Union of Closure with Closure of Complement is Whole Space
Jump to navigation
Jump to search
Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $H \subseteq S$ be a subset of $S$.
Let $H^-$ denote the closure of $H$ in $T$.
Let $S \setminus H$ denote the complement of $H$ relative to $S$.
Then:
- $H^- \cup \left({S \setminus H}\right)^- = S$
Proof
We have that:
- $H^- \cup \left({S \setminus H}\right)^- \subseteq S$
by definition of $S$.
From Union with Relative Complement:
- $H \cup \left({S \setminus H}\right) = S$
From Set is Subset of its Topological Closure:
\(\ds H\) | \(\subseteq\) | \(\ds H^-\) | ||||||||||||
\(\ds \left({S \setminus H}\right)\) | \(\subseteq\) | \(\ds \left({S \setminus H}\right)^-\) |
From Set Union Preserves Subsets:
- $H \subseteq H^-, \left({S \setminus H}\right) \subseteq \left({S \setminus H}\right)^- \implies H \cup \left({S \setminus H}\right) \subseteq H^- \cup \left({S \setminus H}\right)^-$
which means:
- $S \subseteq H^- \cup \left({S \setminus H}\right)^-$
The result follows by definition of equality of sets.
$\blacksquare$