Closure of a subset of a normed space is equal to the set of limits of sequences in $A$ I'm not sure how to show that the closure of a subset $A$ of a normed space $\mathbb{X}$ is equal to the set of limits of sequences in $A$. Could someone help me?
 A: Let $\overline{A}$ denote the closure of $A$ in $(X,\Vert\cdot\Vert)$. We have to prove that $x\in \overline{A}$ if and only if there exists a sequence $\{x_{n}\}_{n}\subset A$ converging to $x$.
Suppose that $x\in\overline{A}$. By definition, it means that any open subset containing $x$ also contains a point of $A$. Let $B_{n}=\{y\in X\mid \Vert x-y\Vert<\frac{1}{n}\}$. As any $B_{n}$ is open (it is an open ball) and contains $x$, it contains a point of $A$. We construct a sequence $\{x_{n}\}_{n}$ of points in $A$ by subsequently taking a point in $B_{n}\cap A$.
Now, suppose that there exists a sequence $\{x_{n}\}_{n}\subset A$ converging to some $x\in X$. By definition, it means that 
$$\forall\epsilon>0,\exists N_{\epsilon}\in\mathbb{N}:\forall n\geq N_{\epsilon}:\Vert x_{n}-x \Vert<\epsilon\tag{1}$$
For any point $x$ contained in an open subset $U$ in a normed space (which is, in particular, a metric space), there exists an open ball centered in this point $x$ and contained in $U$. Here, if this ball is of radius $R>0$, you know by $(1)$ that there exists $x_{N_{R}}$ (since $R>0$) in this ball and, by construction, $x_{N_{R}}\in A$, so that $x_{N_{R}}\in U\cap A$.
Hence, $x\in\overline{A}$.
A: As a base of neighborhoods of $x$ is given by the open balls $B(x,r),$ you have that $$x\in \overline{A}\iff\forall r>0,B(x,r)\cap A\neq\varnothing.$$
Choosing a such element for all $r=\frac{1}{n},$ $n\in\mathbb{N}^*,$ you will get your sequence. 
For the other part, John Ma has given you the idea.
