A well-ordering on a set is an order relation where every nonempty subset has a least element. Well-orderings are total orders, which means that any 2 elements are always comparable (as the pair must have a least element among them). Two well-ordered sets are called isomorphic if there is a strictly monotone bijection between them (=which preserves the order, and thus all properties of this order)

The natural order on the set ℕ is a well-ordering (this is an
expression of the recursion axiom). Thus the concept of
well-ordering generalizes the order relation on ℕ. There are many
well-orderings which are not isomorphic to that of ℕ.

The restriction of a well-ordering to any subset is also a
well-ordering. So any subset of ℕ is well-ordered. In fact the
well-orderings on finite sets are the total orders, and they are
isomorphic as soon as they have the same number of elements.

But there are well-orderings bigger than ℕ too. The description of
any well-ordering is always the same: it starts from the smallest
element (0), then the smallest above 0 (1), then the smallest
above 1 (2), and so on up to infinity; then the smallest after all
natural numbers, then the next one, then the next,... and so on to
infinity, then the next... and so on and so forth, even possibly
beyond any description, but this can stop at any time. See Chaitin's
speech for a more explicit description.

Contrary to the general study of order relations, for
well-orderings we generally need the strict (antireflexive) order.
So we shall write < for the strict order, and ≤ for the large
order.

An initial segment of a well-ordered set E is a subset of E of
the form {x∈E | x<y} for some y∈E..

A fundamental property is that for any two well-ordered sets
E and F, there exists a unique isomorphism that is either between
E and F, or between E and an initial segment of F, or between an
initial segment of E and F.

(The proof is a little difficult technically; it can be helped by
the below construction that replaces Zorn's lemma).

An *ordinal* can be defined as an isomorphism class of
well-ordered sets. The traditional study of ordinals, and thus of
well-orderings, formalizes each ordinal by its only representative
whose strict well-ordering is given by the membership predicate (
∈ ) and that are transitive (x∈A∈E ⇒ x∈E). In such conditions we
have {x∈E | x<y}=y, which may simplify things; the large
well-ordering is then the inclusion. But the proof of existence of
such an ordinal isomorphic to any well-ordered set, requires the
schema of replacement to be proven. Here we shall call ZF-ordinals
the transitive sets well-ordered by ∈.

In fact most properties of well-orderings and their isomorphism
classes can as well be developed without such a representative,
thus without the axiom schema of replacement, though it may be a
little bit more tedious.

Any set of ordinals (as classes of well-ordered sets) is well-ordered by defining "equality" by the isomorphism, and the order between non-isomorphic well-orderings is given by which of both sets is isomorphic to an initial segment of the other.

In the case of ZF-ordinals this all simplifies as the same
predicate ∈ defines the strict well-ordering inside and between
all ordinals.

For any ordinal there is a bigger one obtained by adding an
element after it. In fact, for any well-ordered set E, we can
define the "next" well-ordered set as the set of all initial
segments of E plus E itself. That is E ∪{E} in the case of
ZF-ordinals. And for any set of ordinals we can take their
"union", that is the smallest ordinal that is larger than them
all. This is simply the union as sets in the case of ZF-ordinals.

Among all ordinals in bijection with a given ordinal, there is a
smallest one. Such a smallest ordinal inside a cardinal (an
ordinal not in bijection with any initial segment of itself) will
be chosen to represent this cardinal.

We shall show later that the axiom of choice implies the existence of a well-ordering on all sets. This is a bit difficult to prove but the converse is simple:

If all sets can be well-ordered then the axiom of choice can be deduced in the following way. If ∀x∈E, ∃y∈F, xRy, then with a well-ordering on F we can define f from E to F by f(x) = the smallest y such that xRy.

Thus without the axiom of choice we cannot assume that
ZF-ordinals which are cardinals represent all cardinals.

If we could well-order any set (thus, if we accept the axiom of
choice) then we could simply argue that the class of
cardinal-ordinals does not form a set because this has been shown
for cardinals in general.

For example, ℕ is not the biggest cardinal because P(ℕ) is bigger
and any well-ordering on it gives an ordinal with cardinality
larger than ℕ.

Otherwise the traditional method also uses the axiom schema of
replacement, making it unclear which assumptions are needed to
show that the class of cardinal-ordinals is not a set. However we
can still deduce it from just the powerset axiom, without either
the axiom of choice nor the axioms of replacement, but this is a
little bit more subtle. Here it goes:

For any well-ordered set X, take the set A={R⊂X×X | R is a
well-ordering of a subset of X}. Then the quotient of A by the
relation of isomorphism, is naturally a well-ordered set whose
ordinal is the smallest ordinal with cardinality strictly larger
than X. Because if it was in bijection with a subset of X then it
would define a well-ordering of this subset...

This is the smallest uncountable cardinal-ordinal, thus a
candidate as an intermediate cardinal between ℕ and P(ℕ).

According to the above, it can be explicitly constructed out of
P(ℕ), as the quotient of a well-defined subset of P(ℕ) ≈
P(ℕ×ℕ) by a well-defined equivalence relation. Thus there
exists a surjection from P(ℕ) to it.

Using the axiom of choice it can be represented as a subset of
P(ℕ).

The continuum hypothesis, that is undecidable, is takes the form
of the claim "There is a bijection exists between Aleph1 and
P(ℕ)".

This is a construction of somethings that aims to behave as a
nonstandard set of real numbers (a "nonstandard set of real
numbers" basically means the system of "real numbers" in a model
of set theory or second-order arithmetic whose "set of natural
numbers" is non-standard), but based on ordinals instead of normal
model-theoretical tools.

However the "explicit" character of this construction happens at
the expense of ordinary properties: only the basic operations seem
to be defined (but it does not even seem clear how it works and
whether it will keep working in all cases) but not more
complicated ones: while the multiplication seems to be defined
(?), the decomposition of numbers in prime factors is not (as far
as I know) since this decomposition requires logical tools not
contained in this construction.

So I won't present this construction here.

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.**

The proof will make use of Tarski's Fixed Point
Theorem (3.5).

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(∅) = {∅}

∀

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.

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

∀Then ∀A⊂C, ∀y∈K, ((∃x∈A,y⊂x)⇒y⊂∪A) and ((∀x∈A,x⊂y)⇒∪A⊂y)x∈C, ∀x',z∈K, x' = x∪{φ(x)}⇒((z⊂x⊂x')∨(x⊂z)). Now ifx⊂zand φ(x)∉zthen (∀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.

ThusCis stable byf, thusC=K, thus ∀x,y∈K, (x⊂y ∨ y⊂x).

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

∀

(To be continued).