Characterization of the Closure of a Set Say I have a metric space $(X, d)$ and a set $A\in X$.  I want to prove that if $a \in \overline{A}$ then there is a convergent sequence $\{x_n\}$ that converges to $a$. Could I have any help?
 A: This depends on the definition of the closure you're using. In the case of metric space, you can actually define the closure $\bar{A}$ as the set of limits of sequences of elements of $A$.
If the definition you are using is the general definition (which works for topological spaces too), that is the set of points of which any neighbourhood meets $A$, then the proof that the two formulations are equivalent is the following :
First, suppose $x$ checks the first definition, and let $V$ be a neighbourhood of $A$, and $\epsilon$ such that $B(x,\epsilon) \subset V$. Since $x$ is the limit of a sequence $(x_n)_n$ in $A$, we may find a $N$ such that $x_n \in B(x,\epsilon) \subset V$ for all $n\geq N$.
Conversely, suppose that $x$ checks the second definition. Then chose $x_n \in A$ such that $x_n \in B(x, \frac{1}{n})$, it is then obvious that $(x_n)_n$ converges to $x$.
A: If $a \in \overline{A}$  (the closure of A), then there are two possibilities:


*

*$a \in A$ and it's an isolated point of $A$.

*$a$ is a limit point of $A$.


For the first case, consider the sequence whose terms are all $a$. Since $a \in A$, this sequence is in $A$. It's trivial to show that this sequence converges to $a$.
For the second case, consider a collection of neighborhoods of $a$ of the form: $B(a; \frac{1}{n})$ where $n \in \mathbb{N}$. Each element of this collection is an open ball centered at $a$ whose radius is $\frac{1}{n}$. Since $a$ is a limit point of $A$, for each neighborhood $B(a; \frac{1}{n})$ we can pick $x_n \in A$ for which $x_n \in B(a; \frac{1}{n})$. It's easy to show that $\{x_n\}_{n\in\mathbb{N}}$ is a sequence in $A$ that converges to $a$.
