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.
 A: 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.)
