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

• 1989: Ryszard Engelking: General Topology (revised and completed ed.)

• Mizar article TOPGEN_1:29