# Equivalence of Definitions of T1 Space

## Contents

## Theorem

The following definitions of the concept of **$T_1$ (Fréchet) space** are equivalent:

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

### Definition by Open Sets

$\left({S, \tau}\right)$ is a **Fréchet space** or **$T_1$ space** if and only if:

- $\forall x, y \in S$ such that $x \ne y$, both:
- $\exists U \in \tau: x \in U, y \notin U$

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

### Definition by Closed Points

$\left({S, \tau}\right)$ is a **Fréchet space** or **$T_1$ space** if and only if all points of $S$ are closed in $T$.

## Proof

### 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.

$\Box$

### 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

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

$\blacksquare$

## Sources

- 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*... (previous) ... (next): $\text{I}: \ \S 2$ - 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*: Problems: $\S 2: \ 7$