Closure Equals Union with Derivative
Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $A$ be a subset of $S$.
Then:
- $A^- = A \cup A'$
where
- $A'$ denotes the derivative of $A$
- $A^-$ denotes the closure of $A$.
Proof
Closure Subset of Union
It is to be proved that:
- $A^- \subseteq A \cup A'$
Let $x \in A^-$.
In the case where $x \in A$ then $x \in A \cup A'$ by definition of set union.
Let:
- $(1): \quad x \notin A$
From Characterization of Derivative by Open Sets, to prove $x \in A'$ it is enough to show that:
- for every open set $U$ of $T$:
- if $x \in U$
- then there exists a point $y$ of $T$ such that $y \in A \cap U$ and $x \ne y$.
Let $U$ be an open set of $T$.
Let $x \in U$.
Then by Condition for Point being in Closure:
- $A \cap U \ne \O$
Then by definition of empty set:
- $\exists y \in S: y \in A \cap U$
By definition of set intersection:
- $y \in A$ and $y \in U$
But as $x \notin A$ it follows by definition of set intersection that:
- $x \notin A \cap U$
So by $(1)$:
- $x \ne y$
Thus $y$ fulfils the conditions of the hypothesis, and so:
- $x \in A'$
Hence by definition of set union:
- $x \in A \cup A'$
Thus in all cases:
- $x \in A^- \implies x \in A \cup A'$
and so:
- $A^- \subseteq A \cup A'$
$\Box$
Union Subset of Closure
It is to be proved that:
- $A \cup A' \subseteq A^-$
By Set is Subset of its Topological Closure:
- $A \subseteq A^-$
By Derivative is Included in Closure:
- $A' \subseteq A^-$
Hence by Union of Subsets is Subset:
- $A \cup A' \subseteq A^-$
$\Box$
Hence by definition of set equality:
- $A^- = A \cup A'$
Sources
- Mizar article TOPGEN_1:29