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

This page is beyond the scope of ZFC, and should not be used in anything other than the theory in which it resides.

If you see any proofs that link to this page, please insert this template at the top.

If you believe that the contents of this page can be reworked to allow ZFC, then you can discuss it at the talk page.

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:

\(\displaystyle \left({x \cap \left({\Bbb U \setminus A}\right)}\right)\)

\(=\)

\(\displaystyle \left({\left({x \cap \Bbb U}\right) \setminus A}\right)\)

\(\displaystyle \)

\(=\)

\(\displaystyle \left({x \setminus A}\right)\)

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: $\S 5.24$