Exercise WO.3

Proof. First, if $E=\varnothing$ then it must be that $J=\varnothing$ as well so that they vacuously have the same order type. Otherwise, following the hint, choose $e_0\in E$ and define $h:J\to E$ by

and $h(\alpha )=e_0$ otherwise, noting that this function is uniquely defined by the general principle of recursive definition (Exercise WO.1). We show that $h(\alpha )\le k(\alpha )$ for all $\alpha \in J$ using transfinite induction (see Lemma 10.10.1). So consider $\alpha \in J$ and assume that $h(x)\le k(x)$ for all $x\in S_{\alpha }$. Since $k$ preserves order we have that $h(x)\le k(x)<k(\alpha )$ when $x<\alpha$. In particular, this means that $h(x)\ne k(\alpha )$ for all $x\in S_{\alpha }$ so that $k(\alpha )\in E-h(S_{\alpha })$. Hence $E-h(S_{\alpha })$ is not empty so that $h(S_{\alpha })\ne E$. Thus $h(\alpha )$ is the smallest element of $E-h(S_{\alpha })$ and so $h(\alpha )\le k(\alpha )$ since $k(\alpha )\in E-h(S_{\alpha })$. This completes the induction.

Therefore, for any $\alpha \in J$ and any $x<\alpha$ we have $h(x)\le k(x)<k(\alpha )$ since $k$ preserves order so that $h(x)\ne k(\alpha )$. As in the induction step above, it follows that $h(S_{\alpha })\ne E$. Hence, since $\alpha$ was arbitrary,

for all $\alpha \in J$. It then follows from Exercise WO.2 part (a) that $h$ is order-preserving and maps $J$ onto $E$ or a section of $E$. This clearly shows that $J$ has the order type of $E$ or a section of $E$ as desired. □