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Given $X$ metric space and $E$ is any subset of $X$.

If $x$ is a limit point of $E$, for every $\epsilon>0$, prove that neighborhood of x contains infinitely many element

I use the fact that when we have $x$ as a limit point of $E$, we can find a sequence on $E$ that converges to $x$, and because the sequence on $E$ and subset of neighborhood of $x$, we can conclude that neighborhood of $x$ contain infinitely many elements

Does my argument make sense? I hope you can tell me if I'm wrong guys

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  • $\begingroup$ What if the sequence converging to $x$ is just $x, x, x, \ldots$? $\endgroup$ – mihaild May 21 at 13:07
  • $\begingroup$ I mean it's not a trivial one since we can make any smaller neighborhood of x $\endgroup$ – Prastya Susanto May 21 at 13:26
  • $\begingroup$ Please don't rollback improvement. $\endgroup$ – YuiTo Cheng May 21 at 14:00
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Let $V$ be a neighborhood of $x$. Then there is a $r>0$ such that $B_r(x)\subset V$. Take $N\in\mathbb N$ such that $\frac1N<r$. For each $n\in\mathbb N$, take $x_n\in B_{\frac1{n+N}}(x)\setminus\{x\}$. Then the set $S=\{x_n\,|\,n\in\mathbb N\}$ must be an infinite set; otherwise, let $\varepsilon$ be the smallest distance from $x$ to an element of $S$. Then we would always have $d(x,x_n)\geqslant\varepsilon$, which is impossible, since $d(x,x_n)<\frac1{n+N}$. So, $S$ is an infinite set and $S\subset V$.

In your proof, you did not prove that your sequence has infinitely many distinct elements.

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  • $\begingroup$ Great explanation, thanks Mr! $\endgroup$ – Prastya Susanto May 21 at 13:33

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