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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?

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You 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? –  rschwieb May 3 '12 at 14:54
Since the limit ordinals are exactly the ordinals of the form $\omega \gamma$, this is more or less just (left) division by $\omega$. –  Chris Eagle May 3 '12 at 14:54
@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) –  Martin Sleziak May 4 '12 at 8:30
(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 or class function question‌​. One more ping for @Nils. –  Martin Sleziak May 4 '12 at 8:30
@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 "<insert horrible formula here> defines a well-ordering of the universe". –  Chris Eagle May 4 '12 at 15:41
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1 Answer

up vote 4 down vote accepted

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.

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Thanks, by the way. –  Dejan Govc May 3 '12 at 15:12
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