# Can fineness of topology be decided from closed sets?

Suppose $X$ be a set and $\tau,\tau'$ be two topologies on $X$ and it is known that the two topologies are comparable.

Now, wlg, let's say we manage to prove that any open set $U$ of $\tau$ is open in $\tau'$ as well then we have proved that $\tau'$ is finer than $\tau.$

Now is there any such thing we can do with the closed sets,since closed sets are simply the complements of the open sets.Try this:

Suppose $\tau\subset \tau'$. Let $C$ be any closed set in $\tau$ then $X\backslash C=U$ is open in $\tau$ and hence in $\tau'.$ Then $C$ being the complement of an open set,is closed in $\tau'$ as well.

Conversely,suppose every closed set of $\tau$ is closed in $\tau'$.Let $U$ be any open set in $\tau.$ Then $C=X\backslash U$ is closed in $\tau.$ And since it is closed in $\tau'$ as well,so $U$ is open in $\tau'$. Thus $\tau\subset \tau'$.

From this we can say, Every closed set of $\tau$ is closed in $\tau'$ iff $\tau \subset \tau'$

My idea is correct or is there any delicate mistake in my concept? Thank you.