Singleton is a Subset
From ProofWiki
Theorem
In the following theorem, $x$ must be a set and $A$ can be any class, proper or not.
- $\forall x: \left( \{ x \} \subseteq A \iff x \in A \right)$
Proof
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\) | \(\displaystyle \{ x \} \subseteq A\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\iff\) | \(\displaystyle \forall y \left( y \in \{ x \} \implies y \in A \right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition:Subset | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\iff\) | \(\displaystyle \forall y ( y = x \implies y \in A )\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition:Singleton | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\iff\) | \(\displaystyle x \in A\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Equality implies Substitution |
$\blacksquare$
