How to show that $(\Bbb Q, <)$ contains an order isomorphic copy of $\epsilon_0$ which is the least ordinal satisfies $\omega^\epsilon = \epsilon$? We define $\epsilon_0 = \sup\{\omega, \omega^\omega,\omega^{\omega^\omega},\omega^{\omega^{\omega^\omega}},\ldots\}$. How to find a subset of $ \Bbb Q$, $(A, <_{\Bbb Q})$ that is isomorphic to $\epsilon_0$?
So far, I'm only able to show that there's an order isomophic copy of $\omega^\omega$ in $\Bbb Q$ and don't know how to proceed.
 A: Proceed by induction.
Suppose that $A_\alpha$ is a subset of $\Bbb Q$ of order type $\alpha$, and that $A_\alpha$ is bounded. Let $q>\sup A_\alpha$, then $A_{\alpha+1}=A_\alpha\cup\{q\}$. Clearly $A_{\alpha+1}$ has order type $\alpha+1$ and is bounded.
Now suppose $\delta$ is a limit ordinal, and for all $\alpha<\delta$ we have $A_\alpha$ defined. Pick an increasing sequence $\alpha_n$, and write $\delta=\bigcup[\alpha_n,\alpha_{n+1})$, and let $\beta_n$ be the order type of the $n$-th interval. Then $\beta_n<\delta$, and therefore we can find $A_{\beta_n}$ as a subset of $\Bbb Q$.
We know that $(0,1)\cap\Bbb Q$ is order-isomorphic to $\Bbb Q$, so embed $A_{\beta_n}$ into $(n,n+1)$, and let $A'_\delta$ be the union of those embedded copies. It is easy to see that $A'_\delta$ has order type $\delta$. Since we want it to be bounded (so the induction can continue), we use the order isomorphism mentioned above to map $A'_\delta$ into an isomorphic copy of itself in $(0,1)\cap\Bbb Q$.
A: Any countable totally ordered set $(X,<)$ is isomorphic to a subset of $(\mathbb Q,<)$ :
Pick any bijection $f : \Bbb N \to X$, and define $\phi : X \to \Bbb Q$ blindly by picking for each $n$, any value for $\phi(f(n))$ that is consistent with the previously chosen values (no matter what you do, there will always be enough room, because $\Bbb Q$ with $n-1$ points removed is $n$ open intervals, so $n$ copies of $\Bbb Q$)
Here you could have some almost explicit map if you have an order-preserving map $f : \Bbb Q^\omega \to (0 ; 1)$ by setting $\phi(k) = 1-1/(k+2)$ for $k \in \omega$, and if $\alpha = (a_1, \ldots, a_n) \in \beta^\omega$, $\phi(\alpha) = f((\phi(a_1),\ldots,\phi(a_n)) + \max \lfloor \phi(a_i) \rfloor + 1\in \Bbb Q^\omega)$
In turn, you can make such a map $f: \Bbb Q^\omega \to \Bbb Q$ using the same construction with map a $\mathbb Q^2 \to (0;1)$.
A: A different proof of the fact cited by mercio (every countable linear ordering embeds into ${\bf Q}$) can be formulated as follows:


*

*${\bf Q}$ is the unique (up to isomorphism) countable dense linear ordering without endpoints (this is a very well-known fact, can be shown using back-and-forth method).

*Any linear order $L$ can be extended to a dense linear ordering without endpoints of cardinality $\aleph_0+\lvert L\rvert$ by recursively adding a single point at the end, at the beginning, and between each pair of points which have nothing between. At each step of the construction we add at most $\lvert L\rvert+\aleph_0$ points, so after countably many steps we have added at most $\aleph_0+ \lvert L\rvert$ points, and it's easy to see that we obtain a dense linear ordering.

*Therefore, if $L$ was countable to begin with, the dense linear ordering we obtain in the previous point is countable and hence isomorphic to ${\bf Q}$.

