Equivalence of definitions of a closed set

Question

So I am working through some problems on my own and ran into one I have a question about. It is as follows. Let $(X_1,\tau_1)$ and $(X_2,\tau_2)$ be topological spaces, show that a set is closed iff whenever $\{x_\gamma\}_{\gamma\in\Gamma}\subseteq A$ with $x_\gamma\rightarrow x\in X$ then $x\in A.$ I don't see how to use the hint that $\mathcal{U}(x)=\{U\in\tau_1:x\in U\}$ form a directed set with $U\leq V$ implying $V\subseteq U$. Any help pointing me in the right direction would be appreciated.

Attempt: Suppose that whenever $\{x_\gamma\}_{\gamma\in\Gamma}\subseteq A$ with $x_\gamma\rightarrow x\in X$ then $x\in A.$ For all $x\in X_1$ s.t. $x\not\in A$ (i.e. $x\in A^C$) we have that $x$ is not a limit point of $A$. Hence, $A^C$ is a neighborhood of $x$. Now we only need that $A^C$ is open, which is true since it is a neighborhood of all its points as $x$ was arbitrary. This should give us the first direction. I don't see how to get the other direction based using the provided hint.

Answer 1

Assume A closed. Let n be a net into A that converges to a.
To show a in A use the theorem for closed A
x in A iff for all open U nhood x, U $\cap$ A not empty
and some facts about convergence of nets to show x in A.

Conversely, to prove A is closed, show
if for all open U nhood x, U $\cap$ A not empty, then x in A.
So assume for all open U nhood x, U $\cap$ A not empty.
Use the hint to construct a net into A that converges to x
and with that conclude x in A.

Answer 2

You define $C$ to be closed if it’s complement is open. But the net convergence equivalent is easier proved using the following equivalent definition:

$C$ is closed iff for all $x \in X$: if every open set $O$ that contains $x$ intersects $C$, then $x \in C$.

Given this, suppose $C$ obeys the net convergence condition. Suppose that $x$ is a point such that every open neighbourhood of $x$ intersects $C$. Then we can apply the hint: define $\Gamma$ be the directed set $\mathcal{U}(x)$ under reverse inclusion, and pick (by the assumption on $x$ this is possible) a point $x(U) \in C \cap U$ for every $U \in \mathcal{U}(x)$. Then this defined a net: $x: \Gamma \to X$ where all values lie in $C$ by construction and also $x_\gamma \to x$: let $O$ be any open subset that contains $x$ and note that $O \in \Gamma$ and if $U \ge O$ we have that $U \subseteq O$ and so $x(U) \in U \subseteq O$, so $x(U) \in O$. So this shows convergence. By the net convergence condition we can thus conclude $x \in C$ and we are done with one direction.

Suppose $C$ is closed and let $x: \Gamma \to X$ be any net with values in $C$ such that it converges to $p \in X$. Then every open neighbourhood of $p$ contains a point of the net (by convergence to $p$) so in particular a point of $C$. By the equivalent formulation of closedness we see that $p \in C$ and so $C$ obeys the net convergence condition.