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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.

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closed as off-topic by Math1000, Saad, mrtaurho, Shailesh, Parcly Taxel Mar 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.

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  • $\begingroup$ Thanks. Why does $U$ must exist? $\endgroup$ – Jake Mar 7 at 3:07
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    $\begingroup$ 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 $. $\endgroup$ – Chris Custer Mar 7 at 3:21
  • $\begingroup$ And I'm not entirely clear on why $U$ is open.. $\endgroup$ – Jake Mar 8 at 1:03
  • $\begingroup$ There has to be at least one such open $U$, or else we arrive at the contradiction outlined above. $\endgroup$ – Chris Custer Mar 8 at 1:22

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