Equivalence of Definitions of Ultraconnected Space

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Theorem

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

Definition 1

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

Definition 2

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

$\forall x, y \in S: \set x^- \cap \set y^- \ne \O$

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 $T = \struct {S, \tau}$ 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 $\set x^-$ and $\set y^-$ are closed.

By hypothesis:

$\set x^- \cap \set y^- \ne \O$

That is, $T = \struct {S, \tau}$ is ultraconnected by Definition 2.

$\Box$

Let $T = \struct {S, \tau}$ be ultraconnected by Definition 2:

$\forall x, y \in S: \set x^- \cap \set y^- \ne \O$

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

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

Then:

$\set x^- \cap \set y^- \ne \O$

But then from Topological Closure of Subset is Subset of Topological Closure we have that:

$\set x^- \subseteq V_1^-$

$\set y^- \subseteq V_2^-$

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

So from Intersection is Subset:

$\set x^- \cap \set y^- \subseteq V_1$

$\set x^- \cap \set y^- \subseteq V_2$

from which:

$V_1 \cap V_2 \ne \O$

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 $T = \struct {S, \tau}$ 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.

$\blacksquare$

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