Double Pointed Countable Complement Topology fulfils no Separation Axioms
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Theorem
Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Let $T \times D$ be the double pointed topology on $T$.
Then $T \times D$ is not a $T_0$ space, $T_1$ space, $T_2$ space, $T_3$ space, $T_4$ space or $T_5$ space.
Proof
From Separation Axioms on Double Pointed Topology, we have that:
- $T \times D$ is not a $T_0$ space.
- $T \times D$ is not a $T_1$ space.
- $T \times D$ is not a $T_2$ space.
Also from Separation Axioms on Double Pointed Topology, we have that:
- $T \times D$ is a $T_3$ space if and only if $T$ is a $T_3$ space
- $T \times D$ is a $T_4$ space if and only if $T$ is a $T_4$ space
- $T \times D$ is a $T_5$ space if and only if $T$ is a $T_5$ space
But from Countable Complement Space is not $T_3$, $T_4$ or $T_5$:
- $T$ is not a $T_3$ space
- $T$ is not a $T_4$ space
- $T$ is not a $T_5$ space
Hence the result.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $21$. Double Pointed Countable Complement Topology: $6$