T5 Space is T4 Space
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Theorem
Let $\struct {S, \tau}$ be a $T_5$ space.
Then $\struct {S, \tau}$ is also a $T_4$ space.
Proof
Let $\struct {S, \tau}$ be a $T_5$ space.
From the definition of $T_5$ space:
- $\forall A, B \subseteq S, A^- \cap B = A \cap B^- = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$
where $A^-$ is the closure of $A$ in $T$.
Let $C, D \subseteq S$ be disjoint sets which are closed in $T$.
Thus $C, D \in \map \complement \tau$ from the definition of closed set.
From Topological Closure is Closed:
- $C^- = C, D^- = D$
and so from $C \cap D = \O$:
- $C^- \cap D = C \cap D^- = \O$
Thus from the definition of $T_5$ space:
- $\forall C, D \in \map \complement \tau, C \cap D = \O: \exists U, V \in \tau: C \subseteq U, D \subseteq V$
which is precisely the definition of a $T_4$ 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