Equivalence of Definitions of T1 Space

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

The following two sets of conditions defining a $T_1$ (Fréchet) space are logically equivalent:

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


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

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

Let $x, y \in S$.

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

Thus $(2)$ and $(3)$ give us that $y$ is not a limit point of $x$.

As $x$ and $y$ are any two points in $S$, it follows that $x$ has no limit points.

Thus it holds vacuously that all the limit point of $x$ are in $\left\{{x}\right\}$.

By Closed Set iff Contains all its Limit Points, we have that $\left\{{x}\right\}$ is a closed set in $T$.

This holds for all $x \in S$.

Hence $T = \left({S, \tau}\right)$ is a topological space for which all points are closed.

Definition by Closed Points implies Definition by Open Sets
$T = \left({S, \tau}\right)$ is a topological space for which all points are closed.

Let $x, y \in S$.

By Closed Set iff Contains all its Limit Points, we have that:
 * all the limit points of $x$ are in $\left\{{x}\right\}$
 * all the limit points of $y$ are in $\left\{{y}\right\}$.

It follows by definition of limit point (reversing the above argument) that:
 * $\exists U \in \tau: x \in U, y \notin U$

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