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I am currently studying Topology by working through my lecture notes. During the lecture our Professor gave a remark which states

Let $(X,\mathcal{O})$ be a topological space. If for all $x\in X$ there is a countable family $\mathcal{O}_X\subset\mathcal{O}$ with $U\in\mathcal{O}_x\implies x\in U,\ \forall V\in \mathcal{O},x\in V \exists U\in\mathcal{O}_x: U\subset V$ then an accumulation point implies a convergent subsequence.

As the lecture was not very well structured I am not completly sure if this is even correct. Does anybody of You know that statement to be true? And if so might have a hint on how to proof it? Either way I would be very glad.

Edit: Changed "subseries" to "subsequence"

lyding
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  • What do you mean by "accumulation point implies a convergent subseries"? First of all, an implication "X implies Y" only makes sense when X and Y are statements but neither "accumulation point" nor "a convergent subseries" is a statement. Second of all, a (sub)series only makes sense when there is at least a commutative group structure on $X$ which you have not mentioned. – Adayah Sep 13 '22 at 17:02

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Given a point $x$, a family $\mathcal{O}_x$ with the properties you've described is called a "local base at $x$". A space where at every point there is a countable local base is called a "first countable space". So what you're asking is: Is it true that in a first countable space, every accumulation point of a set is a limit of a sequence from that set? (At least, I think that's the question: "subseries" is a bit of a strange term to use in this context, as "series" usually refers to infinite summation, which doesn't seem to be what is meant).

It is true; for instance, see Accumulation points of sequences as limits of subsequences?.

Chris Eagle
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  • Thanks for the reply, "series" was a tyoo. German is my mother tongue and I frequently mix up sequence and series. I will fix it. – lyding Sep 15 '22 at 12:02