Singleton Equality
From ProofWiki
Theorems
Let $x$ and $y$ be sets.
Then:
- $\left\{{x}\right\} \subseteq \left\{{y}\right\} \iff x = y$
- $\left\{{x}\right\} = \left\{{y}\right\} \iff x = y$
Proof
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \left\{ {x}\right\} \subseteq \left\{ {y}\right\}\) | \(\iff\) | \(\displaystyle x = y\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of Subset | ||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle \forall z: \left({z \in \left\{ {x}\right\} \implies z \in \left\{ {y}\right\} }\right)\) | \(\iff\) | \(\displaystyle \forall z: \left({z = x \implies z = y}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of Singleton | ||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle \forall z: \left({z = x \implies z = y}\right)\) | \(\iff\) | \(\displaystyle x = y\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Equality implies Substitution |
The chained biconditionals in the first part satisfy the first part of the theorem.
Then:
- $x = y \implies \{ x \} = \{ y \}$ (Substitutivity of equality)
- $\{ x \} = \{ y \} \implies \{ x \} \subseteq \{ y \}$ (Theorem???)
- $\{ x \} = \{ y \} \implies x = y$
- $\{ x \} = \{ y \} \iff x = y$
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$\blacksquare$
Source
- Willard Quine: Set Theory and Its Logic (1963): $\S 7.7$
