Definitions

# Ordinal arithmetic

In the mathematical field of set theory, ordinal arithmetic describes the three usual operations on ordinal numbers: addition, multiplication, and exponentiation. Each can be defined in essentially two different ways: either by constructing an explicit well-ordered set which represents the operation or by using transfinite recursion. Cantor normal form provides a standardized way of writing ordinals. The so-called "natural" arithmetical operations retain commutativity at the expense of continuity.

The union of two disjoint well-ordered sets S and T can be well-ordered. The order-type of that union is the ordinal which results from adding the order-types of S and T. If two well-ordered sets are not already disjoint, then they can be replaced by order-isomorphic disjoint sets, e.g. replace S by S × {0} and T by T × {1}. Thus the well-ordered set S is written "to the left" of the well-ordered set T, meaning one defines an order on S $cup$ T in which every element of S is smaller than every element of T. The sets S and T themselves keep the ordering they already have. This addition is associative and generalizes the addition of natural numbers.

The first transfinite ordinal is ω, the set of all natural numbers. Let's try to visualize the ordinal ω + ω: two copies of the natural numbers ordered in the normal fashion and the second copy completely to the right of the first. If we write the second copy as {0' < 1' < 2', ...} then ω + ω looks like

0 < 1 < 2 < 3 < ... < 0' < 1' < 2' < ...
This is different from ω because in ω only 0 does not have a direct predecessor while in ω + ω the two elements 0 and 0' do not have direct predecessors. Here are 3 + ω and ω + 3:
0 < 1 < 2 < 0' < 1' < 2' < ...
0 < 1 < 2 < ... < 0' < 1' < 2'
After relabeling, the former just looks like ω itself while the latter does not: we have 3 + ω = ω. But ω + 3 is not equal to ω since the former has a largest element and the latter does not. So our addition is not commutative.

One can see for example that (ω + 4) + ω = ω + (4 + ω) = ω + ω.

The definition of addition can also be given inductively (the following induction is on β):

• α + 0 = α,
• α + (β + 1) = (α + β) + 1 (here, "+ 1" denotes the successor of an ordinal),
• and if δ is limit then α + δ is the limit of the α + β for all β < δ.

Using this definition, we also see that ω + 3 is a successor ordinal (it is the successor of ω + 2) whereas 3 + ω is the limit of 3 + 0 = 3, 3 + 1 = 4, 3 + 2 = 5, etc., which is just ω.

Zero is an additive identity α + 0 = 0 + α = α.

Addition is associative (α + β) + γ = α + (β + γ).

Addition is strictly increasing and continuous in the right argument:

$alpha < beta Rightarrow gamma + alpha < gamma + beta$
but the analogous relation does not hold for the left argument; instead we only have:
$alpha < beta Rightarrow alpha+gamma le beta+gamma$

Ordinal addition is left-cancellative: if α + β = α + γ, then β = γ. Furthemore, one can define left subtraction for ordinals βα: there is a unique γ such that α = β + γ. On the other hand, right cancellation does not work:

$3+omega = 0+omega = omega$ but $3 neq 0$
Nor does right subtraction, even when βα: there does not exist any γ such that γ + 42 = ω.

## Multiplication

The Cartesian product, S×T, of two well-ordered sets S and T can be well-ordered by a variant of lexicographical order which puts the least significant position first. Effectively, each element of T is replaced by a disjoint copy of S. The order-type of the Cartesian product is the ordinal which results from multiplying the order-types of S and T. Again, this operation is associative and generalizes the multiplication of natural numbers.

Here is ω·2:

00 < 10 < 20 < 30 < ... < 01 < 11 < 21 < 31 < ...
and we see: ω·2 = ω + ω. But 2·ω looks like this:
00 < 10 < 01 < 11 < 02 < 12 < 03 < 13 < ...
and after relabeling, this looks just like ω and so we get 2·ω = ω ≠ ω·2. Hence multiplication of ordinals is not commutative.

