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.
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.