolated.

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This problem is studied in more detail here.

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(Apologies for self-promotion.)