Identity Mapping to Expansion is Closed

Theorem
Let $X$ be a set on which $\vartheta_1$ and $\vartheta_2$ are topologies such that $\vartheta_1 \subseteq \vartheta_2$.

Let $I_X: \left({X, \vartheta_1}\right) \to \left({X, \vartheta_2}\right)$ be the identity mapping from $\left({X, \vartheta_1}\right)$ to $\left({X, \vartheta_2}\right)$.

Then $I_X$ is closed.

Proof
$\vartheta_1 \subseteq \vartheta_2$ means that every open set of $\left({X, \vartheta_1}\right)$ is also an open set of $\left({X, \vartheta_2}\right)$.

Let $A \subseteq X$ be closed in $\left({X, \vartheta_1}\right)$

Then by definition $X \setminus A$ is [Definition:Open Set (Topology)|open]] in $\left({X, \vartheta_1}\right)$.

Then $I_X \left({X \setminus A}\right) = X \setminus A$ is [Definition:Open Set (Topology)|open]] in $\left({X, \vartheta_2}\right)$.

So, by definition, $I_X \left({A}\right) = A$ is closed in $\left({X, \vartheta_2}\right)$.

So for all $A \subseteq X$ closed in $\left({X, \vartheta_1}\right)$, it holds that $I_X \left({A}\right)$ is closed in $\left({X, \vartheta_2}\right)$.

So by definition of closed mapping, the result follows.