convergent subsequence in $S$ versus in $\overline{S}$ I am reading equivalent definition of a precompact set $S\subset H$, where $H$ is a Banach space.
We have $S\subset H$ is precompact iff every sequence in $S$ has a convergent subsequence.
But shouldn't it be stated as iff every sequence in $\overline{S}$ (closure of $S$) has a convergent subsequence? Because precompactness requires $\overline{S}$ to be compact?
My hypothesis: Does it use the fact (which I don't know how to prove, or even whether it is ture or not) that every sequence in $S$ has convergent subsequence iff every sequence in $\overline{S}$ has convergent subsequence?
By the way, I assume 'convergence' here means converges to some element in $H.$ Then $S\subset \overline{S}$ implies one direction of my hypothesis, but the other direction remains unclear to me yet. 
 A: Since $H$ is a metric space, a subset $S$ of $H$ is compact if and only if it is sequentially compact. This means that $S$ is compact if and only if every sequence in $S$ has a convergent subsequence, where the limit is also in $S$.
Now, as you state, a subset $S$ of $H$ is said to be precompact if and only if $\overline{S}$ is compact. Thus $S$ is precompact iff every sequence in $\overline{S}$ has a convergent subsequence, where the limit is also in $\overline{S}$.
It turns out that the above is equivalent to the statement that every sequence in $S$ has a convergent subsequence, where the limit here is not assumed to be in $S$. Let's prove this:
($\implies$): Suppose $S$ is precompact and let $(x_n)$ be a sequence in $S$. Then $(x_n)$ is a sequence in $\overline{S}$, which is compact. Hence there exists a subsequence $(x_{n_k})$ of $(x_n)$ and a point $x\in\overline{S}$ such that $x_{n_k}\to x$.
($\impliedby$): Suppose every subsequence in $S$ has a convergent subsequence. We show that $\overline{S}$ is compact. To do so, we show that every sequence in $\overline{S}$ has a subsequence which converges to a point in $\overline{S}$. Let $(x_n)$ be a sequence in $\overline{S}$. Since every element in $\overline{S}$ is the limit of a sequence in $S$, for each $n$ there exists a sequence $(y_{n,m})_m$ in $S$ such that $y_{n,m}\to x_n$.
Thus, for each $n$, there exists a natural number $M_n$ such that
$$
\|y_{n,m}-x_n\| < 1/n, \quad \text{for all $m\ge M_n$}.
$$
Since the diagonal sequence $(y_{n,M_n})_n$ is a sequence in $S$, our hypothesis gives a subsequence $(y_{n_k,M_{n_k}})_k$ of it that converges to a point $y$.
We show that $x_{n_k}\to y$.
Let $\varepsilon>0$.
Choose a natural number $N$ such that $1/N<\varepsilon/2$ and
$$
\|y_{n_k,M_{n_k}}-y\| < \varepsilon/2, \quad \text{for all $k\ge N$}.
$$
Then if $k\ge N$, we have $n_k\ge k \ge N$ and hence
$$
\|x_{n_k}-y\|
\le \|x_{n_k}-y_{n_k,M_{n_k}}\|+\|y_{n_k,M_{n_k}}-y\|
< 1/n_k+\varepsilon/2
< \varepsilon.
$$
Since $\overline{S}$ is closed, we also have that $y\in\overline{S}$.
This completes the proof.
A: No, you're interested in finding out whether the set $S$ satisfies the property that every sequence in S has a convergent subsquence in S.
It may very well be that S does not satisfy the property (e.g. $S=]0;1]$ in $\mathbb{R}$ since $x_{n}=\frac{1}{n}$ does not converge in S) while obviously $\overline{S}$ does.
And therefore no, it did not mean convergence in $H$ but in $S$ (remember that compacity / precompacity / completeness / etc... are intrinsic properties of the metric subspace $S$).
