Subset of rational numbers isomorphic to $\omega^{\omega+1}$

Question

Does anyone know how to find/prove a rational subset isomorphic to $\omega^{\omega+1}$?

I have couple of ideas but I find it hard to prove.

Answer 1

This can be done less constructively by building the subset using transfinite induction, but here's a concrete example:

If we have nonempty sets of positive rationals $X$ and $Y$, they're totally ordered by the usual ordering, giving us posets $(X,<)$ and $(Y,<)$. We define the "concatenation" $X{\widetilde +}Y$ as $X\cup\{y+\textrm{sup }X\vert y\in Y\}$ (visually, for our purposes this behaves like "$(X,<)$ with $(Y,<)$ appended onto the end".) Since addition is continuous we also have the following: \begin{eqnarray}&&\textrm{sup }(X{\widetilde +}Y)\\=&&\textrm{sup }(X\cup\{y+\textrm{sup }X\vert y\in Y\})\\=&&\textrm{sup }\{y+\textrm{sup }X\vert y\in Y\}\\=&&\textrm{sup }Y+\textrm{sup }X.\end{eqnarray}

When writing an infinite $\widetilde +$-sum $Y_0{\widetilde +}Y_1{\widetilde +}Y_2{\widetilde +}\ldots$ we mean $\bigcup_{n\in\mathbb N}\underbrace{Y_0{\widetilde +}Y_1{\widetilde +}\ldots Y_n}_n$, the set including all its partial sums. Let's also define the "compression function": $c(X)=\{\frac{x}{2}\mid x\in X\}$, this scales a set of rationals (ordered under the usual ordering) by $\frac12$.

Let $X_0=\{1-2^{-n}\mid n\in\mathbb N\}=\{0,\frac12,\frac34,\frac78,\ldots\}$, and define $X_{\nu+1}=c(X_\nu){\widetilde +}c(c(X_\nu)){\widetilde +}c(c(c(X_\nu))){\widetilde +}\ldots$ (for $\nu\in\mathbb N\cup\{\omega\}$), and $X_\omega=c(X_0){\widetilde +}c(c(X_1)){\widetilde +}c(c(c(X_2))){\widetilde +}\ldots$ . Then $(X_{\omega+1},<)$ should have order type $\omega^{\omega+1}$.

---

In order to prove that even $X_\omega$ is well-defined, we first need to show that for all $n<\omega$, $\textrm{sup }X_n=1$. This is trivial for $n=0$, so assume $n>0$ and that it's been proven for the $n-1$ case. We then have $\textrm{sup }c(X_{n-1})=\frac12$, $\textrm{sup }c(c(X_{n-1}))=\frac14$, and in general $\textrm{sup }c^m(X_{n-1})=2^{-m}$ for natural $m$. By repeated application of the supremum property on each partial sum, we obtain $\textrm{sup }X_n=1$: \begin{eqnarray}&&\textrm{sup }X_n \\ =&&\textrm{sup }(c(X_{n-1}){\widetilde +}c(c(X_{n-1})){\widetilde +}c(c(c(X_{n-1})))+\ldots) \\ =&&\textrm{sup }c(X_{n-1})+\textrm{sup }c(c(X_{n-1}))+\textrm{sup }c(c(c(X_{n-1})))+\ldots \\ =&&\frac12+\frac14+\frac18+\ldots=1.\end{eqnarray}

Also $X_n$ is isomorphic to $\omega$ copies of $X_{n-1}$ laid end-to-end, so the order type of $X_n$ is the order type of $X_{n-1}$ multiplied by $\omega$.

By similar logic to the inductive case (except with $\textrm{sup }c^n(X_n)=2^{-n}$) we can show that $\sum_{n<\omega}\textrm{sup }c^n(X_n)$ converges, so $X_\omega$ is bounded, so $X_{\omega+1}$ is well-defined. $X_\omega$ is isomorphic to $\omega+\omega^2+\omega^3+\ldots$, which equals $\omega^\omega$, and similar order type logic applies to find the order type of $X_{\omega+1}$. I also believe this method can be extended for higher ordinals.