Suppose $X$ is an uncountable set and $\prec_1, \prec_2$ are two well orderings on $X$. Show that there is an uncountable $Y \subseteq X$ such that $\prec_1, \prec_2$ agree on $Y$.
How do I show that there is this uncountable $Y \subseteq X$? And the question says to show such a $Y$ exists. Can't I just take $Y=X$?
Please help. Thank you.
You could start by proving a lemma:
Lemma. In any ancountable well-ordered set $W$, there is an uncountable subset $Z$ such that each element of $Z$ has only countably many predecessors in $Z$.
Now, given two well-orderings $\prec_1$, $\prec_2$ of an uncountable set $X$, you can apply the lemma twice to get an uncountable subset $S$ of $X$ such that, for each element $y\in S$, the sets $\{s\in S:s\prec_1 y\}$ and $\{s\in S:s\prec_2 y\}$ are both countable.
Finally, use transfinite recursion to define a function $f:\omega_1\to S$ which is strictly increasing with respect to both orderings of $S$. Let $Y$ be the range of $f$.
Alternatively, there is a Ramsey-type theorem, called the Dushnik–Miller theorem if I remember right, which says that, given an uncountable set $X$ and a partition $\binom X2=K_0\cup K_1$, either there is an infinite set $Y\subseteq X$ such that $\binom Y2\subseteq K_0$, or else there is an uncountable set $Y\subseteq X$ such that $\binom Y2\subseteq K_1$. Applied to your two well-orderings of $X$, the Dushnik–Miller theorem says that either there is an infinite set on which the two orderings are opposites of each other (which is impossible because well-ordering), or else there is an uncountable set on which the two orderings agree, which is what you want.