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I'm just trying to understand better how to see $\alpha^{\beta}$ for an arbitrary ordinal. I've already know that one can think about $\alpha . \beta$ as $\langle \alpha \times \beta, AntiLex\rangle$ such that $AntiLex$ is the antilexicographical order. I want to know whether there is an analogous way (to the product) to think about the exponentiation.

Thanks in advance.

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up vote 8 down vote accepted

$\newcommand{\supp}{\operatorname{supp}}$For a function $f:\beta\to\alpha$ define the support of $f$ to be $\supp(f)=\{\xi\in\beta:f(\xi)\ne 0\}$. Let $$F=\left\{f\in{}^\beta\alpha:\supp(f)\text{ is finite}\right\}\;,$$

and let $\preceq$ be the anti-lexicographic order on $F$; then $\langle F,\preceq\rangle$ is order-isomorphic to $\langle\alpha^\beta,\le\rangle$.

Added: I just discovered that essentially this idea is used in this PDF to discuss certain aspects of powers $\Delta^\Gamma$, where $\Delta$ and $\Gamma$ are linear orders.

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Oh, that's unusual, and yet it goesback to Hausdorff... Typically, for arbitrary posets $\mathbb P,\mathbb Q$, $\mathbb P^{\mathbb Q}$ is taken to be the set of all order preserving maps, partially ordered pointwise (McKenzie uses this, among others). – Andrés E. Caicedo Jan 15 '13 at 1:44
@Andres: I was fascinated: I’d never seen the Hausdorff stuff before. I keep being surprised by how many things he did in some form! – Brian M. Scott Jan 15 '13 at 1:48

Yes, there is such a description. Say that a function $f:\beta\to\alpha$ has finite support iff $\{\gamma<\beta\mid f(\gamma)\ne0\}$ is finite. Then $\alpha^\beta$ is the order type of the set of functions $f:\beta\to\alpha$ that have finite support, ordered by $$ f<g $$ iff, letting $\gamma$ be largest such that $f(\gamma)\ne g(\gamma)$, we have that $f(\gamma)<g(\gamma)$. One can check (by induction on $\beta$, or otherwise), that this definition coincides with the one given by transfinite recursion.

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On the other hand, for cardinals $\alpha$ and $\beta$, cardinal exponentiation $\alpha^\beta$ is defined as the size of the set of all functions $f:\beta\to\alpha$. In general, there are plenty more functions than those with finite support. – Andrés E. Caicedo Jan 15 '13 at 1:26

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