The full problem is as follows:
Let $(X,d)$ be a metric space and $A\subseteq X$. Let $E$ be the set of all $p\in X$ for which there is a sequence $\left \{ p_n \right \}$ with $p_n \in A$ for each $n \in\mathbb{N}$ and $lim_{n\rightarrow \infty }(p_n) = p$. Show that $E$ is the closure of $A$.
My thoughts:
To show that E is the closure of A, I have to show that E contains all limit points of A. But, the way the problem defines $E$, to me it is already said that the limits of arbitrary sequence $p_n$ reside in $E$! So what exactly is there left to prove here? Perhaps I am misunderstanding the goal of this exercise or reading it improperly.
Just need a hint to start. Thanks!
If $x \in E$, then by assumption $x$ is the limit of some sequence in $A$, so $A \cap U$ is nonempty for all neighborhoods $U$ of $x$; so $E \subset$ the closure of $A$. If $x \in$ the closure of $A$, then $A \cap U$ is nonempty for all neighborhoods $U$ of $x$. So for every integer $n \geq 1$ there is some $p_{n} \in A \cap B(x,n^{-1})$. Then $(p_{n})$ by construction is a sequence in $A$ that converges to $x$, so $x \in E$. We are done here.