Condition for Closed Extension Space to be T0 Space
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Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_p = \struct {S^*_p, \tau^*_p}$ be the closed extension space of $T$.
Then $T^*_p$ is a $T_0$ (Kolmogorov) space if and only if $T$ is.
Proof
By definition:
- $\tau^*_p = \set {U \cup \set p: U \in \tau} \cup \set \O$
Let $T = \struct {S, \tau}$ be a $T_0$ space.
Then:
- $\forall x, y \in S$ such that $x \ne y$, either:
- $\exists U \in \tau: x \in U, y \notin U$
- or:
- $\exists U \in \tau: y \in U, x \notin U$
Let $x, y \in S^*_p$ such that $x \ne y, x \ne p, y \ne p$.
Then:
- $\exists U \in \tau: x \in U, y \notin U$
or:
- $\exists U \in \tau: y \in U, x \notin U$
Now let $U^* = U \cup \set p$.
It follows that:
- $x \in U^*, y \notin U^*$
or:
- $y \in U^*, x \notin U^*$
Now consider $x, p \in S^*_p$ such that $x \ne p$.
We have that:
- $\set p \in \tau^*_p: p \in \set p, x \notin \set p$
So we have shown that:
- $\forall x, y \in S^*_p$ such that $x \ne y$, either:
- $\exists U \in \tau^*_p: x \in U, y \notin U$
- or:
- $\exists U \in \tau^*_p: y \in U, x \notin U$
and so $T^*_p$ is a $T_0$ space.
Now suppose that $T = \struct {S, \tau}$ is not a $T_0$ space.
Let $x, y \in S$ such that:
- $\forall U \in \tau: x \in U \implies y \in U$
- $\forall U \in \tau: y \in U \implies x \in U$
Now consider $U^* = U \cup \set p$ for any $U \in \tau$.
Then:
- As $x \in U \implies y \in U$ it follows that $x \in U^* \implies y \in U^*$.
- As $y \in U \implies x \in U$ it follows that $y \in U^* \implies x \in U^*$.
Hence $T^*_p$ is not a $T_0$ space.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $12$. Closed Extension Topology: $21$