Closed Set of Ultraconnected Space is Ultraconnected

Theorem

Let $T = \struct {S, \tau}$ be an ultraconnected topological space.

Let $F \subset S$ be a closed set in $T$.

Then $F$ is ultraconnected.

Proof

Let $A, B$ be two non-empty closed sets in $\struct {F, \tau}$.

By Closed Set in Topological Subspace, $A, B$ are closed in $T$ as well.

By Definition of Ultraconnected Space, $A$ and $B$ are not disjoint.

Since $A$, $B$ are arbitrary, no two non-empty closed sets of $\struct {F, \tau}$ are disjoint.

Hence the result from Definition of Ultraconnected Space.

$\blacksquare$

Also see

• Open Set of Irreducible Space is Irreducible