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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)

Mark
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mark
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    Hi Mark, welcome to Math.SE. Please indicate your own thoughts and progress on the problem. This helps others better tailor an answer to your background and situation. It also shows people you are interested in the question and aren't just looking for someone to do your homework for you - StackExchange is not a homework site. – Ian Miller Mar 24 '16 at 04:24
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2 Answers2

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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.

Alec Rhea
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  • I think this works but there is no reason why it's better to start with what you call surrational numbers. Wouldn't the same thing work starting with the rationnal numbers (allowing cuts of the form $(F,\varnothing)$)? – nombre Jul 06 '17 at 09:51
  • You will never obtain a non-archimedean field by filling cuts in $\mathbb{Q}$, and $\mathbb{Q}_\infty$ is very non-archimedean, and contains no real closed fields. If you try to create a field of fractions for Conways Omnific Integers, it is a real-closed field and in fact the surreal numbers once again. If you are suggesting something else, perhaps I misunderstand your meaning. – Alec Rhea Jul 06 '17 at 13:16
  • If you're suggesting we allow left sets that "contain the whole field" and then order this above all cuts in the field this might work, but I have not explored that formalism. This would also require a prior definition of $\mathbb{Q}$ however, while the above method builds $\mathbb{Q}$ when restricted to $\omega$, and builds a larger non-archimedean non-real closed field when restricted to some other $\delta$-number. – Alec Rhea Jul 06 '17 at 13:35
  • Perhaps it is I who didn't understand what you meant. I just don't think those smallest fields containing all ordinals up to some $\delta$-number are very special regarding the idea of building surreals using them. But it is true that any proper class sized ordered Field will contain those minimal fields so I understand that one might want to use them to build upon. This just doesn't seem to be useful to me. (We must use different notions of cuts: for me, a cut in $X$ is a pair $(A,B)$ where $A < B$, $A$ has no maximum, $B$ has no minimum, and $A \cup B = X$. ($B$ is redundant)). – nombre Jul 06 '17 at 14:13
  • It's considered useful by some algebraists and analysts I've shown it to who are comfortable with constructing the Reals out of the Integers, but less comfortable with the recursive definition of the Surreals given by Conway. The above process is almost logically identical to the process of building $\mathbb{R}$ out of $\mathbb{Z}$, but instead builds $N_0$ out of $O_n$ and builds an analogous 'discrete subring' or 'subfield without roots' along the way. – Alec Rhea Jul 06 '17 at 14:17
  • $\mathbb{Q}_\infty$ is a proper class sized field with no real-closed subfields, so not all proper class sized fields will contain such subfields -- they are unique in being real-closed.

    This construction also shows that all ordered fields are simply phenomenon obtainable by taking well-ordered sets and adding sufficient structure to them, which I found interesting in its own right. Also, we can't build the surreals using restrictions -- we can build a set sized model of them if we restrict at an inaccessible cardinal number, but we only get a proper class using all of $O_n$.

    – Alec Rhea Jul 06 '17 at 14:18
  • I don't understand your argument regarding the fact that some class-sized ordered fields may not contain surrationnals, but the argument I had in mind does not actually work as it is. What's true is that any ordered field without set-sized cofinal subset contains the surrationnals. Do you have an example of a class-sized ordered Field without $\mathbb{Q}_{\infty}$ as a subfield? – nombre Jul 06 '17 at 14:45
  • Ah, I misunderstood you -- I thought you were asserting that proper class sized fields had the real-closed versions of the fields we can build out of $\delta$-numbers. I believe you are correct that any proper class sized ordered field will contain a subfield that is isomorphic to $\mathbb{Q}_\infty$. – Alec Rhea Jul 06 '17 at 14:47
  • Surrationnals can be built from rationnals by adding "full cuts": at each successor step add $(F,\varnothing)$ where $F$ is the current field. With "full cuts", any ordered field can be completed into $No$ by filling every set-sized cut and taking real closures. Full cuts seem easier to understand than other cuts, so the method starting with surrationnals does not seem to be simpler. But I understand that it is easier to grasp than Conway's definition: this makes $No$ look like a regular universal object while hinting at its connection with ordinal numbers. – nombre Jul 06 '17 at 14:48
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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.

goblin GONE
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