# 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

\(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \set x \subseteq \set y\) | ||||||||||

\(\displaystyle \) | \(\leadstoandfrom\) | \(\displaystyle \forall z:\) | \(\displaystyle \paren {z \in \set x \implies z \in \set y}\) | Definition of Subset | |||||||||

\(\displaystyle \) | \(\leadstoandfrom\) | \(\displaystyle \forall z:\) | \(\displaystyle \paren {z = x \implies z = y}\) | Definition of Singleton | |||||||||

\(\displaystyle \) | \(\leadstoandfrom\) | \(\displaystyle \) | \(\displaystyle x = y\) | Equality implies Substitution |

$\Box$

Then:

\(\displaystyle x\) | \(=\) | \(\displaystyle y\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \set x\) | \(=\) | \(\displaystyle \set y\) | Substitutivity of Equality | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \set x\) | \(\subseteq\) | \(\displaystyle \set y\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x\) | \(=\) | \(\displaystyle y\) | by the first part |

$\blacksquare$

## Sources

- 1963: Willard Van Orman Quine:
*Set Theory and Its Logic*: $\S 7.7$