• Theorem 1: Let S(A) be the family of all initial segments of A. S(A) is ordered by set inclusion. The function f that maps an element x to its initial segment is one-to-one and onto; i.e., f:A→S(A). Furthermore f preserves order so A is similar to S(A).

• Theorem 2: Suppose A is a well-ordered set and B is a subset of A such that there exists a similarity mapping f of A into B. Then for every element x of A, x≤f(x).

  Proof: Let D={x|f(x)≤x}. If D=φ, the empty set, then the theorem is true.

  Suppose D is not empty. Then, since A is a well-ordered set, D has a first element d. Since d ∈ D, f(d)≤d. But, since f is a similarity mapping which preserves order f(f(d))≤f(d) and hence f(d) ∈ D. But f(d)≤d so there is a contradiction with the assumption that d is the first element of D. Therefore D must be empty.

• The next theorem is a very useful one.

  Theorem 3: A well-ordered set cannot be similar to one of its initial segments.

  Proof: Let A be a well-ordered set and f a similarity mapping of A onto the initial segment s(x) for some x ∈ A. By definition all elements of s(x) are less than x. Therefore f(x)≤x. But this contradict Theorem 2. Therefore a well-ordered set cannot be similar to one of its initial segments.

• Theorem 4: If A and B are well-ordered sets then either A and B are similar or one is similar to an intial segment of the other.

  Proof: Consider the elements of each set whose initial sets are similar to the initial sets of elements in the other set; i.e., let

  
  S = {x|x∈A and s(x) is similar to s(y) for some y∈B}
  T = {y|y∈B and s(y) is similar to s(x) for some x∈A}
   

  ---

  For example, suppose A = {1,2,3,4} and B = {a,b,c,d,e}. Then S = {1,2,3,4} and T = {a,b,c,d}.

  ---

  Continuation with the proof of Theorem 4. Claim: S is similar to T.

  Proof of Claim: Consider an element of x. By definition there is an element y ∈ B such that s(x) is similar it s(y). This y is unique because otherwise there would be two initial segments of B, say s(y₁) and s(y₂) which were both similar to s(x) and hence similar to each other. But this implies that s(y₁)=s(y₂) and hence y₁=y₂. Likewise x the element of A such that s(x) is similar to y ∈ B is unique. This correspondence defines a one-to- one relationship between the elements of S and T. Thus S and T are similar.

  There are three possibilities:

  • 1. If S=A and T=B then A is similar to B.
  • 2. If S=A and T=s(b) for some b ∈ B, then A is similar to an intial segment of B.
  • 3. If T=B and S=s(a) for some a ∈ A, then B is similar to an initial segment of A.

  The fourth possibility of S=s(a) for some a ∈ A and T=s(b) for some b ∈ B cannot occur. This would imply that a ∈ S, since its initial segment is similar to the initial segment of an element (E S) of B, but since S=s(a) this would have a belonging to its own initial segment. Thus there would be a contradiction.

Definition: The order type of a well-ordered set is the family of well-ordered sets it is similar to. In other words, the order type of a well-ordered set A, denoted ord(A), is the equivalence class A belongs to. If ord(A)=λ then A ∈λ.

There is a particular sequence of ordered sets that are appropriate to use to represent order types. For finite sets the following are "standard" order types:

φ, {φ, {φ}}, {φ,{φ},{φ, {φ}}}, …
 

where φ is used to denote the empty set.

Because this gets tedious to write the following labels are adopted:

φ→ 0
{φ}→ 1
{φ,{φ}}→ 2
{φ,{φ},{φ, {φ}}}→ 3

...................

 

Thus

φ → 0
{0} → 1
{0, 1} → 2
{0, 1, 2} → 3
 

The order type of the natural numbers, {0, 1, 2, 3, … } is denoted as omega, ω.

The Ordering of Ordinals

The ordering of ordinal numbers is achieved through the following construction. If λ and μ are ordinal numbers and A and B are well-ordered sets such that ord(A)=λ and ord(B)=μ then:

λ < μ if A is similar to an initial segment of B
λ = μ if A is similar to B
λ > μ if B is similar to an initial segment of A
λ ≤ μ if λ ≤ μ or λ = μ
λ ≥ μ if λ > μ or λ = μ.

