I've gone though this post and I don't really understand the statement: "A-convergence implies B-convergence, then B-closed implies A-closed"
If $\tau_1\subset \tau_2$, where $\tau_1$ is a topology weaker than $\tau_2$ in some space $X$, then a closed set in $(X, \tau_1)$ is closed in $(X,\tau_2)$
Why is this true? I can't seem to wrap my head around this.
On
If $\tau_1 \subseteq \tau_2$ then $\tau_2$-convergent implies $\tau_1$-convergent, because convergence (sequences/nets/filters) is always defined as "bla converges to $x$" iff "for all open sets containing $x$, something about bla" so if we can do that for all sets in $\tau_2$, we certainly can for $\tau_1$ too.
And $\tau_2$-convergent implies $\tau_1$-convergent means "$\tau_1$-closed implies $\tau_2$-closed": suppose $F$ is $\tau_1$ closed, then suppose a "bla" in $F$ $\tau_2$-converges to $x$, then "bla" $\tau_1$-converges to $x$ so $x \in F$ as $F$ is $\tau_1$-closed, so $F$ is $\tau_2$-closed (which is usually characterised as "for every "bla" on the $F$ that converges to some $x \in X$, $x \in F$".
But of course the direct route is much easier, go via complements. But the convergence "reversal" is good to realise too.
Let $Y\subset X$ be closed in $(X, \tau_1 )$. Then $Y^\complement \in \tau_1$, which implies $Y^\complement \in \tau_2$, so $Y=(Y^\complement)^\complement$ is closed in $(X, \tau_2)$