Ultraconnected iff Closures of Distinct Points Always Intersect

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Theorem

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

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

That is:

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

Let $T = \left({X, \vartheta}\right)$ be a ultraconnected.

Let $x, y \in X$.

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

As $T$ is ultraconnected, it follows that:

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

$\Box$

Now suppose that:

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

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

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

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

$\left\{{x}\right\}^- \subseteq V_1^-$

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

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.

$\blacksquare$