Closed Extension Topology is Topology

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Theorem

Let $T = \left({S, \tau}\right)$ be a topological space.

Let $\tau^*_p$ be the closed extension topology of $\tau$.


Then $\tau^*_p$ is a topology on $S^*_p = S \cup \left\{{p}\right\}$.


Proof

By definition:

$\tau^*_p = \left\{{U \cup \left\{{p}\right\}: U \in \tau}\right\} \cup \left\{{\varnothing}\right\}$


We have that $\varnothing \in \tau^*_p$ by definition.

We also have that $S \in \tau$ so $S \cup \left\{{p}\right\} \in \tau^*_p$.

Now let $U_1, U_2 \in \tau^*_p$.

Then $U_1^* = U_1 \setminus \left\{{p}\right\} \in \tau, U_2^* = U_2 \setminus \left\{{p}\right\} \in \tau$.

Then $p \in U_1$ and $p \in U_2$ and so $p \in U_1 \cap U_2$.

So $U_1 \cap U_2 = \left({U_1^* \cap U_2^*}\right) \cup \left\{{p}\right\} \in \tau^*_p$.


Finally consider $\mathcal U \subseteq \tau^*_p$.

$\displaystyle \bigcup \mathcal U = \bigcup_{U \in \mathcal U} \left({U \setminus \left\{{p}\right\}}\right) \cup \left\{{p}\right\}$

So $\displaystyle \bigcup \mathcal U \in \tau^*_p$.


So $\tau^*_p$ is a topology on $S \cup \left\{{p}\right\}$.

$\blacksquare$


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