Closed Extension Topology is Topology

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Theorem

Let $T = \struct {S, \tau}$ be a topological space.

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


Then $\tau^*_p$ is a topology on $S^*_p = S \cup \set p$.


Proof

By definition:

$\tau^*_p = \set {U \cup \set p: U \in \tau} \cup \set \O$


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

We also have that $S \in \tau$ so $S \cup \set p \in \tau^*_p$.

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

Then $U_1^* = U_1 \setminus \set p \in \tau, U_2^* = U_2 \setminus \set p \in \tau$.

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

So $U_1 \cap U_2 = \paren {U_1^* \cap U_2^*} \cup \set p \in \tau^*_p$.


Finally consider $\UU \subseteq \tau^*_p$.

$\ds \bigcup \UU = \bigcup_{U \mathop \in \UU} \paren {U \setminus \set p} \cup \set p$

So $\ds \bigcup \UU \in \tau^*_p$.


So $\tau^*_p$ is a topology on $S \cup \set p$.

$\blacksquare$


Sources

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