I've come across an exercise as follows: show a surjective order type homomorphism from $\lambda$ to $\eta$ (Hint: use the fact that $(1+\eta+1)\eta=\eta$).
I've figured out a solution I think is valid.
First, pick some arbitrary bijection $f:\mathbb{N}\rightarrow\mathbb{Q}^{\geq0}$, ie, a well ordering on the nonnegative rationals. Define $g:\mathbb{Q}^{\geq0}\rightarrow\mathbb{R}$ as $g(a)=\sum_{x<a, x\in\mathbb{Q}^{\geq0}}\frac{1}{2}^{f^{-1}(x)}$.
For any $x\in\mathbb{Q}^{\geq0}$, let $A_x=[g(x),g(x)+\frac{1}{2}^{f(x)})$. Note that this interval is in the reals.
It's clear (I think) that $[0,g(x))$ is the disjoint union of every $A_y$ for $y<x$. Furthermore, since $\lim_{a\rightarrow\infty} g(a)=2$, we know that every real in $[0,2)$ is contained in some $A_x$.
This gives us a map $h:[0,2)\rightarrow\mathbb{Q}^{\geq 0}$, simply by sending each $y\in[0,2)$ to the $x$ corresponding to the $A_x$ containing $y$.
If $y_1>y_2$, then either
- They're contained in the same $A_x$, so $h(y_1)=h(y_2)$, or
- $y_1$ is contained in $A_{x_1}$, and $y_2$ is contained in $A_{x_2}$, where $x_1>x_2$ (this is obvious by the construction of the $A_x$s), so $h(y_1)>h(y_2)$
Since no $A_x$ is empty, this means $h$ is a surjective homomorphism from $[0,2)$ to $\mathbb{Q}^{\geq0}$. We know $h(0)=0$, since $g(0)=0$. Thus, we restrict the domain of $h$ to $(0,2)$, and recieve $h'$, a surjective homomorphism from $(0,2)$ to $\mathbb{Q}^+$. This is exactly from a set of order type $\lambda$ to order type $\eta$, as desired.
Two questions:
- Is this valid?
- Is there an easier solution/one actually using the hint?