Ordinal Fractions

Question

Is any fraction $\,{x}\big/ {y}\,$ an ordinal number and if so, does ordinal $\,1 = \big\lbrace0,\dots,y - 1\big/y\big\rbrace\,$ instead of $\,\left\lbrace0\right\rbrace\,$?

"If (X, <=) is a well ordered set with ordinal number x, then the set of all ordinals < x is order isomorphic to X. This provides the motivation to define an ordinal as the set of all ordinals less than itself. John von Neumann defined a set x to be an ordinal number iff

1. If y is a member of x, then y is a proper subset of x.

2. If y and z are members of x, then one of the following is true: y = z , y is a member of z, or z is a member of y.

3. If y is a nonempty proper subset of x, then there exists a z member of x such that the y intersection z is empty." (http://mathworld.wolfram.com/OrdinalNumber.html)

Answer 1

Well, $1$ is usually defined to equal $\{0\}$, not $\{0,1\}$, so there's many levels on which I'm not quite following. But if your question is whether fractions of ordinal numbers make sense, the answer is yes - kind of.

In particular, if you're willing to move from the usual Cantorian definitions of addition and multiplication of ordinal numbers to the Hessenberg operations, then it is indeed possible to make sense of ordinal fractions. One way of doing this is to view each ordinal as the corresponding surreal number and to just use surreal arithmetic, which agrees with Hessenberg arithmetic on the ordinals. As far as I know, expressions like $$\frac{\pi}{\omega^2+1}-\sqrt{2\omega}$$ make perfect sense in the surreal numbers. Perhaps someone with more expertise can come along and confirm.

Answer 2

I incidentally answered this question recently when trying to build an alternative construction of the Surreal numbers.

We can build ordinal fractions, but the shortest logical route to such an object is by considering a subclass of $O_n\times O_n\times O_n\times O_n$, where $O_n$ is the class of all ordinals.

As the community wiki post mentions we can consider an isomorphism between $O_n$ and $N_0$ (the Surreal numbers), but this requires prior knowledge of the Surreals to be of any assistance and this is somewhat of a high bar for entry.

For this approach, we must first develop the notion of the Cantor normal form of an ordinal -- this is essentially a unique base-$\omega$ representation of any ordinal $\alpha\in O_n$. It is entirely equivlent to consider the class of all polynomials with $\omega$ as the lone independent variable, coefficients in the natural numbers, and exponents in the ordinals. We use Hessenberg addition to add the exponents when we multiply two 'polynomials', and add them in the expected fashion by 'combining like terms'. For example: $$n\times\omega^0=n\times1=n,$$ $$(\omega^{\omega^4}+\omega^2+1)+(\omega^4+3\omega^2+5)=\omega^{\omega^4}+\omega^4+4\omega^2+6,$$ $$(\omega^{\omega^4}+\omega^2+1)\times(\omega^4+3\omega^2+5)=\omega^{\omega^4+4}+3\omega^{\omega^4+2}+5\omega^{\omega^4}+\omega^6+4\omega^4+8\omega^2+5,$$

and so on. Before we can build ordinal fractions, we must first allow the coefficients on our polynomials to range over all integers instead of just natural numbers. This actually requires some doing, but if you would like to see a rigorous proof that we can legitimately extend the normal forms in this manner I construct them here (https://arxiv.org/abs/1706.08908) beginning on page 26, and I construct ordinal fractions beginning on page 43 -- I also provide a visual interpretation for the structures in question.

Informally, suppose $\alpha,\beta\in O_n$ and consider the subclass of $O_n\times O_n$ given by $$\mathbb{Z}_\infty=\{(\alpha,\beta):\alpha\ \text{and}\ \beta\ \text{share no powers of}\ \omega\ \text{in their cantor normal forms}\}.$$ We will call $\mathbb{Z}_\infty$ the Surintegers, and they are equivalent to allowing our cantor normal forms to have negative integer coefficients -- we interpret the left coordinate as the 'negative part' and the right coordinate as the 'positive part', and when writing them out in 'normal form' the Cantor normal form of the left coordinate picks up negative signs. For example, $$(\omega^{\omega^\omega}+\omega^\delta+5,\omega^4+\omega^2+\omega)=-\omega^{\omega^\omega}-\omega^\delta+\omega^4+\omega^2+\omega-5.$$

Note that if we require that $\alpha<\omega$ and $\beta<\omega$, this is the integers $\mathbb{Z}$ with $(0,n)=n$ and $(n,0)=-n$. Allowing $\alpha$ and $\beta$ to range over all ordinals we obtain a proper class-sized, discretely ordered, commutative ring with identity.

We now build a field of fractions for $\mathbb{Z}_\infty$ -- these will be our 'fractional ordinals'. Let $a\in\mathbb{Z}_\infty$ and $b\in\mathbb{Z}_\infty^+\sim\{0\}$, and consider the subclass of $\mathbb{Z}_\infty\times\mathbb{Z}_\infty^+\sim\{0\}$ given by $$\mathbb{Q}_\infty=\{(a,b):a\ \text{and}\ b\ \text{are coprime}\}.$$ We will call $\mathbb{Q}_\infty$ the Surrational numbers, and they contain the class of all fractional ordinals. For $p\in\mathbb{Q}_\infty$ such that $p=(a,b)=\big((\alpha,\beta),(\gamma,\delta)\big)$, we write $p=\frac{a}{b}=\frac{\beta-\alpha}{\delta-\gamma}$. Setting $\alpha=\gamma=0$, we obtain a simple fractional form for any two ordinals we wish to consider.

If you wish to see the actual ordering and binary operations we use in $\mathbb{Z}_\infty$ and $\mathbb{Q}_\infty$, feel free to look at the above paper -- they are what you would intuitively expect though.

I also 'cut up' set-sized subfields of these Surrational numbers to form real closed fields and then build algebraically closed complex numbers out of these real closed fields, and ultimately I believe we can construct the Surreal numbers and Surcomplex numbers using cuts of all sizes from all real-closed subfields.