Singleton of Element is Subset

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Let $S$ be a set.

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


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

Proof 1

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


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:

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


$\set x \subseteq S$


Sufficient Condition

Let $\set x \subseteq S$.

From the definition of a subset:

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