Closed Set of Ultraconnected Space is Ultraconnected
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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.
$\blacksquare$