Exercise WO.2

Proof. First, for any $\alpha \in J$ and $\beta \in E$, let $S_{\alpha }$ denote the section of $J$ by $\alpha$, and $T_{\beta }$ denote the section of $E$ by $\beta$. To avoid ambiguity, also suppose that $<$ is the well-order on $J$ and $\prec$ is the well-order on $E$. We show that each of these conditions is equivalent to the condition that $h(S_{\alpha })=T_{h(\alpha )}$ for every $\alpha \in J$. Call this condition (iii). This of course also shows that the conditions are equivalent to each other.

First, we show that (i) implies (iii). So suppose that $h$ is order preserving and its image is $E$ or a section of $E$. Consider any $\alpha \in J$ and any $y\in h(S_{\alpha })$ so that there is an $x\in S_{\alpha }$ where $y=h(x)$. Then $x<\alpha$ and $y=h(x)\prec h(\alpha )$ since $h$ preserves order. Therefore $y\in T_{h(\alpha )}$ so that $h(S_{\alpha })\subset T_{h(\alpha )}$ since $y$ was arbitrary. Now consider $y\in T_{h(\alpha )}$ so that $y\prec h(\alpha )$. Since also clearly $y\in E$ (since $T_{h(\alpha )}\subset E$), $y$ is in the image of $h$ if its image is all of $E$. If the image of $h$ is some section of $E$, say $T_{\beta }$, then clearly $h(\alpha )\in T_{\beta }$ since $h(\alpha )$ is obviously in the image of $h$. Hence we have $y\prec h(\alpha )\prec \beta$ so that $y\in T_{\beta }$ and hence in the image of $h$. Since $y$ is in the image of $h$ in either case, there is an $x\in J$ such that $y=h(x)$. Then $h(x)=y\prec h(\alpha )$ so that $x<\alpha$ since $h$ preserves order. Hence $x\in S_{\alpha }$ so that $y\in h(S_{\alpha })$ since $y=h(x)$. This shows that $T_{h(\alpha )}\subset h(S_{\alpha })$ since $y$ was arbitrary. Therefore $h(S_{\alpha })=T_{h(\alpha )}$ so that condition (iii) is true since $\alpha$ was arbitrary.

Next, we show that (iii) implies (i). So suppose that $h(S_{\alpha })=T_{h(\alpha )}$ for all $\alpha \in J$. First, it is easy to see that $h$ preserves order since, if $x,y\in J$ where $x<y$, then we have that $x\in S_y$ so that clearly $h(x)\in h(S_y)=T_{h(y)}$, and hence $h(x)<h(y)$. To show that the image of $h$, i.e. $h(J)$, is either $E$ or a section of $E$, consider the set $E-h(J)$.

Case: $E-h(J)=\varnothing$. Then clearly for any $y\in E$ we must have that $y\in h(J)$ since otherwise it would be that $y\in E-h(J)$. Thus $E\subset h(J)$ since $y$ was arbitrary. Also clearly $h(J)\subset E$ since $E$ is the range of $h$. This shows that $h(J)=E$.

Case: $E-h(J)\ne \varnothing$. Then clearly $E-h(J)$ is a nonempty subset of $E$ so that it has a smallest element $\beta$ since $E$ is well-ordered, noting that clearly $\beta \notin h(J)$. We claim that $h(J)=T_{\beta }$. So consider any $y\in h(J)$ so that there is an $x\in J$ where $y=h(x)$. Suppose for a moment that $\beta \preccurlyeq y$. Now it cannot be that $\beta =y$ since $y\in h(J)$ but $\beta \notin h(J)$, and so $\beta \prec y$. But then $\beta \in T_y=T_{h(x)}=h(S_x)$ since $x\in J$. Then $\beta$ is in the image of $h$ since clearly $h(S_x)\subset h(J)$. As this contradicts the fact that $\beta \notin h(J)$, it must be that $y\prec \beta$ so that $y\in T_{\beta }$. This shows that $h(J)\subset T_{\beta }$ since $y$ was arbitrary. Suppose now that $y\in T_{\beta }$ so that $y\prec \beta$. Since $\beta$ is the smallest element of $E-h(J)$, it follows that $y\notin E-h(J)$. Since clearly $y\in E$ (since $T_{\beta }\subset E$), it must be that $y\in h(J)$. This shows that $T_{\beta }\subset h(J)$ since $y$ was arbitrary. Hence we have shown that $h(J)=T_{\beta }$.

