Power Set of Doubleton
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Theorem
Let $x, y$ be distinct objects.
Then the power set of the doubleton $\set {x, y}$ is:
- $\powerset {\set {x, y} } = \set {\O, \set x, \set y, \set {x, y} }$
Proof
By definition of a subset:
- $\set x , \set y, \set {x, y} \subseteq \set {x, y}$
Let $A \subseteq \set {x, y}$:
- $A \ne \set x, \set y, \set {x, y}$
From set equality:
- $\set {x, y} \nsubseteq A$
From Doubleton of Elements is Subset:
- either $x \notin A$ or $y \notin A$.
Without loss of generality assume that $x \notin A$.
From Intersection With Singleton is Disjoint if Not Element:
- $A \cap \set x = \O$
From Subset of Set Difference iff Disjoint Set:
- $A \subseteq \set {x, y} \setminus \set x$
From Set Difference of Doubleton and Singleton is Singleton:
- $A \subseteq \set y$
From set equality:
- $\set y \nsubseteq A$
From Singleton of Element is Subset:
- $y \notin A$.
From Intersection With Singleton is Disjoint if Not Element:
- $A \cap \set y = \O$
From Subset of Set Difference iff Disjoint Set:
- $A \subseteq \set y \setminus \set y$
From Set Difference with Self is Empty Set:
- $A \subseteq \O$
From Empty Set is Subset of All Sets:
- $\O \subseteq A$
From set equality:
- $A = \O$
It follows that:
- $\powerset {\set {x, y} } = \set {\O, \set x, \set y, \set {x, y} }$
$\blacksquare$