# Equivalence of Definitions of Ultraconnected Space

## Theorem

The following definitions of the concept of Ultraconnected Space are equivalent:

### Definition 1

A topological space $T = \left({S, \tau}\right)$ is ultraconnected if and only if no two non-empty closed sets are disjoint.

### Definition 2

A topological space $T = \left({S, \tau}\right)$ is ultraconnected if and only if the closures of every distinct pair elements of $S$ are not disjoint:

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

### Definition 3

A topological space $T = \left({S, \tau}\right)$ is ultraconnected if and only if every closed set of $T$ is connected.

## Proof

### $(1)$ iff $(2)$

Let $T = \left({S, \tau}\right)$ be ultraconnected by Definition 1:

no two non-empty closed sets of $T$ are disjoint.

Let $x, y \in S$.

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

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

That is, $T = \left({S, \tau}\right)$ is ultraconnected by Definition 2.

$\Box$

Let $T = \left({S, \tau}\right)$ be ultraconnected by Definition 2:

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

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 Topological Closure of Subset is Subset of Topological 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 from Intersection is Subset:

$\left\{{x}\right\}^- \cap \left\{{y}\right\}^- \subseteq V_1$
$\left\{{x}\right\}^- \cap \left\{{y}\right\}^- \subseteq V_2$

from which:

$V_1 \cap V_2 \ne \varnothing$

As $V_1$ and $V_2$ are arbitrary, it follows that $T$ is ultraconnected by Definition 1.

$\blacksquare$

### Closed Sets Intersect implies Closed Sets are Connected

Let $T = \left({S, \tau}\right)$ be ultraconnected in the sense that:

no two non-empty closed sets of $T$ are disjoint.

Let $F \subseteq S$ be an arbitrary closed set of $T$.

Aiming for a contradiction, suppose $F$ is not connected.

Then there exist non-empty closed set $G, H$ in $F$ that are disjoint (and whose union is $F$).

By Closed Set in Closed Subspace, $G$ and $H$ are closed in $T$.

Because $G \cap H = \varnothing$, $T$ is not ultraconnected.

Thus $F$ is connected.

As $F$ is arbitrary, this applies to all closed set of $T$.

Thus $T = \left({S, \tau}\right)$ is ultraconnected in the sense that:

every closed set of $T$ is connected..

$\Box$

### Closed Sets are Connected implies Closed Sets Intersect

Let $T = \left({S, \tau}\right)$ be ultraconnected in the sense that:

every closed set of $T$ is connected.

Let $G$ and $H$ be closed sets of $T$.

Then their union $G \cup H$ is closed in $T$.

By assumption, $G \cup H$ is connected.

By Closed Set in Closed Subspace, $G$ and $H$ are closed sets of $G \cup H$.

Because $G \cup H$ is connected, $G \cap H$ is non-empty.

Because $G$ and $H$ were arbitrary, this applies to intersection of all such closed sets

Thus $T = \left({S, \tau}\right)$ is ultraconnected in the sense that:

no two non-empty closed sets of $T$ are disjoint.

$\blacksquare$