# 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:

\(\displaystyle S\) | \(=\) | \(\displaystyle A \cup B\) | |||||||||||

\(\displaystyle \) | \(\subseteq\) | \(\displaystyle A \cup B^-\) | 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 $S$.

$\blacksquare$