Distributivity partially holds for ordinal arithmetic: R(S + T) = RS + RT. However, the other distributive law (T + U)R = TR + UR is not generally true: (1 + 1) ·ω = 2·ω = ω while 1·ω + 1·ω = ω + ω which is different. Therefore, the ordinal numbers do not form a ring.

The definition of multiplication can also be given inductively (the following induction is on β):

• α·0 = 0,
• α·(β + 1) = (α·β) + α,
• and if δ is limit then α·δ is the limit of the α·β for all β < δ.

The main properties of the product are:

α·0 = 0·α = 0.

One is a multiplicative identity α·1 = 1·α = α.

Multiplication is associative (α·βγ = α·(β·γ).

Multiplication is strictly increasing and continuous in the right argument:
(α < β and γ > 0) $Rightarrow$ γ·α < γ·β,
but if one reverses the arguments the inequality does not work:
for example, 1 < 2 but 1·ω = 2·ω = ω.

Instead one gets αβ $Rightarrow$ α·γβ·γ.

There is a left cancellation law: If α > 0 and α·β = α·γ, then β = γ.
The example above shows that right cancellation does not work.

α·β = 0 $Rightarrow$ α = 0 or β = 0.

α·(β + γ) = α·β + α·γ (distributive law on the left). On the other hand, there is no distributive law on the right.

e.g. (ω + 1)·2 = ω + 1 + ω + 1 = ω + ω + 1 = ω·2 + 1 which is not ω·2 + 2.

There is also left division with remainder: for all α and β, if β > 0, then there are unique γ and δ such that α = β·γ + δ and δ < β. (This doesn't mean the ordinals are a Euclidean domain, however, since they aren't even a ring, and the Euclidean "norm" is ordinal-valued.)

Right division does not work: there is no α such that α·ω ≤ ωω ≤ (α + 1)·ω.

## Exponentiation

Exponentiation of well ordered sets is defined as follows. If the exponent is a finite set, the power is the product of iterated multiplication. For instance, ω2 = ω·ω using the operation of ordinal multiplication.

To generalize this to the case when the exponent is an infinite ordinal requires a different viewpoint. Note that ω·ω can be visualized as the set of functions from 2 = {0,1} to ω = {0,1,2,...}, ordered lexicographically with the least significant position first:

(0,0) < (1,0) < (2,0) < (3,0) < ... < (0,1) < (1,1) < (2,1) < (3,1) < ... < (0,2) < (1,2) < (2,2) < ...
Here for brevity, we have replaced the function {(0,k), (1,m)} by the ordered pair (k, m).

Similarly, for any finite exponent n, $omega^n$ can be visualized as the set of functions from n (the domain) to the natural numbers (the range). These functions can be abbreviated as n-tuples of natural numbers.

For $omega^omega$, we might try to visualize the set of infinite sequences of natural numbers. However, if we try to use any effectively defined ordering on this set, we find it is not well-ordered. Using the variant lexicographical ordering again, we restrict the set of sequences to those for which only a finite number of elements of the sequence are different from zero. This is naturally motivated as the limit of the finite powers of the base (similar to the concept of coproduct in algebra).

The lexicographical order on this set is a well ordering that resembles the ordering of natural numbers written in decimal notation, except with digit positions reversed, and with arbitrary natural numbers instead of just the digits 0-9:

(0,0,0,...) < (1,0,0,0,...) < (2,0,0,0,...) < ... <
(0,1,0,0,0,...) < (1,1,0,0,0,...) < (2,1,0,0,0,...) < ... <
(0,2,0,0,0,...) < (1,2,0,0,0,...) < (2,2,0,0,0,...)
< ... <
(0,0,1,0,0,0,...) < (1,0,1,0,0,0,...) < (2,0,1,0,0,0,...)
< ...

In general, any well ordered set B can be raised to the power of another well ordered set E, resulting in another well ordered set, the power BE: each elements is a function from E to B such that only a finite number of elements of the domain E map to an element larger than the least element of the range B; the order is lexicographic with the least significant position first.

We find $1^omega = 1$, $2^omega = omega$, $2^\left\{omega+1\right\} = omega cdot 2 = omega + omega$.

