## An alternative to Zorn's lemma

This page is a draft. The bulk of it was moved to 4.6 The following theorem is independent of the axiom of choice, but
the results in set theory traditionally deduced from Zorn's lemma
can instead be deduced from this together with an elementary
expression of the axiom of choice.

**Theorem**. *Let **φ a function whose domain D is a
set of sets, such that*∀*x*∈*D, **φ*(*x*)∉*x.
Then there exists a unique subset K*⊂*D satisfying the
following properties*

- The restriction φ
_{|K} is injective
- ∀x∈K, x⊂φ[K]

- The binary relation on K defined by (x,y)↦(φ(x)∈y)
is a strict well-ordering
- φ[K]∉D.

**Proof.**

**Note : I'm not satisfied with this proof that is quite long
and complicated. I hope to find a more elegant proof (through
more interesting intermediate steps) in the future. In particular I'll surely re-write it as a consequence of Tarski's Fixed Point
Theorem.**

Let *f* be the function from P(D) to itself defined by:

- ∀
*x*∈*D*, *f*({*x*}) = {*x*∪{φ(*x*)}}
if *x*∪{φ(*x*)}∈ D, or ∅ otherwise
- For any other A⊂D, f(A) = {∪A} if ∪A ∈ D, or ∅ otherwise. In
particular, f(∅) = {∅}

Then let *K* be the smallest subset of *D* stable by *f*
(such that ∀*A*⊂*K*, f(*A*)⊂*K*). By the fixed
point theorem, ∀*x*∈*D*, *x*∈*K*⇔(x=∅∨(∃*y* ∈ *K*, *x*
= *y*∪{φ(*y*)})∨(∃*A* ⊂
K, x = ∪A)).

∀*x*∈*K*,∀*A*⊂*x*, (∀*y*,*y'*∈*K*,
(*y*⊂*A* ∧ (*y*∪{φ(*y*)}= *y*'⊂*x*))⇒*y*'⊂*A*))⇒*A*=*x*
because

If (∀*y*,*y*'∈*K*, ((*y*⊂*A*)∧
*y*∪{φ(*y*)}= y'⊂x)⇒y'⊂A)), then the set C={z∈K|
z⊂x⇒z⊂A}is stable by f. Thus C=K. But x∈K, thus x⊂x⇒x⊂A. Finally
A=x.

∀*x,y*∈*K*, (*x*⊂y ∨ *y*⊂*x*) because

Let us show that *C*={*x*∈*K*|∀*y*∈*K*,
*x*⊂*y*∨*y*⊂*x*}is stable by *f*. The
stability by unions is straightforward :

∀*A*⊂*C*, ∀*y*∈*K*, ((∃*x*∈*A*,
*y*⊂*x*)⇒*y*⊂∪*A* ) and ((∀*x*∈A, *x*⊂*y*)⇒∪*A*⊂*y*)

Then ∀*x*∈*C*, ∀*x*',*z*∈*K*, x' =
x∪{φ(x)}⇒((*z*⊂*x*⊂*x*')∨(*x*⊂*z*)). Now
if *x*⊂*z* and φ(*x*)∉*z* then (∀y,y'∈K, (y⊂x
∧ y∪{φ(y)}= y'⊂z)⇒(x=y'∨x⊄y')⇒y'⊂x), so that x=z according to the
above, and z⊂x'. In the other case of x⊂z and φ(x)∈z we have x'⊂z.
Finally x'∈C.

Thus *C* is stable by *f*, thus *C*=*K*,
thus ∀*x,y*∈*K*, (x⊂y ∨ y⊂x).

∀ *x,y*∈*K* , (x⊂y ∧ *x*∪{φ(*x*)}∉*K*)
⇒x=y because

∀z,z*'*∈*K*, (*z*⊂*x* ∧
(z∪{φ(z)}= z'⊂*y*))⇒ (z'=x ∨ x⊄z') ⇒ *z*'⊂*x*.

The restriction φ_{|K} is injective because

∀*x*⊂*y*∈*K*, φ(x)=φ(y)⇒φ(x)∉y ⇒(*x*∪{φ(*x*)}∉*K*
∨ *y* = *x*)⇒x=y.

(To be continued).

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