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Please, I don't understand the concept of ordinal numbers; for example, since they describe infinities, how can I possibly tell which set has ordinal $\omega+1,\omega+2$ or even $\omega.2$ or $\omega^2$?

Thanks in advance.

amWhy
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arsene stein
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1 Answers1

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I like to imagine these ordinals embedded in a familiar place: $\Bbb Q$, the set of rational numbers. A good representative of $\omega$ is the sequence:

$$0,\frac12,\frac23,\frac34,\ldots$$

Conveniently enough, all of these numbers are less than $1$, so if you want a set representing the ordinal $\omega+1$:

$$0,\frac12,\frac23,\frac34,\ldots,1$$

Similary, $2\omega$:

$$0,\frac12,\frac23,\frac34,\ldots,1,1+\frac12,1+\frac23,1+\frac34\ldots$$

In this way, we can embed $n\omega$ into the rational numbers for any $n$. Playing this game in the spaces between every pair of natural numbers, we get $\omega^2$.

Can we go further? Yes, indeed. The function $x\mapsto\frac{x}{x+1}$ maps the interval $[0,\infty)$ injectively and monotically into the interval $[0,1)$. Applying that transformation to our $\omega^2$ sequence compresses the whole thing into the unit interval. Copying the whole thing, by adding $1$ to each number, then gives us $2\omega^2$. Placing copies in each interval, we can get $3\omega^2, 4\omega^2,\ldots \omega^3$. Fun, isn't it?

This process can be repeated as many times as you like, to obtain embeddings in the rational numbers of $\omega^n$ for any $n$. As for $\omega^\omega$.... if there's a good way to embed that into $\Bbb Q$, I'd love to see it!

G Tony Jacobs
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    We can embed $\omega^\omega$ in $\Bbb Q$. Put a copy of $\omega$ in $[0,1)$ like you did above, then put $\omega^2$ in $[1,2)$, put $\omega^3$ in $[2,3)$ and so on. Now we can apply your transformation to put it all in $[0,1)$ and keep going. – Ross Millikan Jan 01 '18 at 20:07
  • Wonderful! Thank you :) I like how this construction doesn't even use all of the rationals (at least, I'm pretty sure it doesn't....) – G Tony Jacobs Jan 01 '18 at 20:09
  • No, it doesn't use them all. Certainly after you pack it all into $[0,1)$ there are a lot left. Even before that, it can't use them all because the usual order on the rationals is not a well order. There must be a successor for each one. The successor to $\frac 12$ must be some positive distance above it, so there are countably many skipped right there. – Ross Millikan Jan 01 '18 at 20:21
  • If you keep pushing it, are there any that will never get touched? (That's not a very well-articulated question, I know...) – G Tony Jacobs Jan 01 '18 at 20:22
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    In fact, any countable linear order type can be realized as a subset of the rational numbers in their usual ordering. See here, for example. However, doing this in some explicit way (after specifying what this might mean) gets you tied into knots, so to speak, even when just dealing with ordinals --- see Embedding ordinals in $\mathbb{Q}$. – Dave L. Renfro Jan 01 '18 at 20:23
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    I don't know how to justify that a particular rational is not used at any stage of the construction. You can certainly decide to avoid a specific one without any problem. More difficult is to define a generic process, which we have sketched, and prove that a particular one is never used. You might well be able to do that based on the prime factorization of the denominator, but it would take careful analysis of the algorithm to show it. – Ross Millikan Jan 01 '18 at 20:25
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    It is easy to find some that won't get touched in the construction of $\omega^\omega$. If our first $\omega$ is represented by $(1,2,3,4,5,\ldots)$ it gets shifted to $(\frac 12, \frac 23, \frac 34, \ldots, \frac n{n+1},\ldots)$. If we then compress $\omega^\omega$ to $[0,1)$ the first bit goes $(\frac 13, \frac 25, \frac 37,\ldots \frac n{2n+1}\ldots)$ so any rational in $[0,\frac 13)$ gets skipped. It is if you want to do bigger ordinals with many more compression steps that I don't know how to find one that won't get hit. – Ross Millikan Jan 02 '18 at 00:53
  • Yes, it's certainly true that the end of the interval near 0 fills up much more slowly. It's in the limit, if you keep repeating the process, where I'm curious. – G Tony Jacobs Jan 02 '18 at 00:55
  • I should really just ask this as a new question, and formulate it clearly, lol – G Tony Jacobs Jan 02 '18 at 00:56
  • Here: https://math.stackexchange.com/questions/2588251/will-every-rational-number-eventually-be-in-this-set – G Tony Jacobs Jan 02 '18 at 01:16
  • But isn't $\omega^{\omega}$ uncountable? How can we hope to embed it in $\mathbb{Q}$? – asdf Apr 20 '18 at 14:14
  • You might be thinking of cardinal exponentiation, not ordinal exponentiation. They're quite different. With ordinal exponentiation, $\omega^\omega$ is still countable, as is $\omega^{\omega^\omega}$, and even $\omega^{\omega^\ldots}$. – G Tony Jacobs Apr 20 '18 at 17:59