It is well known that the sum $\alpha+\beta$ of two ordinals $\alpha,\beta$ can be defined "geometrically" as the order type of the sum $(\{0\}\times \alpha)\cup(\{1\}\times\beta)$ endowed with the lexicographic order.

Also the product $\alpha\cdot\beta$ of ordinals $\alpha,\beta$ is the order type of the Cartesian product $\beta\times\alpha$ endowed with the lexicographic order.

What about the exponentiation of ordinals?

Does $\alpha^\beta$ have some nice "geometric'' or combinatorial model?

Maybe as some set of (partial) functions endowed with a suitable well-order?


Ordinal exponentiation is a special case of linear order exponentiation. For any linear order $L$, element $a\in L$, and ordinal $\beta$ we can define the $\beta$th power of $L$ at $a$, which I'll call "$L_a^\beta$," as the set of functions $f:\beta\rightarrow L$ such that all but finitely many $\alpha\in\beta$ have $f(\alpha)=a$. The ordering on this set is given by looking at the last point of difference: $$f\trianglelefteq g\iff f=g\mbox{ or } f(\max\{x:f(x)\not=g(x)\})<g(\max\{x:f(x)\not=g(x)\}).$$

For ordinals $\alpha,\beta$ we have $\alpha^\beta=\alpha_0^\beta$. Rosenstein's book treats this in some detail (and is generally an awesome book all-around - it's a huge tragedy that it's so hard to find).

What I like about this definition is that it very nicely complements the definition of cardinal exponentiation: they start with the same basic idea of counting functions between sets, but ordinal exponentiation is the "finite support" version.

  • $\begingroup$ @nombre Oh god, of course you're right, I don't know what I was thinking. $\endgroup$ Apr 25 '20 at 23:11
  • $\begingroup$ It's not well-ordered. $(1,0,0,...) > (0,1,0,...) > ... > (0,...,0,1,0,...) > ...$. Maybe replace min with max? $\endgroup$ Apr 27 '20 at 9:02
  • $\begingroup$ @MonroeEskew It would make sense to use $\max$ instead if you want to generalize ordinal exponentiation, but it won't be well-ordered if $L$ isn't anyways. $\endgroup$
    – nombre
    Apr 28 '20 at 9:27
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    $\begingroup$ @MonroeEskew Derp, fixed - thanks! $\endgroup$ Apr 29 '20 at 13:04
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    $\begingroup$ Do you need $\beta$ to be well-ordered? It looks to me like this definition still makes sense when $\beta$ is any linear order, so you've really defined a way of exponentiating with two linear orders . . . right? $\endgroup$
    – Will Brian
    Apr 29 '20 at 13:30

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