T1 Space is T0 Space

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Theorem

Let $\left({S, \tau}\right)$ be a Fréchet ($T_1$) space.


Then $\left({S, \tau}\right)$ is also a Kolmogorov ($T_0$) space.


Proof

Let $\left({S, \tau}\right)$ be a $T_1$ space.

Let $x, y \in S: x \ne y$.

From the definition of $T_1$ space:

Both
$\exists U \in \tau: x \in U, y \notin U$
and:
$\exists V \in \tau: y \in V, x \notin V$


From the Rule of Simplification:

$\exists U \in \tau: x \in U, y \notin U$


From the Rule of Addition:

Either
$\exists U \in \tau: x \in U, y \notin U$
or:
$\exists V \in \tau: y \in V, x \notin V$

which is precisely the definition of a Kolmogorov ($T_0$) space.

$\blacksquare$


Sources