In a metric space $X$, a subset $S \subset X$ is relatively sequentially compact if and only if its closure $\overline{S}$ is sequentially compact.

Question

Prove that in a metric space $X$, a subset $S \subset X$ is relatively sequentially compact if and only if its closure $\overline{S}$ is sequentially compact.

The terms relatively sequentially compact and sequentially compact are defined as follows for a general topological space $X$:

• A subset $S \subset X$ is sequentially compact if each sequence in $S$ has a convergent subsequence in $X$ with a limit in $S$.

• A subset $S \subset X$ is relatively sequentially compact if each sequence in $S$ has a convergent subsequence in $X$.

My attempt for proving the first implication is the following:

Suppose $S$ is relatively sequentially compact. Suppose $(x_n)_n \subset \overline{S}$ with $x_n \in S \subset \overline{S}$, for all $n$. Since $S$ is relatively sequentially compact and $\overline{S}$ is closed, the sequence $(x_n)_n$ has a convergent subsequence with limit in $\overline{S}$.

I have difficulties with completing the proof of this first implication. I think the converse implication is clear:

Suppose $\overline{S}$ is sequentially compact. Then, in particular, any sequence $(x_n)_n \subset S$ has a converging subsequence with limit in $\overline{S} \subset X$.

Any help, solutions or comments are highly appreciated.

Answer 1

The adherence $\bar S$ is the subset of limit points of $S$.

Suppose that $S$ is relatively sequentially compact and $(y_n)$ a sequence of elements of $\bar S$. There exists a sequence $(x^n_m\in S)$ which converges towards $y_n$. Choose $m(n)$ such that $d(y_n,x^n_{m(n)})<1/n$. You can extract a sequence $(x^{f(n)}_{m(f(n)}$ which converges towards $y\in \bar S$ For every $c>0$ there exists $N>0$ such that $n>N$ implies that $d(y,x^{f(n)}_{m(f(n))})<c/2$ and $1/n<c/2$. $d(y,y_{f(n)})\leq d(y,x^{f(n)}_{m(f(n))})+d(x^{f(n)}_{m(f(n))},y_{f(n)})\leq c/2+c/2=c$. We deduce that $(y_{f(n)})$ converges towards $y$.