Suppose we have a metric space $(X,d)$
We say that $Y\subset X$ is pre compact if every sequence in $Y$ has a convergent subsequence. Notice that generally the subsequence might not converge in $Y$.
Now suppose that $Y$ is a non-closed and pre compact subset of $X$. Is it then true in general metric spaces that $cl(Y)$ must be a compact subset of $X$ ? If not, are there any naturally assumptions that can be added to obtain the implication?
My thought was that we might have a sequence $$\left\{y_{n}\right\}_{n=1}^{\infty}\subseteq cl(Y)\setminus Y$$ with no convergent subsequence.
Let $y_n \in \bar Y$. Then by definition of $\bar Y$, there are $x_n \in Y$ so that $d(x_n , y_n) < \frac 1n$. Then as $Y$ is precompact, there is $x\in X$ so that $x_n \to x$. Thus $x\in \bar Y$. Also $y_n \to x$. Thus $\bar Y$ is compact.