# Question of an isomorphism of $\epsilon_ 0$ and a subset of the rationals.

I don't know if this question is appropriated for this site. Anyway, I'm searching for an isomorphism of order $f:K \longrightarrow \epsilon_o$, such that $(K, \leq)$ is a subset(proper or not) of $(\mathbb{Q}, \leq)$ and $\epsilon_o = \sup\{\omega, \omega^{\omega}, \omega^{\omega^{\omega}}, ... \}$. Actually, I'm not finding neither an isomorphism between $K$ and $\omega^{\omega}$. Thanks in advanced.

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Do you want an explicit one written down? Or just proof that there is one? –  Jason DeVito Jul 1 '12 at 2:37
an explicit one –  Yamauti Jul 1 '12 at 2:39
I am having doubts that an explicit one can be written. You can probably write down such isomorphism for $\alpha<\varepsilon_0$, though. –  Asaf Karagila Jul 1 '12 at 6:28
@Yamauti: It would really depend on the course. You may have theorems in your arsenal which allow you to quickly answer this. It is also possible that what seemed easy was actually pretty hard. –  Asaf Karagila Jul 18 '12 at 23:05
This question seems to be related: Embedding ordinals in $\mathbb Q$. A (non-constructive) proof of the fact that all countable ordinals embed in $\mathbb Q$ is given, e.g., in this answer. –  Martin Sleziak Sep 22 '12 at 8:25

Here is a method for constructing a subset of the rationals order isomorphic to $\varepsilon_0$.

First, you need a fundamental sequence for each limit ordinal $\alpha$ less than $\varepsilon_0$. (A fundamental sequence for $\alpha$ is an increasing sequence of ordinals whose supremum is $\alpha$.) Here is the standard definition in case you don't know it. This is an inductive definition over $\alpha$. Represent $\alpha$ in Cantor normal form,

$\alpha = \omega^{\alpha_1} + \omega^{\alpha_2} + \ldots + \omega^{\alpha_n}$.

If $\alpha_n = \beta + 1$, define $\alpha [i] = \omega^{\alpha_1} + \omega^{\alpha_2} + \ldots + \omega^{\beta} * i$.

If $\alpha_n$ is a limit ordinal, define $\alpha [i] = \omega^{\alpha_1} + \omega^{\alpha_2} + \ldots + \omega^{\alpha_n [i]}$.

For $\varepsilon_0$, we can use the fundamental sequence $\varepsilon_0[0] = 1, \varepsilon_0[1] = \omega, \varepsilon_0[2] = \omega^\omega, \varepsilon_0[3] = \omega^{\omega^\omega}, \ldots$ .

Now that we have fundamental sequences, we can start the construction. We will associate ordinals to various points in the interval [0, 1]. First, as a setup, place 0 at 0, and $\varepsilon_0$ at 1 (Note: 1 will not be included in $K$; I'm just placing $\varepsilon_0$ temporarily.) Next, we iterate the following rule $\omega$ times:

RULE: For each limit ordinal $\alpha$ we have already placed, we place the fundamental sequence for that $\alpha$ in the interval between $\alpha$ and the previously placed ordinal.

For the first iteration, there is just one limit ordinal, $\varepsilon_0$. We have placed $\varepsilon_0$ at 1, and the previous ordinal is at 0, so we are to place the fundamental sequence for $\varepsilon_0$ in the interval [0, 1]. We can choose any infinite sequence of points in [0, 1]; we will use the sequence 1/2, 3/4, 7/8, ... . So place $\varepsilon_0[0]$ at 1/2, $\varepsilon_0[1]$ at 3/4, $\varepsilon_0[2]$ at 7/8, and so on.

For the second iteration, we now have infinitely many limit ordinals placed; for each limit ordinal, we apply the previous procedure. For example, take $\varepsilon_0[3] = \omega^{\omega^\omega}$, which was placed at 15/16. The previous ordinal was placed at 7/8, so we want to insert the fundamental sequence for $\omega^{\omega^\omega}$ into the interval [7/8, 15/16]. Now, there is a slight problem; we have $\omega^{\omega^\omega}[0] = \omega$ and $\omega^{\omega^\omega}[1] = \omega^\omega$, neither of which are greater than the previously placed ordinal, $\omega^\omega$. So we simply drop any elements of the fundamental sequence that are not greater than the previously placed ordinal. So, we place $\omega^{\omega^\omega}[2] = \omega^{\omega^2}$ at 15/16 - 1/32, $\omega^{\omega^\omega}[3] = \omega^{\omega^3}$ at 15/16 - 1/64, and $\omega^{\omega^\omega}[n] = \omega^{\omega^n}$ at $\frac{15}{16} - \frac{1}{2^{n+3}}$. We repeat this procedure for all limit ordinals previously placed.

After iterating the rule $\omega$ times, we will have placed every ordinal $\le \varepsilon_0$. So we set $K$ equal to the set of all placed points excluding 1, and we have a set of order type $\varepsilon_0$.

This may seem more complicated than what you wanted, but I believe it is as simple as can be expected for an ordinal as complicated as $\varepsilon_0$.

The same procedure can be used for any ordinal of the form $\omega^\alpha$ for which we can define a collection of fundamental sequences; and we can define a collection of fundamental sequences for any ordinal which we can define an ordinal notation. So, for example, we can explicitly define a subset of the rationals of order type $\omega^{CK}_1$, by using Kleene's $O$ to define fundamental sequences. (This subset will not be recursive, of course.)

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