# Equivalence of Definitions of Ultraconnected Space/1 iff 3

## Contents

## Theorem

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

### Closed Sets Intersect

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

### Closed Sets are Connected

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

## Proof

### Closed Sets Intersect implies Closed Sets are Connected

Let $T = \struct {S, \tau}$ 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 $G \cap H = \O$, $T$ is not ultraconnected.

This is a contradiction.

Thus $F$ is connected.

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

Thus $T = \struct {S, \tau}$ is ultraconnected in the sense that:

- every closed set of $T$ is connected..

$\Box$

### Closed Sets are Connected implies Closed Sets Intersect

Let $T = \struct {S, \tau}$ 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 $T = \struct {S, \tau}$ is ultraconnected in the sense that:

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

$\blacksquare$