Closedness of a set by using convergent nets I am currently studying about nets, so it is all new to me. There is one thing that I could not find anywhere, so I try ask here. (If you know a reference, please let me know.)
Let $X$ be a topological space, and let $A\subseteq X$. Is it true that $A$ is closed if (and only if) for every net $(x_i)_{i\in I}$ in $A$ is convergent in $A$? This certainly does if we were working with sequences. I am mainly interested in the "if" part.
What I know is that

Suppose $x\in X$. Then $x\in\overline{A}$ if and only if there is a net in $A$ that converges to $x$.

If $M$ is a collection of limit points of every net in $A$, then $M\subseteq \overline{A}$. How to see that $A$ is closed here?
 A: Your quoted statement tells you that if $x$ is the limit of a net in $A$, then $x\in\operatorname{cl}A$. Thus, if every limit of a net in $A$ is a point of $A$, then $\operatorname{cl}A\subseteq A$, and hence $A=\operatorname{cl}A$, and $A$ is therefore closed.
A: Let $x\in \overline A$ and consider the net $\{x_U\}_{U\in I}$ where  $x_U\in A\cap U,$ and $U$ is a neighborhood of $x,$, belonging to the directed set of neighborhoods of $x$ ordered by reverse inclusion. Then, $(x_U)\to x$ and so $x\in A$ by hypothesis, so $A$ is closed.
A: What is not true is that every net in $A$ converges, but that if some net from $A$ converges to some $x \in X$, we know that $x \in A$.
Put otherwise: a convergent net with terms in $A$ cannot have a limit outside of $A$. For nets, this is an iff: if $A$ is closed it holds (consider $X\setminus A$ which is open etc.) and if it holds for all nets (which is hard to check, as there are many such nets with terms in $A$ ..) then $A$ must be closed.
For sequences we only have one direction ($A$ closed then no sequence from $A$ can converge to an $x$ outside of $A$) but not always the other way around. We do have this converse in metric spaces, and other first countable spaces, where sequences suffice to describe the topology.
