Relation between infinite order type $\Theta$ and $\omega$ I want to understand in case if $\Theta$ is an arbitrary infinite order type, why we have either $\omega \preceq \Theta$ or $\omega \preceq\Theta^*$.
Where $\Theta^*$ is reverse of order type $\Theta$. 
 A: Here is one possible argument. We show that if $\Theta$ is infinite, either $\omega$ or $\omega^*$ embeds into $\Theta$.
If $\Theta$ contains no minimal element then $\omega^*\prec\Theta$. If it contains no maximal element then $\omega\prec\Theta$. Let $m,M$ be the minimal and maximal elements, and define $I_0 = [m,M]$. Since $\Theta$ is infinite, we can pick some $x_1$ such that $m < x_1 < M$. One of the intervals $[m,x_1],[x_1,M]$ is infinite; choose one arbitrarily, and call it $I_1$. In this way, we can define an infinite sequence of intervals $I_0 \supset I_1 \supset \cdots$.
One of the following must have infinitely often: $I_{t+1}$ is the left sub-interval of $I_t$, or $I_{t+1}$ is the right sub-interval of $I_t$. In the first case, $\omega^* \prec \Theta$ (consider the right endpoints of the intervals $I_{t+1}$), and in the second, $\omega \prec \Theta$ (consider the left endpoints of the intervals $I_{t+1}$).
A: Let $<$ be a linear order on an infinite set $S$. For $p\in S$ let $p^+=\{q\in S: p<q\}$ and $p^-=\{q\in S:q<p\}.$ 
Now WLOG let $p_0\in S$ such that $p_0^+$ is infinite. (If no such $p_0$ exists, work with the reverse-order $<^*.$)
Suppose  there exists $q\in p_n^+$ such that $q^+$ is infinite. Then let $p_{n+1}$ be some such $q.$ 
If this does not produce an infinite strictly increasing sequence $(p_n)_{n\in \omega},$ then there exists $n_0$ such that $p_{n_0}^+$ is infinite, but such that  $q^+$ is finite for all $q\in p_{n_0}^+.$
In that case, for any $q\in p_{n_0}^+$ the set $  p_{n_0}^+\cap q^-$ is infinite, so take $q_0\in p_{n_0}^+,$ and choose $q_{n+1}\in p_{n_0}^+\cap q_n^-.$ This gives a strictly decreasing sequence $(q_n)_{n\in \omega}.$
