Power Set of Singleton

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 $\set x$.

$\blacksquare$