Equivalence of Definitions of Connected Topological Space/No Separation iff No Union of Closed Sets

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

Then:
 * $T$ admits no separation


 * $T$ has no two disjoint non-empty closed sets whose union is $S$.
 * $T$ has no two disjoint non-empty closed sets whose union is $S$.

Proof
From the definition of logical equivalence it follows that the statement can be expressed as:


 * $T$ admits a separation


 * there exist two closed sets of $T$ which form a (set) partition of $S$.
 * there exist two closed sets of $T$ which form a (set) partition of $S$.

By definition, a separation of $T$ is a (set) partition of $S$ by $A, B$ which are open in $T$.

From Complements of Components of Two-Component Partition form Partition:
 * $A \mid B$ is a (set) partition of $S$ $\complement_S \left({A}\right) \mid \complement_S \left({B}\right)$ is a (set) partition of $S$.

Hence the result by definition of closed set.