Characterization of T3 Space
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Theorem
Let $T = \struct{S, \tau}$ be a topological space.
The following statements are equivalent:
- $(1) \quad T$ is a $T_3$ space
- $(2) \quad \forall F \subseteq S : S \setminus F \in \tau, y \in S \setminus F : \exists V \in \tau : y \in V, V^- \cap F = \O$
- $(3) \quad \forall U \in \tau, y \in U : \exists V \in \tau : y \in V, V^- \subseteq U$
where $V^-$ denotes the closure of $V$ in $T$
Proof
Statement $(1)$ implies Statement $(2)$
Let $T$ be a $T_3$ space.
Let $F \subseteq S : S \setminus F \in \tau$.
Let $y \in S \setminus F$.
By definition of $T_3$ space:
- $\exists V, W \in \tau : y \in V, F \subseteq W : V \cap W = \O$
From Subset of Set Difference iff Disjoint Set:
- $V \subseteq S \setminus W$
By definition of closed set:
- $S \setminus W$ is closed in $T$
From Closure of Subset of Closed Set of Topological Space is Subset:
- $V^- \subseteq S \setminus W$
From Set Difference with Subset is Superset of Set Difference:
- $S \setminus W \subseteq S \setminus F$
From Subset Relation is Transitive:
- $V^- \subseteq S \setminus F$
From Subset of Set Difference iff Disjoint Set:
- $V^- \cap F = \O$
The result follows.
$\Box$
Statement $(2)$ implies Statement $(3)$
Let $T$ satisfy:
- $\forall F \subseteq S : S \setminus F \in \tau, y \in S \setminus F : \exists V \in \tau : y \in V, V^- \cap F = \O$
Let $U \in \tau$.
Let $y \in U$.
Let $F = S \setminus U$.
From Set Difference with Set Difference:
- $S \setminus F = U$
Hence:
- $S \setminus F \in \tau$
and
- $y \in S \setminus F$
From $(2)$:
- $\exists V \in \tau : y \in V, V^- \cap F = \O$
Hence:
- $V^- \cap S \setminus U = \O$
From Subset of Set Difference iff Disjoint Set:
- $V^- \subseteq U$
Since $U$ and $y$ were arbitrary, we have shown:
- $\forall U \in \tau, y \in U : \exists V \in \tau : y \in V, V^- \subseteq U$
$\Box$
Statement $(3)$ implies Statement $(1)$
Let $T$ satisfy:
- $\forall U \in S, y \in U : \exists V \in \tau : y \in V, V^- \subseteq U$
Let $U \in \tau$.
Let $y \in U$.
From $(3)$:
- $\exists V \in \tau : y \in V, V^- \subseteq U$
Let $N = V^-$.
By definition of closure:
- $V \subseteq N$
From Topological Closure is Closed:
- $S \ N \in \tau$
Hence:
- $\exists N \subseteq S : S \setminus N \in \tau : \exists V \in \tau : y \in V \subseteq N \subseteq U$
Since $U$ and $y$ were arbitrary, we have shown:
- $\forall U \in \tau, y \in U : \exists N \subseteq S : S \setminus N \in \tau : \exists V \in \tau : y \in V \subseteq N \subseteq U$
By definition, $T$ is a $T_3$ space
$\blacksquare$