T5 Space is T4 Space

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Theorem

Let $\left({S, \tau}\right)$ be a $T_5$ space.


Then $\left({S, \tau}\right)$ is also a $T_4$ space.


Proof

Let $\left({S, \tau}\right)$ be a $T_5$ space.


From the definition of $T_5$ space:

$\forall A, B \subseteq S, A^- \cap B = A \cap B^- = \varnothing: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \varnothing$

where $A^-$ is the closure of $A$ in $T$.


Let $C, D \subseteq S$ be disjoint sets which are closed in $T$.

Thus $C, D \in \complement \left({\tau}\right)$ from the definition of closed set.

From Topological Closure is Closed:

$C^- = C, D^- = D$

and so from $C \cap D = \varnothing$:

$C^- \cap D = C \cap D^- = \varnothing$


Thus from the definition of $T_5$ space:

$\forall C, D \in \complement \left({\tau}\right), C \cap D = \varnothing: \exists U, V \in \tau: C \subseteq U, D \subseteq V$

which is precisely the definition of a $T_4$ space.

$\blacksquare$


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