Difficulty understanding set theory problem in Munkres The following problem is from the set theory part of Munkres Topology, specifically from supplementary problems on well-ordering. The problem states:
Let $J$ and $E$ be well-ordered sets; let $h: J \rightarrow E$. Show the following two statements are equivalent:
(i) $h$ is order preserving and its image is $E$ or a section of $E$
(ii) $h(\alpha) = \text {smallest} \ [E - h(S_\alpha)]$ for all $\alpha$
[Hint: Show that each of these conditions implies that $h(S_\alpha)$ is a section of $E$; conclude that it must be the section by $h(\alpha)$.] 
I can't see a way to begin tackling this question. The previous exercise was to prove the general principle of recursive definition, but I don't see it helping with this question. I'm absolutely stuck so if anyone would care to give a hint, I would be very grateful.
 A: $\text{(i)}\Longrightarrow\text{(ii)}$: Let $\alpha\in J$. Then we have that:
$$h[S_\alpha]=\{h(a)|\;a\in J\text{ and }a<_J\alpha\}$$
Since $h:J\longrightarrow E$ is order preserving, and $a<_J\alpha$ if, and only if $h(a)<_Eh(\alpha)$,we can wirte:
$$h[S_\alpha]=\{h(a)|\;a\in J\text{ and }h(a)<_Eh(\alpha)\}$$
Thus $h[S_\alpha]\subseteq S_{h(\alpha)}$. Suppose that $S_{h(\alpha)}\not= h[S_\alpha]$, that is, that $h[S_\alpha]$ is a proper subset of $S_{h(\alpha)}$, or in other words, that there exists $b\in S_{h(\alpha)}$ that is not the image of any element of $S_\alpha$. Since $\text{Im}(h)$ is a section of $E$ or the whole set $E$, we can conclude that $b\in\text{ Im}(h)$, for it is an element of $S_{h(\alpha)}$ $-$ for instance, if $\text{Im}(h)=S_\beta$, for some $\beta\in E$, certainly $h(\alpha)\in\text{Im}(h)=S_\beta$, so $h(\alpha)<\beta$, and if $b\in S_{h(\alpha)}$, then $b<h(\alpha)<\beta$, so $b\in S_\beta=\text{Im}(h)$, as we wanted $-$. But if $b\in\text{Im}(h)$, then there exists some $a\in J$ such that $h(a)=b$. We know that $a\not\in S_\alpha$, so $a\geq_J\alpha$. But this is absurd; $h$ is order preserving, so we should have that $h(a)=b\geq_E h(\alpha)$, but also $h(\alpha)>_Eb\;$$-$since $b\in S_{h(\alpha)}-$. Therefore, we obtained that:
$$h[S_\alpha]=\{b|\;b\in E\text{ and }b<h(\alpha)\}=S_{h(\alpha)}$$
And from the previous observation:
$$E\setminus h[S_\alpha]=E\setminus S_{h(\alpha)}=\{x|\;x\in E\text{ but }x\not\in S_{h_\alpha}\}=\{x|\;x\in E\text{ and }h(\alpha)\leq_Ex\}$$
And clearly, the first element of this set, in the sense of $<_E$ is $h(\alpha)$
$\text{(ii)}\Longrightarrow\text{(i)}$: Let $\alpha,\beta\in J$ with $\alpha<_J\beta$. We want to prove that $h(\alpha)<_Eh(\beta)$. Since $\alpha<_J\beta$, we have that $S_\alpha\subset S_\beta$, and $h[S_\alpha]\subseteq h[S_\beta]$, so $E\setminus h[S_\beta]\subseteq E\setminus h[S_\alpha]$. This means that the first element of the first set will be greater or equal than the first element of the other; that is, $h(\alpha)\leq_Eh(\beta)$. We have to prove that, indeed, $h(\alpha)\not=h(\beta)$. Suppose, on the contrary, that $h(\alpha)=h(\beta)$. That means that the set:
$$\mathcal{S}=\{x\in J\;|\;\exists y\in J\text{ such that }x<_Jy\text{ and }h(x)=h(y)\}$$
Is not empty, so it has a $<_J$-minimal element, denote it by $\gamma$. By hypothesis, for all $\eta<_J\gamma$, $h(\eta)<_Eh(\theta)$ for all $\theta>_J\eta$. By definition of $\gamma$, there exists at least one element of $E$ greater than $\gamma\;$$-$ at least the one whose image via $h$ coincides with $h(\gamma)-$; call it $\delta$.
By the previous observations, we have that $S_\gamma\subset S_\delta$, and $h[S_\gamma]\subset h[S_\delta]$ because no element of $S_\gamma$ belongs to $\mathcal{S}$, that is, $h[S_\gamma]\not=h[S_\delta]$; otherwise we would have an element of $S_\delta\setminus S_\gamma$ whose image coincides with the image of an element of $S_\gamma$, which is absurd. That implies that $E\setminus{h[S_\delta]}\subset E\setminus{h[S_\gamma]}$, or in other words, that $h(\gamma)<_Eh(\delta)$, which is absurd.
In the end, we can conclude that $h$ is a order preserving function.
Now we have to prove that $\text{Im}(h)$ is a section of $E$ or the whole set $E$.
If $\text{Im}(h)=E$ there is nothing to prove
If $\text{Im}(h)\subset E$, we have to prove that $\text{Im}(h)$ is a section of $E$. To do so, take $\alpha\in \text{Im}(h)$ and $\beta\in E$ with $\beta<_E\alpha$. Our goal is to prove that $\beta\in\text{Im}(h)$, so $\text{Im}(h)$ is an initial segment of $E$, and therefore, a section of $E$, since $<_E$ is a well-ordering.
Since $\alpha\in\text{Im}(h)$, there exists some $a\in J$ such that $\alpha=h(a)=\text{ the first element of }E\setminus h[S_a]$. Note that $\beta\not\in\text{Im}(h)$ implies that $\beta\not\in h[S_a]$, so $\beta\in E\setminus h[S_a]$ and $\beta<_E\alpha=h(a)$, which is absurd, since this contradicts the choice of $\alpha$ as the least element in $B\setminus h[S_a]$ in the sense of $<_E$.
Note: This exercise is much easier using ordinals, and applying its properties; for each well-ordered set $\langle A,<_A\rangle$, there exists an unique ordinal $\alpha$ such that $\langle A, <_A\rangle\cong\langle\alpha,\in_\alpha\rangle$, and all the questions about the well-ordered structure $\langle A,<_A\rangle$ can be translated into questions about $\langle\alpha,\in_\alpha\rangle$. In some of the literature this is known as Mirimanoff-Von Neumann theorem, for instance, c.f. Theorem 3.1., page 111, in Introduction to Set Theory, 3rd ed. by Hrbacek and Jech. Didn't use any of those tools since I wasn't sure if any of this is treated at all in Munkres' book.
