Epsilon Induction

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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:

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

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