Epsilon Induction
Theorem
Let $A$ be a class.
Let $\Bbb U$ denote the universe.
- $\left({\forall x: \left({x \subseteq A \implies x \in A}\right)}\right) \implies A = \Bbb U$
Proof
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Suppose that
- $\forall x: \left({x \subseteq A \implies x \in A}\right)$
In particular: to results, although most are up only for sets.
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Consider $\Bbb U \setminus A$.
Aiming for a contradiction, suppose that:
- $\Bbb U \setminus A \ne \varnothing$
Then by Axiom of Foundation (Strong Form), we have that:
- $\exists x \notin A: \left({x \cap \left({\Bbb U \setminus A}\right)}\right) = \varnothing$
But:
| \(\displaystyle \left({x \cap \left({\Bbb U \setminus A}\right)}\right)\) | \(=\) | \(\displaystyle \left({\left({x \cap \Bbb U}\right) \setminus A}\right)\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \left({x \setminus A}\right)\) | $\quad$ | $\quad$ |
Thus:
- $x \setminus A = \varnothing$
So from Set Difference with Superset is Empty Set:
- $x \subseteq A$
Thus by hypothesis:
- $x \in A$
contradicting the fact that $x \notin A$.
Therefore we can conclude that
- $\left({\Bbb U \setminus A}\right) = \varnothing$
and so from Set Difference with Superset is Empty Set:
- $\Bbb U \subseteq A$
Furthermore, by definition of universe:
- $A \subseteq \Bbb U$
so by definition of set equality:
- $A = \Bbb U$
$\blacksquare$
Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 5.24$
