# metric space and limit point

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

• What if the sequence converging to $x$ is just $x, x, x, \ldots$? – mihaild May 21 at 13:07
• I mean it's not a trivial one since we can make any smaller neighborhood of x – Prastya Susanto May 21 at 13:26
• Please don't rollback improvement. – YuiTo Cheng May 21 at 14:00

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 $$\frac1N. For each $$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$$.