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
\(\ds \) | \(\) | \(\ds \set x \subseteq A\) | ||||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds \forall y: \paren {y \in \set x \implies y \in A}\) | 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 |
$\blacksquare$
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:
- $x \in \set x \implies x \in S$
$\blacksquare$