Equivalence of Definitions of Connected Topological Space/No Clopen Sets implies No Union of Separated Sets
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:
\(\ds S\) | \(=\) | \(\ds A \cup B\) | ||||||||||||
\(\ds \) | \(\subseteq\) | \(\ds A \cup B^-\) | Set is Subset of its Topological Closure | |||||||||||
\(\ds \) | \(\subseteq\) | \(\ds S\) | 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 $S$.
$\blacksquare$