Therefore in every case either the image of $h$ is $E$ or a section of $E$ as desired. This completes the proof of (i).

Now we show that (ii) implies (iii). So suppose that $h(\alpha )$ is the smallest element of $E-h(S_{\alpha })$ for every $\alpha \in J$. First, we show that $h$ is injective. So consider any $x,y\in J$ where $x\ne y$. We can assume without loss of generality that $x<y$ so that $x\in S_y$ and hence $h(x)\in h(S_y)$. However, since we have that $h(y)$ is the smallest element of $E-h(S_y)$, clearly $h(y)\notin h(S_y)$. Therefore we have that $h(x)\ne h(y)$ so that $h$ is injective.

Now consider any $\alpha \in J$ so that clearly $h(\alpha )$ is the smallest element of $E-h(S_{\alpha })$. Suppose that $y\in h(S_{\alpha })$ so that there is an $x\in S_{\alpha }$ where $y=h(x)$, and therefore $x<\alpha$. Consider the possibility that $h(\alpha )\preccurlyeq h(x)=y$. It cannot be that $h(\alpha )=h(x)=y$ since $x\ne \alpha$ and $h$ is injective, so it must be that $h(\alpha )\prec h(x)$. It then follows that $h(\alpha )\notin E-h(S_x)$ since $h(x)$ is the smallest element of $E-h(S_x)$. Thus $h(\alpha )\in h(S_x)$ since clearly $h(\alpha )\in E$. It then follows from the fact that $h$ is injective that $\alpha \in S_x$ so that we have $\alpha <x<\alpha$, which is clearly a contradiction. So it must be that $y=h(x)\prec h(\alpha )$ so that $y\in T_{h(\alpha )}$. This shows that $h(S_{\alpha })\subset T_{h(\alpha )}$ since $y$ was arbitrary.

Now suppose that $y\in T_{h(\alpha )}$ so that $y\prec h(\alpha )$. Since $h(\alpha )$ is the smallest element of $E-h(S_{\alpha })$, it follows that $y\notin E-h(S_{\alpha })$. Since clearly $y\in E$, it must be that $y\in h(S_{\alpha })$. This shows that $T_{h(\alpha )}\subset h(S_{\alpha })$ since $y$ was arbitrary, and hence $h(S_{\alpha })=T_{h(\alpha )}$, which shows (iii) since $\alpha$ was arbitrary.

Lastly, we show that (iii) implies (ii). So suppose that $h(S_{\alpha })=T_{h(\alpha )}$ for every $\alpha \in J$ and consider any such $\alpha$. Clearly we have that $h(\alpha )\in E$ but $h(\alpha )\notin T_{h(\alpha )}=h(S_{\alpha })$ so that $h(\alpha )\in E-h(S_{\alpha })$. Suppose for the moment that $h(\alpha )$ is not the smallest element of $E-h(S_{\alpha })$ so that there is a $\beta \in E-h(S_{\alpha })$ where $\beta \prec h(\alpha )$. Then $\beta \in T_{h(\alpha )}$ so that it must be that $\beta \notin E-T_{h(\alpha )}=E-h(S_{\alpha })$ since $h(S_{\alpha })=T_{h(\alpha )}$. Clearly, this is a contradiction so it must be that $h(\alpha )$ really is the smallest element of $E-h(S_{\alpha })$, which shows (ii) since $\alpha$ was arbitrary. □