Equivalence of Definitions of Ultraconnected Space

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

Then $T$ is ultraconnected iff the closures of every distinct pair of points of $S$ are not disjoint.

That is:
 * $\forall x, y \in S: \left\{{x}\right\}^- \cap \left\{{y}\right\}^- \ne \varnothing$

Proof
Let $T = \left({S, \tau}\right)$ be ultraconnected.

Let $x, y \in S$.

By Topological Closure is Closed, both$\left\{{x}\right\}^-$ and $\left\{{y}\right\}^-$ are closed.

As $T$ is ultraconnected, it follows that:
 * $\left\{{x}\right\}^- \cap \left\{{y}\right\}^- \ne \varnothing$

Now suppose that:


 * $\forall x, y \in S: \left\{{x}\right\}^- \cap \left\{{y}\right\}^- \ne \varnothing$

Let $V_1, V_2$ be closed sets of $T$.

Let $x \in V_1, y \in V_2$.

Then $\left\{{x}\right\}^- \cap \left\{{y}\right\}^- \ne \varnothing$.

But then from Closure of Subset is Subset of Closure we have that:
 * $\left\{{x}\right\}^- \subseteq V_1^-$


 * $\left\{{y}\right\}^- \subseteq V_2^-$

But from Closed Set Equals its Closure $V_1^- = V_1, V_2^- = V_2$.

So it follows from Intersection Subset that:
 * $\left\{{x}\right\}^- \cap \left\{{y}\right\}^- \subseteq V_1$
 * $\left\{{x}\right\}^- \cap \left\{{y}\right\}^- \subseteq V_2$

from which it follows:
 * $V_1 \cap V_2 \ne \varnothing$

As $V_1$ and $V_2$ are arbitrary, it follows that $T$ is ultraconnected by definition.