Exponentiation on order types How is exponentiation defined on order types?
We know that $2^\omega=\omega$. What is $2^{\omega^*}$? Is it $\omega^*$? $\eta$? $\lambda$? I'm guessing $\eta$, but I'm not sure.
$\omega$ is the order type of the positive integers. $\omega^*$ is the order type of the negative integers. $\eta$ is the order type of the rationals; it is the same as the order type of the dyadics (numbers with finite decimal expansions in binary)—indeed, it's the same as the order type of any countable dense set without endpoints. $\lambda$ is the order type of the reals.
 A: (Getting another old question off the unanswered list.)
In his Grundzüge einer Theorie der geordneten Mengen, Mathematische Annalen 65 (1908), Felix Hausdorff defined a notion of exponentiation for order types that for well orders agrees with ordinal exponentiation. Let $\langle I,\le_I\rangle$ and $\langle X,\le_X\rangle$ be linear orders. For $J\subseteq I$ let $\le_J$ be the order on $J$ induced by $\le_I$.
Fix a base point $x_0\in X$. Let ${^I}\!X$ be the set of functions $\xi:I\to X$. (This is often written $X^I$, but I will be using $X^I$ for order exponentiation.) For $\xi\in{^I}\!X$ let $\operatorname{supp}(\xi)=\{i\in I:\xi(i)\ne x_0\}$, and define
$$X^I=\left\{\xi\in{^I}\!X:\ge_{\operatorname{supp}(\xi)}\text{well-orders }\operatorname{supp}(\xi)\right\}\,.$$
Note the reversal of the order: $\xi\in{^I}\!X$ belongs to $X^I$ iff $\le_{\operatorname{supp}(\xi)}$ inversely well-orders $\operatorname{supp}(\xi)$, i.e., iff each non-empty subset of $\operatorname{supp}(\xi)$ has a greatest (or last) element with respect to $\le_I$. For $\xi,\eta\in X^I$ let $\operatorname{diff}(\xi,\eta)=\{i\in I:\xi(i)\ne\eta(i)\}$, and observe that $\operatorname{diff}(\xi,\eta)\subseteq\operatorname{supp}(\xi)\cup\operatorname{supp}(\eta)$, which is well-ordered by $\ge_I$. Thus, if $\xi\ne\eta$, $\operatorname{diff}(\xi,\eta)$ has a largest element with respect to $\le_I$, and we set $i_{\xi,\eta}=i_{\eta,\xi}=\max\limits_{\le_I}\operatorname{diff}(\xi,\eta)$. Finally, set $\xi\preceq\eta$ iff either

*

*$\xi=\eta$, or

*$\xi\ne\eta$, and $\xi(i_{\xi,\eta})<_X\eta(i_{\xi,\eta})$;

it is easy to check that $\preceq$, the antilexicographic order on $X^I$, is a linear order. (Note that the order $\preceq$ does depend on the choice of base point, a fact not reflected in the notation.) Moreover, if $\alpha$ and $\beta$ are ordinals, $\le_X$ well-orders $X$ in type $\alpha$, $x_0=\min\limits_{\le_X}X$, and $\le_I$ well-orders $I$ in type $\beta$, then $\preceq$ well-orders $X^I$ in type $\alpha^\beta$ (ordinal exponentiation).

Both W.C. Holland, S. Kuhlmann, S.H. McCleary, Lexicographic Exponentiation of Chains [PDF] and Trevor Green Univ. of Saskatchewan Masters Thesis Properties of Chain Products and Ehrenfeucht-Fraïssé Games on Chains [PostScript file] begin by covering this material (and more) in English.

Now apply this definition to the exponential $2^{\omega^*}$. Since $\omega^*$ is inversely well-ordered, the underlying set of the exponential is the entire set ${^{\omega^*}}2$. It doesn’t really matter whether we take $0$ or $1$ to be the base point, but it’s convenient to make $0$ the base point, so that $\xi\in 2^{\omega^*}$ is simply the indicator (characteristic) function of $\operatorname{supp}(\xi)$, and every subset of $\omega$ is the support of some $\xi\in 2^{\omega^*}$.
$\left\langle 2^{\omega^*},\preceq\right\rangle$ is then clearly order-isomorphic to $\left\langle{^\omega}2,\le_{\text{lex}}\right\rangle$, the set of infinite binary strings ordered lexicographically. We can get a better picture of that order by thinking of each $\xi\in{^\omega}2$ as the binary expansion of a real number $\varphi(\xi)\in[0,1]$ via the surjection
$$\varphi:{^\omega}2\to[0,1]:\xi\mapsto\sum_{k\in\omega}2^{-(\xi(k)+1)}\,.$$
Let $D$ be the set of dyadic rationals in $(0,1)$, and let $x\in[0,1]$. If $x\notin D$, there is a unique $\xi_x\in{^\omega}2$ such that $\varphi(\xi_x)=x$, and if $x\in D$, there are a unique $\xi_x\in{^\omega}2$ and $n_x\in\omega$ such that $\varphi(\xi_x)=x$, $\xi_x(n_x)=0$, and $\xi_x(k)=1$ for all $k>n_x$. If $x\in D$ we define $\xi_x^+\in{^\omega}2$ by setting $\xi_x^+(k)=\xi_x(k)$ for $k<n_x$, $\xi_x^+(n_x)=1$, and $\xi_x^+(k)=0$ for $k>n_x$; $\xi_x^+$ is the immediate successor of $\xi_x$ in the linear order $\left\langle{^\omega}2,\le_{\text{lex}}\right\rangle$. Note that $\{\xi_x:x\in[0,1]\}\cup\{\xi_x^+:x\in D\}={^\omega}2$.
It’s not hard to check that for all $x,y\in[0,1]$, $x<y$ iff $\xi_x<_{\text{lex}}\xi_y$, so that $\varphi\upharpoonright\{\xi_x:x\in[0,1]\}$ is an order-isomorphism. It follows that $\left\langle{^\omega}2,\le_{\text{lex}}\right\rangle$ (and hence $\left\langle 2^{\omega^*},\preceq\right\rangle$) is order-isomorphic to the linear order obtained from $\left\langle[0,1],\le\right\rangle$ by splitting each $x\in D$ into two adjacent points.
More formally, let $Y=[0,1]\times\{0,1\}$, let $X=\{\langle x,i\rangle\in Y:i=0\text{ or }x\in D\}$, and let $\unlhd$ be the lexicographic order on $X$; then $\left\langle{^\omega}2,\le_{\text{lex}}\right\rangle$ (and hence $\left\langle 2^{\omega^*},\preceq\right\rangle$) is order-isomorphic to $\langle X,\unlhd\rangle$. The point $\langle x,0\rangle$ in $X$ corresponds to the sequence $\xi_x$, and for $x\in D$ the point $\langle x,1\rangle$ in $X$ corresponds to $\xi_x^+$.
Added: As Akiva Weinberger suggested in a comment below, $\left\langle{^\omega}2,\le_{\text{lex}}\right\rangle$ is in fact order-isomorphic to $C$, the middle-thirds Cantor set, with the usual order. This is immediate from the well-known fact that
$$\psi:{^\omega}\{0,2\}\to C:\xi\mapsto\sum_{k\in\omega}3^{-(\xi(k)+1)}$$
is an order isomorphism from $\left\langle{^\omega}\{0,2\},\le_{\text{lex}}\right\rangle$ to $\langle C,\le\rangle$, since $\left\langle{^\omega}\{0,2\},\le_{\text{lex}}\right\rangle$ is clearly order-isomorphic to $\left\langle{^\omega}2,\le_{\text{lex}}\right\rangle$.