The order type of the power BE is the ordinal which results from applying ordinal exponentiation to the order type of the base B and the order type of the exponent E.

The definition of exponentiation can also be given inductively (the following induction is on β, the exponent):

• α0 = 1,
• αβ+1 = (αβα, and
• if δ is limit, then αδ is the limit of the αβ for all β < δ.

Properties of ordinal exponentiation (shared by exponentiation of natural numbers since they are the finite ordinals):

1. α0 = 1.
2. If 0 < α, then 0α = 0.
3. 1α = 1.
4. α1 = α.
5. αβ·αγ = αβ + γ.
6. (αβ)γ = αβ·γ.

But there are α, β, and γ for which

(α·β)γαγ·βγ. For instance, (ω·2)2 = ω2·2 ≠ ω2·4.

Ordinal exponentiation is strictly increasing and continuous in the right argument:

1. If γ > 1 and α < β, then γα < γβ.
2. If α < β, then αγβγ. Note, for instance, that 2 < 3 and yet 2ω = 3ω = ω.

Although different bases raised to the same power may yield the same result, it is impossible to obtain the same ordinal by raising the same base to two different powers unless the base is 0 or 1. That is, if α > 1 and αβ = αγ, then β = γ.

For all α and β, if β > 1 and α > 0 then there exist unique γ, δ, and ρ such that

α = βγ·δ + ρ
such that 0 < δ < β and ρ < βγ.

Warning: Ordinal exponentiation is quite different from cardinal exponentiation. For example, the ordinal exponentiation 2ω = ω, but the cardinal exponentiation $2^\left\{aleph_0\right\}$ is the cardinality of the continuum which is much larger than $aleph_0$. To avoid confusing ordinal exponentiation with cardinal exponentiation, one can use symbols for ordinals (e.g. ω) in the former and symbols for cardinals (e.g. $aleph_0$) in the latter.

## Cantor normal form

Ordinal numbers present a rich arithmetic. Every ordinal number α can be uniquely written as $omega^\left\{beta_1\right\} c_1 + omega^\left\{beta_2\right\}c_2 + cdots + omega^\left\{beta_k\right\}c_k$, where k is a natural number, $c_1, c_2, ldots, c_k$ are positive integers, and $beta_1 > beta_2 > ldots > beta_k$ are ordinal numbers (we allow $beta_k=0$). This decomposition of α is called the Cantor normal form of α, and can be considered the positional base-ω numeral system. The highest exponent $beta_1$ is called the degree of $alpha$, and satisfies $beta_1lealpha$. The equality $beta_1=alpha$ applies if and only if $alpha=omega^alpha$. In that case Cantor normal form does not express the ordinal in terms of smaller ones; this can happen as explained below.

A minor variation of Cantor normal form, which is usually slightly easier to work with, is to set all the numbers ci equal to 1 and allow the exponents to be equal. In other words, every ordinal number α can be uniquely written as $omega^\left\{beta_1\right\} + omega^\left\{beta_2\right\} + cdots + omega^\left\{beta_k\right\}$, where k is a natural number, and $beta_1 ge beta_2 ge ldots ge beta_k ge 0$ are ordinal numbers.

The Cantor normal form allows us to uniquely express—and order—the ordinals α which are built from the natural numbers by a finite number of arithmetical operations of addition, multiplication and “raising ω to the power of”: in other words, assuming

The set of finite arithmetical expressions of this form is called ε0 (epsilon nought). It is the smallest ordinal that does not have a finite arithmetical expression, and the smallest ordinal such that $varepsilon_0 = omega^\left\{varepsilon_0\right\}$, i.e. in Cantor normal form the exponent is not smaller than the ordinal itself. It is the limit of the sequence $omega$, $omega^omega$, $omega^\left\{omega^omega\right\}$, etc. The ordinal ε0 is important for various reasons in arithmetic (essentially because it measures the proof-theoretic strength of the first-order Peano arithmetic: that is, Peano's axioms can show transfinite induction up to any ordinal less than ε0 but not up to ε0 itself).

