# Show that a set of all limit points of subsequences is closed [closed]

Assume $$(x_n)_{n\in\mathbb{N}}\in\mathbb{R}^\mathbb{N}$$ and consider the set: $$E:=\{x\in\mathbb{R}, \text{there exists a subsequence of (a_n)_{n\in\mathbb{N}} converging to x.}\}$$ Show that $$E$$ is closed.

I can think about three situations, two of which are easy to prove. The first, $$E$$ is empty. The second, $$E$$ has one element. The third, $$E$$ has more than one element. I'm struggling to prove the third.

I think that I might be misunderstanding what the set E is.

## closed as off-topic by Math1000, Saad, mrtaurho, Shailesh, Parcly TaxelMar 10 at 3:39

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Assuming $$x\not\in E$$, there must be an open set $$U$$ with no element of $$(x_n)$$ contained in $$U$$. That is, $$U\cap (x_n)=\emptyset$$. Hence $$U\cap E=\emptyset$$. Hence $$\Bbb R\setminus E$$ is open.
• Thanks. Why does $U$ must exist? – Jake Mar 7 at 3:07
• Well, if not then for every $U_n=(x-\frac1n,x+\frac1n)$, we have an element $a_n$ of $(x_n)$ with $a_n\in U_n$. So $(a_n)$ is a subsequence of $(x_n)$ with $a_n\to x$. So $x\in E\,\Rightarrow \Leftarrow$. – Chris Custer Mar 7 at 3:21
• And I'm not entirely clear on why $U$ is open.. – Jake Mar 8 at 1:03
• There has to be at least one such open $U$, or else we arrive at the contradiction outlined above. – Chris Custer Mar 8 at 1:22