TITLE: How to show that $(\Bbb Q, <)$ contains an order isomorphic copy of $\epsilon_0$ which is the least ordinal satisfies $\omega^\epsilon = \epsilon$?
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: 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 ${\bf Q}$.