Singleton of Element is Subset

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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:

$\set x = \set {y \in S: y = x}$

From Subset of Set with Propositional Function:

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

Hence:

$\set x \subseteq S$

$\Box$


Sufficient Condition

Let $\set x \subseteq S$.

From the definition of a subset:

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

$\blacksquare$