Equivalence of Definitions of Ultraconnected Space/1 iff 3

Theorem
Let $X$ be a topological space.


 * $(1):\quad$ $X$ is ultraconnected
 * $(2):\quad$ Every closed subset of $X$ is connected

1 implies 2
Let $X$ be ultraconnected.

Let $F \subseteq X$ be an closed subset.

Support $F$ not connected.

Then there exist nonempty closed subsets $G,H$ of $F$ that are disjoint (and whose union is $F$).

By Closed Set in Closed Subspace, $G$ and $H$ are closed subsets of $X$.

Because $G\cap H = \varnothing$, $X$ is not irreducible.

This is a contradiction.

Thus $F$ is connected.

2 implies 1
Let $G$ and $H$ be closed subsets of $X$.

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

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

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

Because $G\cup H$ is connected, $G\cap H$ is nonempty.

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

Also see

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