Closed Set in Coarser Topology is Closed in Finer Topology

Theorem

Let $X$ be a set.

Let $\tau_1$ and $\tau_2$ be topologies on $X$ such that $\tau_1 \subseteq \tau_2$.

That is, $\tau_1$ is coarser than $\tau_2$.

Let $C$ be a closed set in $\struct {X, \tau_1}$.

Then $C$ is closed in $\struct {X, \tau_2}$.

Let $C$ be a closed set in $\struct {X, \tau_1}$.

Then by definition of closed set:

$X \setminus C$ is open in $\struct {X, \tau_1}$.

Hence by definition of open set:

$X \setminus C \in \tau_1$

By definition of subset:

$X \setminus C \in \tau_2$

So by definition of open set:

$X \setminus C$ is open in $\struct {X, \tau_2}$.

So $C$ is closed in $\struct {X, \tau_2}$ by definition.

$\blacksquare$