Let $(X,\mathcal T)$ a topological space. I know that if $\mathcal T$ is metrizable, and $E\subset X$, then $E$ is closed $\iff$ $E$ is sequentially. I was wondering if this is true on every topological spaces or in metrizable space only ?
I guess that if $E$ is closed, then it's sequentially closed. Indeed, let $(x_n)$ a sequence of $E$ that converge in $E$ and denote $x$ it's limit. Then we can easily show that $x\in \bar E=E$ where the last equality hold because $E$ is closed.
For the converse, I suppose that $E$ is not closed, i.e. there is $x\in \bar E$ s.t. $x\notin E$. I know that for all open set that contain $x$, we have that $U\cap E\neq \emptyset$. But I can't construct a sequence that converge to $x$. I also know that if $X$ is not metrizable, then there is $x\in \bar E$ s.t. no sequence from $E$ converge to $x$. But unfortunately, this doesn't contradict the fact that $E$ is sequentially closed.
So may be this property is not true ? If it's not true, what would be the "weakest" condition on the topology to make this property true ? Hausdorff maybe ?