Define a function $f:\mathbf{ON}\to\mathbf{ON}$ as follows: for each ordinal number $\alpha$ let $$f(\alpha)=\operatorname{ord(}\lbrace \beta<\alpha|\text{ }\beta\textrm{ is a limit ordinal}\rbrace).$$ This function seems useful, as it seems to describe in some sense "how many levels" an ordinal number has, which leads me to think it might have been studied before. Is there some standard name/notation for this function?
What is this function on ordinals called?
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0You got me wondering if anything wonky can happen if you define a function with a domain which is a proper class... Is that set-theoretically kosher? – 2012-05-03
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0Since the limit ordinals are exactly the ordinals of the form $\omega \gamma$, this is more or less just (left) division by $\omega$. – 2012-05-03
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0@ChrisEagle: right, that does make sense. Is this standard, though? I'm not sure I've seen division of ordinals defined in Jech's Set Theory, for example. – 2012-05-03
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0I don't know how standard ordinal division is. It was in the course and book from which I first learnt about ordinals, and is described in the wikipedia article on [ordinal arithmetic](http://en.wikipedia.org/wiki/Ordinal_arithmetic). – 2012-05-03
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0@rschwieb: I second you. At least in ZFC functions between proper classes don't make sense, because to my knowledge we cannot even make precise the notion of a proper class using ZFC only. – 2012-05-03
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1@rschwieb: You're correct that we cannot define class functions in ZFC in the sense that a class function is not a *set* of ordered pairs. However, there is usually an easy way to circumvent this. E.g. we can work with restrictions of the function $F$ to subsets of the class of all ordinals. Or instead of calling this a function we try to write down a formula $P(x,y)$ show that for each ordinal $\alpha$ there exists unique ordinal $\beta$ such that $P(\alpha,\beta)$. I guess such consideration are usually omitted in textbooks, since they would make things look complicated and (continued) – 2012-05-04
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1(continued) it is assumed that the reader is experienced enough to mend this by himself. There was also some discussion about this at MSE: [How do we justify functions on the Ordinals](http://math.stackexchange.com/questions/137319/how-do-we-justify-functions-on-the-ordinals) or [class function question](http://math.stackexchange.com/questions/84486/class-function-question). One more ping for @Nils. – 2012-05-04
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0@Martin Sleziak: Above all, thanks for the links. I have yet to study them in detail. But I'd like to get one thing straight so I just state it as something I think is true, and I would really appreciate any criticism: Inside any model of ZF, using ZF alone, you cannot define what a proper class is. You need some meta-language to actually talk about it. – 2012-05-04
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0@Nils: I think that there are people at this site that are much more competent to answer your question from the last comment than me; but I think you're correct; at least for the reason that in the language of ZF every object is a set, so we cannot speak about classes at all. – 2012-05-04
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1@Nils: Yes. So the language of ZFC can't express things like "there is a class with such and such properties". One thing it can do, given an explicit formula defining a class, is say "the class defined by this formula has such and such properties". This is often enough for what one wants to do. So for example ZF+V=L can't prove (or even say) "there's a definable well-ordering of the universe", but it can prove "
defines a well-ordering of the universe". – 2012-05-04 -
0@Martin,Chris: Thanks for your responses! – 2012-05-05
1 Answers
An ordinal $\alpha$ is a limit if and only if it can be written as a product $\alpha=\omega \gamma$ for some ordinal $\gamma$. Thus your function $f$ is almost the same thing as division by $\omega$ (ignoring the remainder).
In detail: for any ordinals $\alpha$, $\beta$, with $\beta>0$, there exists a unique pair of ordinals $\gamma$, $\delta$ such that $\alpha=\beta \gamma + \delta$ and $\delta <\beta$. As in the finite case, we call this division by $\beta$, and call $\gamma$ the quotient and $\delta$ the remainder. If the remainder on dividing $\alpha$ by $\omega$ is $0$, then $f(\alpha)$ is the same as the quotient. If the remainder is nonzero, then $f(\alpha)$ is the quotient plus one.
The above assumes that zero is a limit ordinal. If not, then $f(\alpha)$ will be one smaller whenever it is finite and nonzero.