Characterization of Closure by Open Sets
Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $A$ be a subset of $S$.
Let $x$ be a point of $T$.
Let $A^-$ denote the closure of $A$.
Then $x \in A^-$ if and only if:
- for every open set $U$ of $T$:
- $x \in U \implies A \cap U \ne \O$
Proof
Sufficient Condition
Let $x \in A^-$.
Aiming for a contradiction, suppose there exists an open set $U$ of $T$ such that:
- $x \in U$ and $A \cap U = \O$
We have that $U$ is open in $T$.
So by definition of closed set, $\relcomp S U$ is closed in $T$.
Then:
\(\ds A \cap U\) | \(=\) | \(\ds \O\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds A\) | \(\subseteq\) | \(\ds \relcomp S U\) | Empty Intersection iff Subset of Complement | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds A^-\) | \(\subseteq\) | \(\ds \relcomp S U\) | Definition 3 of Closure (Topology): $\relcomp S U$ is closed | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds A^- \cap U\) | \(=\) | \(\ds \O\) | Empty Intersection iff Subset of Complement |
But we have:
- $x \in A^-$
and also:
- $x \in U$
and thus by definition of set intersection:
- $x \in A^- \cap U$
This contradicts $A^- \cap U = \O$
Hence by Proof by Contradiction the assumption that there exists an open set $U$ of $T$ such that $x \in U$ and $A \cap U = \O$ was false.
So for every open set $U$ of $T$:
- $x \in U \implies A \cap U \ne \O$
$\Box$
Necessary Condition
Let $x$ be such that for every open set $U$ of $T$:
- $x \in U \implies A \cap U \ne \O$
Aiming for a contradiction, suppose $x \notin A^-$.
Then:
- $x \in \relcomp S {A^-}$
Then by assumption:
- $A \cap \relcomp S {A^-} \ne \O$
as $\relcomp S {A^-}$ is open.
By definition of complement:
- $A \cap \relcomp S {A^-} = \O$
So by Empty Intersection iff Subset of Complement:
- $A \nsubseteq A^-$
From Set is Subset of its Topological Closure:
- $A \subseteq A^-$
But from Set Complement inverts Subsets:
- $\relcomp S {A^-} \subseteq \relcomp S A$
from which by Empty Intersection iff Subset of Complement:
- $A \cap \relcomp S {A^-} = \O$
Hence by Proof by Contradiction the assumption that $x \notin A^-$ was false.
So $x \in A^-$.
$\blacksquare$
Sources
- Mizar article TOPS_1:12