Derivative of Derivative is Subset of Derivative in T1 Space

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Theorem

Let $T = \struct {S, \tau}$ be a $T_1$ topological space.

Let $A$ be a subset of $S$.


Then

$A' ' \subseteq A'$

where

$A'$ denotes the derivative of $A$


Proof

Let:

$(1): \quad x \in A' '$

Aiming for a contradiction, suppose $x \notin A'$.

Then by Characterization of Derivative by Open Sets there exists an open subset $G$ of $T$ such that:

$(2): \quad x \in G$

and:

$(3): \quad \lnot \exists y: y \in A \cap G \land x \ne y$

By definition of $T_1$ space:

$\set x$ is closed.

Then by Open Set minus Closed Set is Open:

$(4): \quad G \setminus \set x$ is open.

By $(1)$, $(2)$, and Characterization of Derivative by Open Sets there exists a point $y$ of $T$ such that:

$(5): \quad y \in A' \cap G$

and:

$(6): \quad x \ne y$

Then by definition of intersection:

$y \in A'$

Then by $(6)$ and definition of set difference:

$(7): \quad y \in A' \setminus \set x$

By definition of intersection and $(5)$:

$y \in G$

By $(6)$ and definition of singleton:

$y \notin \set x$

Then by definition of set difference:

$(8): \quad y \in G \setminus \set x$

We will prove:

$(9): \quad G \cap \paren {A \setminus \set x} = \O$
Aiming for a contradiction, suppose:
$G \cap \paren {A \setminus \set x} \ne \O$
Then by definition of the empty set there exists $g$ such that:
$g \in G \cap \paren {A \setminus \set x}$
Hence by definition of intersection:
$g \in G$
and:
$g \in A \setminus \set x$
Then by definition of set difference:
$g \in A$
Hence by definition of intersection:
$g \in A \cap G$
Then by $(3)$:
$x = g$
This by definition of singleton contradicts with $g \notin \set x$ obtained by definition of set difference.
Thus:
$G \cap \paren {A \setminus \set x} = \O$

Define $U = G \setminus \set x$ as an open set by $(4)$.

By $(5)$ and definition of set difference:

$y \in A'$

Then by $(8)$ and Characterization of Derivative by Open Sets there exists a point $q$ of $T$ such that:

$(10): \quad q \in A \cap U$

and:

$(11): \quad y \ne q$

By $(10)$ and definition of intersection:

$q \in A$

By $(11)$ and definition of singleton:

$q \notin \set y$

Then by definition of set difference

$(12): \quad q \in A \setminus \set y$

By definition of intersection:

$q \in U$

Then by $(12)$ and by definition of set difference

$q \ne x$

and:

$q \in A$

Then by definition of set difference:

$q \in A \setminus \set x$

and :

$q \in G$

Hence this contradicts with $(9)$ by definition of intersection.

Thus the result by Proof by Contradiction.

$\blacksquare$


Sources

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