Relations between convergence in nets and topologies. I want to prove that given a net $S$ and topologies $T$ and $T'$, then $T\subset T'\iff$ when $S$ is convergent in $T'$ is also convergent in $T$.
I'm proceeding this way: First I'd like to show that if $T\subset T'$ then if $S\to x$ in $T'$ it happens in $T$ as well, which means that I must show that for every $U$ that satisfies $x\in U\subset T\exists N$ such that $S(n)\in U'$ when $n>N$.
Now, take $U$ as before, since $T'$ is finer than $T$, this means that $\exists U'\subset T'$ such that $x\in U'\subset U$, and since $S$ converge in $T'$ we know that there is $N$ for which $S(n)\in U'$ when $n\geq N$, we conclude by means of the former inclusion that $S(n)\in U$ when $n\geq N$.
Proving the reciprocal: Seems immediate, but is it?. $S\to x$ in $T'$ implies that $S\to x$ in $T$. Let $B,B'$ be basis for $T,T'$ and take $U_B$ such that $x\in U_B\subset B$. Given that $S(n)\in U'_B$ for $n\geq N$ for some $N$ implies that $S(n)\in U_B$, then follows that $U'_B\subset U_b$. Then there is $U'_b$ such that $x\in U'_b\subset U_b$, which means that $T'$ is finer than $T$.
Is this proof right?.
 A: The first part can be shorter: assume the net $(x_s)_{s \in S}$ in $X$ converges to $x$ in $(X,T')$ and $T \subset T'$ is a coarser topology. Let $O$ be an open neighbourhood of $x$ in $(X,T)$. Then $O \in T'$ as well, and so there exists $s_0 \in S$ such that for all $s \ge s_0$ we have that $x_s \in O$, by the definition of convergence of the net in $(X,T')$. So the net converges to $x$ in $(X,T)$ as well.
There is no need to state the definition twice, and you just start with an arbitrary open set from $T$ that contains $x$, and show the net is eventually in it.
The reverse is really a bit more tricky. I think we need first to refine the statement a bit. The right hand side should say, I think, "When a net $S$ in $X$ converges in $T'$ to $x$, then $S$ converges to the same $x$ in $T$ as well". This is in fact what was shown in the short proof above.
It's slightly more convenient to work with closed sets, so I'll use this lemma: in a topological space $(X,T)$, $C$ is closed iff for every net with points from $C$ that converges in $(X,T)$ to some $x \in X$, $x \in C$ as well.
Now suppose that $C$ is closed in $(X,T)$. We want to show that $C$ is closed in $(X,T')$ as well, so (using the lemma), suppose we have a net $S$ from $C$ that converges in $(X,T')$ to some $x$ in $X$. Then the assumption on net convergence (the refined version) that we start with, gives us that $S$ also converges to $x$ in $(X,T)$. As $C$ is closed in $(X,T)$ and the net is from $C$, we know that $x \in C$, as required.
OTOH, if $X = \{0,1\}$ in the indiscrete and the Sierpiński topology, then in both topologies all nets converge (but not always to the same points), so topologies can be different but have the same convergent nets. Still the class of pairs $(n,x)$ (where $n$ a a net on $X$ converging to $x \in X$ under $T$) is different for distinct topologies.
