Then every neighbourhood $U$ of $x$ contains a point of $A$.
So I don't see it happening unless $X$ is a metric space, but the proof is for any topological space.
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Lemon
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What's your question? – bof Oct 23 '14 at 03:02
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what does $x_n\to x$ mean? – bof Oct 23 '14 at 03:02
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It means $x_n$ converges to $x$ – Lemon Oct 23 '14 at 03:03
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I am asking why is the greyed line true. – Lemon Oct 23 '14 at 03:03
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Yes, I know. What does "$x_n$ converges to $x$" mean? Definition? – bof Oct 23 '14 at 03:03
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1https://proofwiki.org/wiki/Definition:Convergent_Sequence/Topology – aram Oct 23 '14 at 03:04
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Duh, thanks. I know what convergence of a sequence means. It think for jip to write out the definition in words would be a good start to understanding why the greyed line is true. It is pretty much immediate from the definition. – bof Oct 23 '14 at 03:07
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@Aram, oh so there is no math, this is just definition... – Lemon Oct 23 '14 at 03:07
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1Definition are part of math. Understanding the definitions is a good start to understanding math. – bof Oct 23 '14 at 03:09
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Well no, I know it make sense if we put a metric on it, but kind of hard for me to swallow if it just in a general topological space. – Lemon Oct 23 '14 at 03:15
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That follows immediately from the definition of convergence of a sequence.
Let $X$ be any topological space, $\sigma=\langle x_k:k\in\Bbb N\rangle$ a sequence in $X$, and $x\in X$. Then by definition $\sigma$ converges to to $x$ if and only if for each nbhd $U$ of $x$ there is an $n_U\in\Bbb N$ such that $x_k\in U$ for all $k\ge n_U$.
In your case we’re assuming that $\langle x_n:n\in\Bbb N\rangle\to x$, so by definition every nbhd $U$ of $x$ not only contains at least one $x_n$, but actually contains all terms of the sequence from some point on. Since by hypothesis the sequence lies entirely in $A$, every nbhd of $x$ therefore must contain at least one point of $A$.
Brian M. Scott
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To be honest, there is an intermediate step which is that a point x is in thr adherence of A if and only if every nbhd of x has a non empty intersection with A. Once you've shown this the sequence lemma follows. – Statistic Dean Sep 27 '20 at 22:43
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