T3 1/2 Space is T3 Space
Theorem
Let $T$ be a $T_{3 \frac 1 2}$ space.
Then $T$ is also a $T_3$ space.
Proof
Let $T = \struct {S, \tau}$ be a $T_{3 \frac 1 2}$ space.
From the definition of $T_{3 \frac 1 2}$ space:
- For any closed set $F \subseteq S$ and any point $y \in S$ such that $y \notin F$, there exists an Urysohn function for $F$ and $\set y$.
Let $F \subseteq S$ be a closed set in $T$ and let $y \in \relcomp S F$.
An Urysohn function for $F$ and $\set y$ is a continuous mapping $f: S \to \closedint 0 1$ where:
- $\forall a \in F: \map f a = 0$
- $\map f y = 1$
Let:
- $U = \map {f^{-1} } 0$
- $V = \map {f^{-1} } 1$
As $f$ is continuous, both $U$ and $V$ are open in $T$.
Suppose $x \in U \cap V$.
Then we would have:
- $x \in U \implies \map f x = 0$
- $x \in V \implies \map f x = 1$
which contradicts the many-to-one nature of a mapping.
So $U \cap V = \O$.
Thus we have:
- $\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$
That is, for any closed set $F \subseteq S$ and any point $y \in S$ such that $y \notin F$ there exist disjoint open sets $U, V \in \tau$ such that $F \subseteq U$, $y \in V$.
which is precisely the definition of a $T_3$ space.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Completely Regular Spaces