Power Set of Singleton
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Theorem
Let $x$ be an object.
Then the power set of the singleton $\set x$ is:
- $\powerset {\set x} = \set {\O, \set x}$
Proof
From Empty Set is Subset of All Sets:
- $\O \in \powerset {\set x}$
Let $A \in \powerset {\set x}$ such that $A \ne \O$
That is:
\(\ds \) | \(\) | \(\ds A \subseteq \set x \land A \ne \O\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \) | \(\) | \(\ds A \subseteq \set x \land \exists y : y \in A\) | Definition of Empty Set | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \) | \(\) | \(\ds A \subseteq \set x \land \exists y : y \in A \land y \in \set x\) | Definition of Subset | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \) | \(\) | \(\ds A \subseteq \set x \land \exists y : y \in A \land y = x\) | Definition of Singleton | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \) | \(\) | \(\ds A \subseteq \set x \land x \in A\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \) | \(\) | \(\ds A \subseteq \set x \land \set x \subseteq A\) | Singleton of Element is Subset | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \) | \(\) | \(\ds A = \set x\) | Definition of Set Equality |
So a subset of $\set x$ is either $\O$ or $\set x$.
$\blacksquare$