Metric Space is T4/Proof 2
Theorem
Let $M = \struct {A, d}$ be a metric space.
Then $M$ is a $T_4$ space.
Proof
We have that a metric space is fully $T_4$.
Then we have that a fully $T_4$ space is $T_4$.
$\blacksquare$
Let $M = \struct {A, d}$ be a metric space.
Then $M$ is a $T_4$ space.
We have that a metric space is fully $T_4$.
Then we have that a fully $T_4$ space is $T_4$.
$\blacksquare$