The Cantor normal form also allows us to compute sums and products of ordinals: to compute the sum, for example, one needs merely know that $omega^\left\{beta\right\} c+omega^\left\{beta\text{'}\right\} c\text{'}$ is $omega^\left\{beta\text{'}\right\}c\text{'}$ if $beta\text{'}>beta$ (if $beta\text{'}=beta$ one can obviously rewrite this as $omega^\left\{beta\right\} \left(c+c\text{'}\right)$, and if

To compare two ordinals written in Cantor normal form, first compare $beta_1$, then $c_1$, then $beta_2$, then $c_2$, etc.. At the first difference, the ordinal which has the larger component is the larger ordinal. If they are the same until one terminates before the other, then the one which terminates first is smaller.

## Natural operations

The natural sum and natural product operations on ordinals were defined in 1906 by Gerhard Hessenberg, and are sometimes called the Hessenberg sum (or product). (Jacobsthal defined a natural power operation in 1907, which is very rarely used.) They are also sometimes called the Conway operations, as they are just the addition and multiplication (restricted to ordinals) of Conway's field of surreal numbers. They have the advantage that they are associative and commutative, and natural product distributes over natural sum. The cost of making these operations commutative is that they lose the continuity in the right argument which is a property of the ordinary sum and product. The natural sum of α and β is sometimes denoted by α#β, and the natural product by a sort of doubled × sign. To define the natural sum of two ordinals, consider once again the disjoint union S$cup$T of two well-ordered sets having these order types. Start by putting a partial order on this disjoint union by taking the orders on S and T separately but imposing no relation between S and T. Now consider the order types of all well-orders which extend this partial order: the least upper bound of all these ordinals (which is, actually, not merely a least upper bound but actually a greatest element) is the natural sum. Also, inductively, we can define the natural sum of α and β (by simultaneous induction on α and β) as the smallest ordinal greater than the natural sum of α and γ for all γ<β and of γ and β for all γ<α.

The natural sum is associative and commutative: it is always greater or equal to the usual sum, but it may be greater. For example, the natural sum of ω and 1 is ω+1 (the usual sum), but this is also the natural sum of 1 and ω.

To define the natural product of two ordinals, consider once again the cartesian product S×T of two well-ordered sets having these order types. Start by putting a partial order on this cartesian product by using just the product order (compare two pairs if and only if each of the two coordinates is comparable). Now consider the order types of all well-orders which extend this partial order: the least upper bound of all these ordinals (which is, actually, not merely a least upper bound but actually a greatest element) is the natural product. There is also an inductive definition of the natural product (by mutual induction), but it is somewhat tedious to write down and we will not do so (see the article on surreal numbers for the definition in that context, which, however, uses Conway subtraction, something which obviously cannot be defined on ordinals).

The natural product is associative and commutative and distributes over the natural sum: it is always greater or equal to the usual product, but it may be greater. For example, the natural product of ω and 2 is ω·2 (the usual product), but this is also the natural product of 2 and ω.

Yet another way to define the natural sum and product of two ordinals α and β is to use the Cantor normal form: one can find a sequence of ordinals γ1 > ... > γn and two sequences (k1, ... , kn) and (j1, ... , jn) of natural numbers (including zero, but satisfying ki+ji >0 for all i) such that

• $alpha = omega^\left\{gamma_1\right\}cdot k_1 + cdots +omega^\left\{gamma_n\right\}cdot k_n$
• $beta = omega^\left\{gamma_1\right\}cdot j_1 + cdots +omega^\left\{gamma_n\right\}cdot j_n$

and defines

• $alpha #beta = omega^\left\{gamma_1\right\}cdot \left(k_1+j_1\right) + cdots +omega^\left\{gamma_n\right\}cdot \left(k_n+j_n\right)$

## References

• Jech, Thomas, 2003. Set Theory: The Third Millennium Edition, Revised and Expanded. Springer. ISBN 3-540-44085-2.
• Kunen, Kenneth, 1980. Set Theory: An Introduction to Independence Proofs. Elsevier. ISBN 0-444-86839-9.