T1 Space is T0 Space

Theorem
Let $\left({X, \vartheta}\right)$ be a Fréchet ($T_1$) space.

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

Proof
Let $x, y \in X: x \ne y$, where $\left({X, \vartheta}\right)$ is a $T_1$ space.

From the definition of $T_1$ space:
 * Both
 * $\exists U \in \vartheta: x \in U, y \notin U$
 * and:
 * $\exists V \in \vartheta: y \in V, x \notin V$

From the rule of simplification:


 * $\exists U \in \vartheta: x \in U, y \notin U$

From the rule of addition:


 * Either


 * $\exists U \in \vartheta: x \in U, y \notin U$
 * or:
 * $\exists V \in \vartheta: y \in V, x \notin V$

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