Metric Spaces: closure of a set is the set of all limits of sequences in that set I am studying metric spaces and got confused about many different ways of defining the closure.
Let $S$ be a subset of $M.$ Then, the closure of $S$ is $ \{x \in M : \forall \epsilon>0, \ \  B(x,\epsilon) \cap S \neq \emptyset \}$
Also, the lecturer mentioned something about converging sequences. Is it true that the closure of $S$ is  $\{x \in M: \text{there exist a sequence} \ (x_n) \in S \ \text{such that} \  d(x_n,x) \rightarrow 0 \ \text{as} \ n \rightarrow \infty\}$
Is this true? and do we need $x_n \neq x$ in the above definition? I remember learning in real analysis that $p$ is a limit point if and only if there exist a sequence $x_n$ converging to $p$ but $x_n \neq p$. Then, since closure includes points that are intuitively 'in' the set and not the limit points, we don't need the condition that $x_n \neq x$.
Have I understood this correctly?
Thanks
 A: The property of a point $x$ that

for all $\varepsilon>0$, the intersection $B(x,\varepsilon)\cap S$ is non-empty

is equivalent to

every neighborhood $U$ of $x$ intersects $S$

since $B(x,\varepsilon)$ is a neighborhood, and every neighborhood $U$ contains a ball $B(x,\varepsilon)$ for some $\varepsilon>0$. These characterizations define $x$ as an adherent point, and one way to define the closure $\overline S$ of a set $S$ is as the set of all adherent points of $S$.
In a metric space $M$ a point $x$ is an adherent point of $S$ if and only if there is a sequence $(x_n)_n$ in $S$ converging to $x$. The if-part is trivial. For the only if-part, we can construct a sequence in $S$ by choosing a point $x_n$ in every ball $B(x,1/n)$. So in a metric space the closure is indeed the set of all limits of sequences in $S$.
Be careful not to confuse the terms limit of a sequence and limit point of a set. A limit point of a set $S$ in a space $X$ is a point $x$ such that every neighborhood $U$ of $x$ contains a point of $S$ distinct from $x$, that is, $U\cap S\setminus\{x\}\ne \emptyset$. In that case, if $X$ is metric, one can construct a sequence $(x_n)_n\subseteq S$ converging to $x$ such that $x_n\ne x$ for each $n$.
