Equivalence of Definitions of Ultraconnected Space/1 iff 3

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.


 * $(1): \quad$ $T$ is ultraconnected
 * $(2): \quad$ Every closed set of $T$ is connected

1 implies 2
Let $T$ be ultraconnected.

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

$F$ is not connected.

Then there exist nonempty 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 irreducible.

This is a contradiction.

Thus $F$ is connected.

2 implies 1
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 nonempty.

Because $G$ and $H$ were arbitrary, $X$ is ultraconnected.

Also see

 * Closed Subset of Ultraconnected Space is Ultraconnected
 * Space is Irreducible iff Open Subsets are Connected, whose proof is almost the same