Epsilon Induction

Formulation 1
Let $A$ be a class. Let $\Bbb U$ denote the universe.
 * $\paren {\forall x: \paren {x \subseteq A \implies x \in A} } \implies A = \Bbb U$

Formulation 2
Let $\map P x$ be a well-formed formula containing $x$ as a free variable.

Suppose that, from the truth of $\map P x$ for every member of a set $a$, the truth of $\map P a$ follows.

That is:
 * $\paren {\forall x \in a : \map P x} \implies \map P a$

Then $\map P x$ is true for every set.

Proof of Formulation 1
Suppose that
 * $\forall x: \paren {x \subseteq A \implies x \in A}$

Consider $\Bbb U \setminus A$.


 * $\Bbb U \setminus A \ne \O$
 * $\Bbb U \setminus A \ne \O$

Then by Axiom of Foundation (Strong Form), we have that:
 * $\exists x \notin A: x \cap \paren {\Bbb U \setminus A} = \O$

But:

Thus:
 * $x \setminus A = \O$

So from Set Difference with Superset is Empty Set:
 * $x \subseteq A$

Thus :
 * $x \in A$

contradicting the fact that $x \notin A$.

Therefore we can conclude that
 * $\Bbb U \setminus A = \O$

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:
 * $A = \Bbb U$

Proof of Formulation 2
there is a set $b$ for which $\neg \map P b$.

Define $\map f x = \set {y \in x : \neg \map P y}$.

From the contrapositive of the assumption:


 * $\neg \map P x \implies \exists y \in x : \neg \map P y$

it follows that $\map f x$ is non-empty whenever $\neg \map P x$.

Define $\map g p = \bigcup \set{f(q) : q \in p}$.

Whenever $\exists x \in p : \neg \map P x$:
 * $\exists y \in \map g p : \neg \map P y$

By the Principle of Recursive Definition, define a sequence $c_n$ such that:
 * $c_0 = \set{b}$
 * $c_{n+1} = \map g {c_n}$

Let $c = \bigcup \set{c_n : n \in \omega}$.

$c$ is non-empty, as $b \in c_0 \subseteq c$.

By the Axiom of Foundation, $\exists x_0 \in c : \paren{\forall x \in d : x \notin c}$.

By construction of $c_n$:
 * $\neg \map P {x_0}$

so by the contrapositive of the assumption:
 * $\exists y_0 \in x_0 : \neg \map P {y_0}$

But:
 * $x_0 \in c_{n_0}$

for some natural number $n_0$.

Then:
 * $y_0 \in c_{n_0 + 1}$

Therefore:
 * $y_0 \in c$

a contradiction.

Thus, by Reductio ad Absurdum, $\map P x$ is true for all $x$.