# Non-Trivial Ultraconnected Space is not T1

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## 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.

$\blacksquare$

## Sources

- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*(2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $4$: Connectedness