Two well orders on an uncountable set We have two well orders $\preceq_1, \preceq_2$ on an uncountable set $X$. 
Why would there be a set $Y\subseteq X$ such that $Y\restriction_{\preceq_1} = Y\restriction_{\preceq_2}$ and Y is uncountable?
I'm trying and trying to come up with a proof but I don't see why that would be true at all. 
Say, I want to do it by induction. Then sure, I can choose two elements of $X$ such that they are in the same relation both with respect to $\preceq_1$ and $\preceq_2$. But why can I find a third element like this? Or $n$-th? Or $\gamma$-th... Perhaps there are no such.
 A: First, there exists $X_1\subset X$ so that $(X_1,\le_1)$ is isomorphic to $\omega_1$. Now there exists $X_2\subset X_1$ so $(X_2,\le_2)$ is isomorphic to $\omega_1$. And hence $(X_2,\le_1)$ is also isomorphic to $\omega_1$, since any uncountable subset of $\omega_1$ is isomorphic to $\omega_1$. 
Say $Y$ is a maximal subset of $X_2$ such that the two orders agree on $Y$. Say $Z$ is the set of all $z\in X_2$ such that $z\le_1 y$ for some $y\in Y$. If $Y$ is countable then $Z$ is countable. So there exists $x\in X_2$ such that $x\notin Z$ and also $x>_2y$ for all $y\in Y$, so $Y$ was not maximal after all.
A: Let $A=\big\{\{x,y\}\in[X]^2:\preceq_1\text{ and }\preceq_2\text{ agree on }x\text{ and }y\big\}$, and let $B=[X]^2\setminus A$. By the Dushnik-Miller theorem either 


*

*there is an uncountable $Y\subseteq X$ such that $[Y]^2\subseteq A$, or 

*there is an infinite $Y\subseteq X$ such that $[Y]^2\subseteq B$. 


Use the fact that $\preceq_1$ and $\preceq_2$ are well-orders to show that the second alternative is impossible.
