Equivalence of Definitions of Connected Topological Space/No Clopen Sets implies No Union of Separated Sets

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Theorem

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

Let the only clopen sets of $T$ be $S$ and $\O$.


Then there are no two non-empty separated sets of $T$ whose union is $S$.


Proof

Suppose $A$ and $B$ are separated subsets of $T$ such that $A \cup B = S$.

By definition of separated sets:

$A \cap B^- = \O$


Then:

\(\displaystyle S\) \(=\) \(\displaystyle A \cup B\)
\(\displaystyle \) \(\subseteq\) \(\displaystyle A \cup B^-\) Set is Subset of its Topological Closure
\(\displaystyle \) \(\subseteq\) \(\displaystyle S\) by definition of $S$

Hence $A = S \setminus B^-$.


From Topological Closure is Closed, $B^-$ is closed in $T$.

Thus $A$ is open in $T$.


Also by definition of separated sets:

$A^- \cap B = \O$

Hence, by the same reasoning, $B$ must also be open.


But:

$A \cap B \subseteq A \cap B^- = \O$

and $A \cup B = S$, by assumption.

So:

$A = S \setminus B$ and $B = S \setminus A$

and we conclude that both $A$ and $B$ are clopen.

Therefore, by hypothesis, one of them must be $S$ and the other must be $\O$.

That is, there are no two non-empty separated sets of $T$ whose union is $S$.

$\blacksquare$