Jtdylktuy

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 ${x_gamma}_{gammainGamma}subseteq A$ with $x_gammarightarrow xin X$ then $xin A.$ I don't see how to use the hint that $mathcal{U}(x)={Uintau_1:xin U}$ form a directed set with $Uleq V$ implying $Vsubseteq U$. Any help pointing me in the right direction would be appreciated.

Attempt: Suppose that whenever ${x_gamma}_{gammainGamma}subseteq A$ with $x_gammarightarrow xin X$ then $xin A.$ For all $xin X_1$ s.t. $xnotin A$
(i.e. $xin A^C$) 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.

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.

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.