# 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

- $\forall x: \left({x \subseteq A \implies x \in A}\right)$

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*: $\S 5.24$