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?.

  • $\begingroup$ did not analyse whole proof, but it seems good (to me) $\endgroup$ – Nikos M. Jun 28 '14 at 3:22

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.


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