# Non-Trivial Ultraconnected Space is not T1

## Theorem

Let $T = \struct {S, \tau}$ be a topological space which is ultraconnected.

If $S$ has more than one element, then $T$ is not a $T_1$ (Fréchet) space.

That is, if $T$ is a $T_1$ (Fréchet) space with more than one element, it is not ultraconnected.

## Proof

$T = \struct {S, \tau}$ be ultraconnected.

Thus by definition:

$(1): \quad \forall x, y \in S: \set x^- \cap \set y^- \ne \O$

Let $a, b \in S$ such that $a \ne b$.

Aiming for a contradiction, suppose $T$ is a $T_1$ (Fréchet) space.

By definition of $T_1$ Space, $\set a$ and $\set b$ are closed.

From Closed Set Equals its Closure we have that $\set a^- = \set a$ and $\set b^- = \set b$.

It immediately follows that:

$\set a^- \cap \set b^- = \O$

But that contradicts $(1)$ above.

The result follows by Proof by Contradiction.

$\blacksquare$