# Identity Mapping to Expansion is Closed

## Theorem

Let $S$ be a set on which $\tau_1$ and $\tau_2$ are topologies such that:

- $\tau_1 \subseteq \tau_2$

that is, such that $\tau_2$ is an expansion of $\tau_1$.

Let $I_X: \struct {S, \tau_1} \to \struct {S, \tau_2}$ be the identity mapping from $\struct {S, \tau_1}$ to $\struct {S, \tau_2}$.

Then $I_S$ is closed.

## Proof

$\tau_1 \subseteq \tau_2$ means that every open set of $\struct {S, \tau_1}$ is also an open set of $\struct {S, \tau_2}$.

Let $A \subseteq S$ be closed in $\struct {S, \tau_1}$

Then by definition $S \setminus A$ is open in $\struct {S, \tau_1}$.

Then $\map {I_S} {S \setminus A} = S \setminus A$ is open in $\struct {S, \tau_2}$.

So, by definition, $\map {I_S} A = A$ is closed in $\struct {S, \tau_2}$.

So for all $A \subseteq S$ closed in $\struct {S, \tau_1}$, it holds that $\map {I_S} A$ is closed in $\struct {S, \tau_2}$.

So by definition of closed mapping, the result follows.

$\blacksquare$

## Sources

- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*(2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces