Either-Or Topology is not T3
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Theorem
Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is not a $T_3$ space.
Proof
Consider the set $\set 0$ which is closed from Closed Sets of Either-Or Topology.
Let $x = \dfrac 1 2$.
Let $U \in \tau$ such that $\set 0 \subseteq U$.
Then by definition $\openint {-1} 1 \subseteq U$ and so $x \in U$.
So we have found a closed set of $T$ and a point in $S$ which do not satisfy the conditions for $T$ to be a $T_3$ 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: $17$. Either-Or Topology: $1$