Either-Or Topology is T5

From ProofWiki
Jump to navigation Jump to search


Let $T = \left({S, \tau}\right)$ be the either-or space.

Then $T$ is a $T_5$ space.


Let $A, B \subseteq S$ such that $A \cap B \ne \varnothing$.

Suppose neither $A$ nor $B$ contain $0$.

Then $A$ and $B$ are both open in $T$.

From Limit Points of Either-Or Topology, the only limit point that either $A$ or $B$ may have is $0$.

So either $A^- = A \cup \left\{{0}\right\}$ or $A^- = A$, where $A^-$ is the closure of $A$

Similarly for $B$.

So $A^- \cap B = \varnothing$, and so are separated, and such that:

$\exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \varnothing$

where $U = A$ and $V = B$, fulfilling the condition for a $T_5$ space.

Suppose one of $A$ and $B$ contains $0$.

Without loss of generality, suppose $0 \in A$.

Then for $A$ and $B$ to be separated, $0 \notin B^-$ and so $B^-$ has to be $\left\{{-1}\right\}$ or $\left\{{1}\right\}$ or their union.

In any of these cases we have that $B$ and $S \setminus B$ are disjoint open sets containing $B$ and $A$ respectively.

So in all cases we see that if $A$ and $B$ are separated, then:

$\exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \varnothing$

where $U = A$ and $V = B$.

and so by definition $T$ is a $T_5$ space.