Vtyjkyuk

Let $(X, tau)$ be a topological space, and let $overline (-) : mathcalP(X) to mathcalP(X)$ be defined as

$$
overlineS := x : exists(x_alpha)_alpha in Lambda subseteq S text net s.t. x_alpha to x.
$$

I'm trying to prove that $overlineA cup B subseteq overlineA cup overlineB$, since I've already seen the other inclusion.

Let $x in overlineA cup B$ so that we have $(x_alpha)_alpha in Lambda subseteq A cup B$ with $x_alpha to x$. My idea was to adapt a similar proof for sequences, where one can assume there are infinite terms of a sequence in one of the sets, and take an appropriate subsequence. I have first proved that $x in A$ or $x in B$ for $Lambda$ finite, and now am tackling the infinite case.

Concretely, now we can assume that infinitely elements of (the image of) the net are on $A$. I want to prove that $Gamma = alpha in Lambda : x_alpha in A $ and the inclusion $Gamma hookrightarrow Lambda$ will give a subnet of the original, thus concluding that $x in overlineA$. However, that does not seem immediate (or correct, even) if we do not assume $Gamma$ cofinal, which will not happen always.

Would you mind providing any hints on whether I am on the right track, and if so on how can I finish my argument? Thanks in advance.

The best idea is to assume $x in overlineA cup B$ giving us a net $n: Lambda to X$ such that $n to x$ and for all $alpha in Lambda$ we have $n(alpha) in A cup B$, and then define $Lambda(A) = alpha in Lambda: n(alpha) in A$ and likewise $Lambda(B)$ and by definition we have that $Lambda = Lambda(A) cup Lambda(B)$, (not disjoint necessarily, of course, but that's OK).

Can it be true that neither is cofinal in $Lambda$? (Hint : no) and then you're done (a cofinal subset gives another (sub)net using inclusion, as you suggested ) What does it mean to be not cofinal?? Get a contradiction from the definition of a directed set.