Identity Mapping to Expansion is Closed

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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 $I_X: \left({S, \tau_1}\right) \to \left({S, \tau_2}\right)$ be the identity mapping from $\left({S, \tau_1}\right)$ to $\left({S, \tau_2}\right)$.


Then $I_S$ is closed.


Proof

$\tau_1 \subseteq \tau_2$ means that every open set of $\left({S, \tau_1}\right)$ is also an open set of $\left({S, \tau_2}\right)$.

Let $A \subseteq S$ be closed in $\left({S, \tau_1}\right)$

Then by definition $S \setminus A$ is open in $\left({S, \tau_1}\right)$.

Then $I_S \left({S \setminus A}\right) = S \setminus A$ is open in $\left({S, \tau_2}\right)$.

So, by definition, $I_S \left({A}\right) = A$ is closed in $\left({S, \tau_2}\right)$.

So for all $A \subseteq S$ closed in $\left({S, \tau_1}\right)$, it holds that $I_S \left({A}\right)$ is closed in $\left({S, \tau_2}\right)$.

So by definition of closed mapping, the result follows.

$\blacksquare$


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