Find a subset of real numbers with standard order isomorphic to $ω^{ω^ω}$

Question

I have to find such subset isomorphic to $ω^{ω^ω}$, $ω$ being the ordinal number. I know of a way to make one isomorphic to $ω^ω$, but I don't know how to continue from here.

Answer 1

In the comments @Alma Arjuna wrote

Simply divide $\mathbb R$ into countably many intervals and find in each one a subset isomorphic to ${\omega}^{\omega}$ (notice every interval is isomorphic to the whole $\mathbb R$).

and @spaceisdarkgreen replied with

that only gets you ${\omega}^{\omega + 1}.$

The reason is that the method Alma Arjuna described gives a representation of $\sup \{ {\omega}^{\omega} \cdot n:\; n < \omega \} = {\omega}^{\omega} \cdot \omega = {\omega}^{\omega + 1}.$

Nonetheless, by continuing to apply this method over and over again, we can reach ${\omega}^{{\omega}^{\omega}},$ but it will take a while. For more details about this method, see my answer to: Intuition for $\omega^\omega$, where I described in detail how one can obtain a representation of ${\omega}^{\omega}$ as a collection of real numbers with their usual ordering.

Reaching ${\omega}^{{\omega}^{\omega}}$ starting with ${\omega}^{\omega}$

If for each positive integer $n$ we put a representation of ${\omega}^{\omega}$ in the interval $[n,\,n+1),$ then we'll get a representation of ${\omega}^{\omega} \cdot \omega = {\omega}^{\omega + 1}.$ Next, for each positive integer $n,$ put a representation of ${\omega}^{\omega + 1}$ in the interval $[n,\,n+1)$ -- this gives a representation of ${\omega}^{\omega + 1} \cdot \omega = {\omega}^{\omega + 2}.$ Again, for each positive integer $n,$ put a representation of ${\omega}^{\omega + 2}$ in the interval $[n,\,n+1)$ -- this gives a representation of ${\omega}^{\omega + 2} \cdot \omega = {\omega}^{\omega + 3}.$ Continuing in this manner, for each positive integer $n$ we'll get a representation of ${\omega}^{\omega + n}.$ Now, for each positive integer $n,$ put a copy of ${\omega}^{\omega + n}$ in the interval $[n,\,n+1)$ -- this gives a representation of ${\omega}^{\omega + \omega} = {\omega}^{\omega \cdot 2}.$

At this point repeat everything we've done, beginning with ${\omega}^{\omega \cdot 2}$ instead of beginning with ${\omega}^{\omega},$ and we'll get a representation of ${\omega}^{\omega \cdot 2 \, + \, \omega} = {\omega}^{\omega \cdot 3}.$ Do this same repeating process beginning with ${\omega}^{\omega \cdot 3}$ instead of beginning with ${\omega}^{\omega},$ and we'll get a representation of ${\omega}^{\omega \cdot 3 \, + \, \omega} = {\omega}^{\omega \cdot 4}.$ Continuing in this manner gives, for each positive integer $n,$ a representation of ${\omega}^{\omega \cdot n}.$

Next, for each positive integer $n,$ put a copy of ${\omega}^{\omega \cdot n}$ in the interval $[n,\,n+1)$ to get a representation of ${\omega}^{\omega \cdot \omega} = {\omega}^{\omega ^2}.$ Now, if we repeat EVERYTHING we've done thus far, beginning with ${\omega}^{\omega ^2}$ instead of beginning with ${\omega}^{\omega},$ then we'll get a representation of ${\omega}^{\omega ^ 2 \cdot \omega} = {\omega}^{\omega ^ 3}.$ Doing the last sentence's "EVERYTHING again" procedure, this time beginning with ${\omega}^{\omega ^ 3},$ gives a representation of ${\omega}^{\omega ^ 4}.$ At this point we see how (but perhaps we're starting to get lost in the reiteratives of the rewinds of the repetitions of starting over again $\ldots),$ for each positive integer $n,$ to get a representation of ${\omega}^{\omega ^ n}.$ Finally, for each positive integer $n,$ put a copy of ${\omega}^{\omega ^ n}$ in the interval $[n,\,n+1)$ to get a representation of ${\omega}^{{\omega}^{\omega}}.$

It's worth mentioning that trying to reach ${\omega}^{{\omega}^{{\omega}^{\omega}}}$ by this method will take MUCH MUCH more work than what I did above, and it will take HUGELY INCREDIBLY more work still to reach ${\omega}^{{\omega}^{{\omega}^{{\omega}^{\omega}}}}.$

Answer 2

Can we use the fact that $\alpha=ω^{ω^ω}$ is countable?

Let $h(\beta) \in \mathbb N$ be an enumeration of all the ordinals less than or equal $\alpha$. Obviously, this enumeration is not order preserving, but that does not matter because we will use $h$ only as an auxiliary function.

Define a mapping $f$ to the reals by transfinite induction starting with $f(0)=0$. Let $f_0(\beta) = \sup_{\lambda < \beta}f(\lambda)$ and define

$$f(\beta) = f_0(\beta) + 2^{-h(\beta)} \qquad 0 < \beta \le \alpha$$

This definition works regardless of whether $\beta$ is a successor ordinal or a limit ordinal.

It is clear that $f$ is order preserving and that $$f(\alpha) \le f(0) + \sum_{\lambda=1}^\alpha 2^{-h(\lambda)} = 0 +\sum_{k \in \mathbb N^+} 2^{-k}=1$$

This shows that the image of $f$ is contained in $[0,1] \subset \mathbb R$.

If this argument is correct, we can a use a similar construction for much larger countable ordinals like $\epsilon_0$ in which the tower of exponents has height $\omega$.