Equivalence of Definitions of T0 Space

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Definition by Open Sets implies Definition by Limit Points
Let $T = \left({S, \tau}\right)$ be a topological space for which:


 * $\forall x, y \in S$, either:
 * $(1): \quad \exists U \in \tau: x \in U, y \notin U$
 * $(2): \quad \exists U \in \tau: y \in U, x \notin U$

Let $x, y \in S$ such that $x \ne y$.

By the definition of limit point of a point:
 * $(3): \quad x$ is a limit point of $y$ if every open set $U \in \tau$ such that $x \in U$ contains $y$.
 * $(4): \quad y$ is a limit point of $x$ if every open set $U \in \tau$ such that $y \in U$ contains $x$.

Thus:
 * If $(3)$ holds then $(1)$ is false.
 * If $(4)$ holds then $(2)$ is false.

So both $(3)$ and $(4)$ can not both hold at once.

Hence $T = \left({S, \tau}\right)$ is a topological space for which no two points can be limit points of each other.

Definition by Limit Points implies Definition by Open Sets
Let $T = \left({S, \tau}\right)$ be a topological space for which no two points can be limit points of each other.

Let $x, y \in S$ such that $x \ne y$.

Suppose that:
 * $(1): \quad \nexists U \in \tau: x \in U, y \notin U$

and
 * $(2): \quad \nexists U \in \tau: y \in U, x \notin U$

Expressing this in different terms:
 * $(3): \quad$ Every open set $U \in \tau$ such that $x \in U$ contains $y$.
 * $(4): \quad$ Every open set $U \in \tau$ such that $y \in U$ contains $x$.

That is, by definition:
 * $x$ is a limit point of $y$ and $y$ is a limit point of $x$.

From this contradiction we deduce that:


 * $\forall x, y \in S$, either:
 * $\exists U \in \tau: x \in U, y \notin U$
 * or:
 * $\exists U \in \tau: y \in U, x \notin U$