• Theorem 5: Any set of ordinals is well-ordered by ≤.

• Theorem 6: Let s(λ ) be the set of all ordinals less than λ . Then ord(s(λ ))=λ .

• Theorem 7: Every ordinal has an immediate successor.

• Definition of a limit element: An element of a well-ordered set is a limit element if it does not have an immediate predecessor and it is not the first element. A limit number is an ordinal which is a limit element of the ordinals.

• Theorem 8: For any infinite ordinal λ there exists ordinal numbers μ and n such that μ is a limit number and n is a finite ordinal.

Addition of Ordinals

Definition of concatenation of well-ordered sets: Let A and B be well-ordered sets. Then (A ; B) is the union of A and B ordered by the relation R where any two elements x and y are ordered according their relation in A or B if they both belong to A or both belong to B and x Definition of addition of ordinals: For any two ordinals λ and μ let A and B be well-ordered sets such that ord(A)=λ and ord(B)=μ . Then

λ +μ = ord((A ; B)).
 

For this definition to be proper it must be established that λ+μ is independent of the representative A and B used in the construction.

Addition is not necessarily commutative; λ+μ is not necessarily the same as μ+λ . For example, 1+ω is not the same as ω+1; i.e., ω+1 = ord({0,1,2,...; 0}) but 1+ω is ord({0,0,1,2,3,...})=ω.

Addition is associative; (λ +μ)+ ν = λ +(μ + ν). There also exists an additive identity; i.e., the empty set φ.

• Theorem 9: The immediate successor of any ordinal λ is λ +1.

Multiplication of Ordinals

Definition of the cartesian product of well-ordered sets:

Let A and B be well-ordered sets. Then (A X B) is the cartesian product of A and B, order pairs (a,b) where a ∈ A and b ∈ B, ordered by the relation R where any two elements (x,y) and (z,w) are ordered according to

(x,y) ≤ (z,w) if y≤w in B or y=w and x≤z in A.
 

This is called a reverse lexicographic ordering.

Definition of multiplication of ordinals:

Definition of multiplication of ordinals: For any two ordinals λ and μ let A and B be well-ordered sets such that ord(A)=λ and ord(B)=μ . Then

λ*μ= ord((A × B)).
 

Again, for this definition to be proper it must be established that λ*μ is independent of the representative A and B used in the construction.

Multiplication is not commutative. For example, 2*ω is not the same as ω*2; i.e., 2=ord({a,b}) and ω=ord({1,2,..}) so

2*ω = ord({(a,1),(b,1),(a,2),(b,2),...}) = ω
but ω*2 = ord({(1,a),(2,a)...; (1,b),(2,b),...})
 

However, multiplication is associative.

In the folowing let card(A) stand for the cardinality of a set A.

• Theorem 10: Let A and B be well ordered sets. Then card(A)≤card(B) implies that ord(A)≤ord(B), and ord(A)≤ord(B) implies that card(A)≤card(B).

The order of the ordinals is 0,1,2,3,..., then ω,ω+1,ω+2,... then ω²,ω²+1, ω²+2,... to ω³ and so on to ω^ω.

From there the sequence goes on to ω to the power of ω^ω.

Next comes (ω to the power of ω^ω) to the power of ω
and on to (ω to the power of ω to the power of ω … .

But there is always a successor to any ordinal.

Definition of Exponentiation of Ordinals

Definition of functions from one well-ordered set to another: Let A and B be well-ordered sets. Then A^B is the set of all functions from A to B.

The Burali-Forti Paradox: The concept of the set of all ordinal numbers is contradictory.

Proof not given.

Modeling ordinal arithmetic in the programming language LISP is convenient because a list is exactly a well-ordered set. For any list one can construct the corresponding ordinal by creating a list that begins with "0" and ends with a cardinal that is one less than the length of the list. The middle of the ordinal list is just the ellipsis "…". Infinite ordinals are ones in which the … has no successor (other than the delimiters ";" and ")".