Equivalence of Definitions of Ultraconnected Space/1 iff 3

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Closed Sets Intersect

A topological space $T = \left({S, \tau}\right)$ 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 $T = \left({S, \tau}\right)$ 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.

This is a contradiction.

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.

$\blacksquare$


Also see