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I want to prove that if $(X,\prec)$ and $(Y,<)$ are well-ordered sets that $X$ must be isomorphic to an initial segment of $Y$ or vice versa. I am trying to do this by defining the function: $$f=\{\langle x,y \rangle\in X\times Y: \hat x \cong \hat y\}$$ I have already proven that it is an isomorphism between its domain and its range, but I'm having difficulty in proving that either dom $f=X$ or ran $f=Y$. I have tried assuming that dom $f\neq X$, which implies that there is an element $x\in X$ such that $\hat x$ is not isomorphic to any initial segment of $Y$, but to no avail. Could anyone give me a hint?

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    $\begingroup$ Try considering the least such $x \in X$. $\endgroup$ – universalset Nov 7 '14 at 12:06
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Assume that $\operatorname{dom}f\neq X$ and $\operatorname{im}f\neq Y$, then let $x\in X-\operatorname{dom}f$ and $y \in Y-\operatorname{im}f$ be minimal. Show now that $f\cup \{(x,y)\}$ is again an isomorphism of initial segments contradicting the maximality of $f$.

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