Equivalence of Definitions of Connected Topological Space/No Union of Closed Sets implies No Subsets with Empty Boundary
Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $T$ have no two disjoint non-empty closed sets whose union is $S$.
Then the only subsets of $S$ whose boundary is empty are $S$ and $\O$.
Proof
Let $H \subseteq S$ be a non-empty subset whose boundary $\partial H$ is empty.
Thus:
\(\ds \partial H\) | \(=\) | \(\ds \O\) | by hypothesis | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds H^- \cap \paren {S \setminus H}^-\) | \(=\) | \(\ds \O\) | Boundary is Intersection of Closure with Closure of Complement |
From Topological Closure is Closed, both $H^-$ and $\paren {S \setminus H}^-$ are closed sets of $T$.
From Union of Closure with Closure of Complement is Whole Space:
- $H^- \cup \paren {S \setminus H}^- = S$
Thus $H^-$ and $\paren {S \setminus H}^-$ are two disjoint closed sets of $T$ whose union is $S$.
Hence, by hypothesis, one of them must be empty.
Suppose $H$ is not empty.
It must therefore follow that:
- $S \setminus H = \O$
Therefore $H = S$.
Thus the only subsets of $S$ whose boundary is empty are $S$ and $\O$.
$\blacksquare$