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