I extracted from here to ask a question~
In Naive Set Theory, Halmos phrases the comparability theorem as follows:
The assertion is that if $\langle X, \leqslant_X \rangle$ and $\langle Y, \leqslant_Y \rangle$ are well ordered sets, then either $X$ and $Y$ are similar, or one of them is similar to an initial segment of the other.
My questions are about the following proof:
We assume that $X$ and $Y$ are non-empty well ordered sets such that neither is similar to an initial segment of the other; we proceed to prove that under these circumstances $X$ must be similar to $Y$. Suppose that $a \in X$ and that $t$ is a sequence of type $a$ in $Y$; in other words $t$ is a function from $s(a)$ into $Y$. Let $f(t)$ be the least of the proper upper bounds of the range of $t$ in $Y$, if there are any; in the contrary case, let $f(t)$ be the least element of $Y$. In the terminology of the transfinite recursion theorem, the function $f$ thereby determined is a sequence function of type $X$ in $Y$. Let $U$ be the function that the transfinite recursion theorem associates with this situation. An easy argument (by transfinite induction) shows that, for each $a \in X$, the function $U$ maps the initial segment determined by $a$ in $X$ one-to-one onto the initial segment determined by $U(a)$ in $Y$. This implies $U$ is a similarity, and the proof is complete.
My question is how to show that $U(x)<U(a)$ for all $x<a$!!
It is necessary to show that $U$ is mapping $s(a)$ onto $s(U(a))$!!