Singleton Equality

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Theorems

Let $x$ and $y$ be sets.

Then:

$\set x \subseteq \set y \iff x = y$
$\set x = \set y \iff x = y$


Proof

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

$\Box$


Then:

\(\ds x\) \(=\) \(\ds y\)
\(\ds \leadsto \ \ \) \(\ds \set x\) \(=\) \(\ds \set y\) Substitutivity of Equality
\(\ds \leadsto \ \ \) \(\ds \set x\) \(\subseteq\) \(\ds \set y\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds y\) by the first part

$\blacksquare$


Sources