# Singleton of Element is Subset

## Theorem

Let $S$ be a set.

Let $\set x$ be the singleton of $x$.

Then:

$x \in S \iff \set x \subseteq S$

## Proof 1

 $\displaystyle$  $\displaystyle \set x \subseteq A$ $\displaystyle$ $\leadstoandfrom$ $\displaystyle \forall y: \paren {y \in \set x \implies y \in A}$ Definition of Subset $\displaystyle$ $\leadstoandfrom$ $\displaystyle \forall y: \paren {y = x \implies y \in A}$ Definition of Singleton $\displaystyle$ $\leadstoandfrom$ $\displaystyle x \in A$ Equality implies Substitution

$\blacksquare$

## Proof 2

### Necessary Condition

Let $x \in S$.

We have:

$\set x = \set {y \in S: y = x}$
$\set {x \in S: \map P x} \subseteq S$

Hence:

$\set x \subseteq S$

$\Box$

### Sufficient Condition

Let $\set x \subseteq S$.

From the definition of a subset:

$x \in \set x \implies x \in S$

$\blacksquare$