Exercise 6.5.11

Proof. By Lemma 6.1.2 there is an $a\in \alpha$ such that $\beta =\{x\in \alpha \mid x/mo>a\}.\; Also,\; since$ a\in \alpha $and$ \alpha $is\; an\; ordinal,$ a$is\; an\; ordinal\; as\; well\; by\; Lemma 6.2.8.\; Finally,\; by\; the\; comments\; after\; Theorem 6.2.10,$ \beta $is\; simply\; the\; ordinal$ a$since$ \beta =\{x\in \alpha \mid x/mo>a\}and\; each$ x$in\; that\; set\; is\; an\; ordinal\; (by\; Lemma 6.2.8\; since\; each$ x\in \alpha $and$ \alpha $is\; an\; ordinal).\; \square $$

Proof. ( $\to$) First suppose that $\alpha =\sup <!--\; FUNCTION\; APPLICATION\; -->A$. Then by the remarks following the proof of Theorem 6.2.6 in the text $\alpha$ is an upper bound of $A$ and if $\beta$ is an upper bound of $A$ then $\alpha \le \beta$. This last statement is simply the contrapositive of the statement that $\beta /mo>\alpha implies\; that$ \beta $is$not an upper bound of $A$ and hence is logically equivalent.

( $\leftarrow$) We show that an ordinal $\alpha$ with the least upper bound property for $A$ is unique, which suffices to show the result since if $\beta$ has this property then $\beta =\sup <!--\; FUNCTION\; APPLICATION\; -->A$ since $\sup <!--\; FUNCTION\; APPLICATION\; -->A$ does as well (by what was just shown above) and the ordinal having this property is unique.

So suppose that ordinals $\alpha$ and $\beta$ both have the least upper bound property for $A$ but that $\alpha \ne \beta$. Without loss of generality we can assume then that $\alpha /mo>\beta .\; But\; then,\; since$ \beta $has\; the\; least\; upper\; bound\; property,$ \alpha $cannot\; be\; an\; upper\; bound\; of$ A$,\; which\; contradicts\; the\; fact\; that$ \alpha $also\; has\; the\; least\; upper\; bound\; property!\; Hence\; it\; has\; to\; be\; that$ \alpha =\beta $,\; which\; shows\; the\; uniquness.\; \square$

Proof. Consider any $x\in E_{\alpha }$, noting that $x/mo>b.$

If $x$ is the greatest element of $E_{\alpha }$ then let $\mathrm{\Delta }x=b-x$, noting that $\mathrm{\Delta }x/mo>0since$ b/mo>x.\; We\; also\; have$$

$$x+\mathrm{\Delta }x=x+b-x=b\le b.$$

Then consider any $y\in 𝑸$ where $x/mo>y/mo>x+\mathrm{\Delta }xso\; that\; clearly$ y\notin E_{\alpha }$since\; otherwise$ x$would\; not\; be\; the\; greatest\; element\; of$ E_{\alpha }$.$

On the other hand if $x$ is not the greatest element of $E_{\alpha }$ then let $f$ be the isomorphism between $\alpha$ and $E_{\alpha }$ and let $\beta =f^{-1}(x)$. It follows that $\beta$ is not the greatest element of $\alpha$ so that $\beta +1\in \alpha$ as well. Then let $\mathrm{\Delta }x=f(\beta +1)-x$, noting that $\mathrm{\Delta }x/mo>0since$ f(\beta +1)/mo>f(\beta )=xsince$ f$is\; an\; isomorphism.\; We\; also\; have$$

$$x+\mathrm{\Delta }x=x+f(\beta +1)-x=f(\beta +1)/mo>b$$

since $f(\beta +1)\in E_{\alpha }$. Now consider any $y\in 𝑸$ where $f(\beta )=x/mo>y/mo>x+\mathrm{\Delta }x=f(\beta +1).\; If\; it\; were\; the\; case\; that$ y\in E_{\alpha }$then\; we\; would\; have\; that$ \beta /mo>f^{-1}(y)/mo>\beta +1since$ f$is\; an\; isomorphism,\; which\; is\; impossible\; since$ \beta $is\; an\; ordinal.\; So\; it\; must\; be\; that$ y\notin E_{\alpha }$as\; desired.\; \square $$

Proof. If it were the case that $x+\mathrm{\Delta }x/mo>ythen\; we\; have\; that$ x/mo>y/mo>x+\mathrm{\Delta }x,\; which\; is\; in\; direct\; contradiction\; to\; Theorem 3since$ y\in E_{\alpha }$.\; \square $$

Proof. So first consider any $y\in F_{\alpha }$ so that $y=\mathit{𝑝𝑥}+q$ for some $x\in E_{\alpha }$. Since $E_{\alpha }$ is an embedding $x\in 𝑸$ so that clearly $y=\mathit{𝑝𝑥}+q\in 𝑸$ as well since $p,q\in 𝑸$. Hence since $y$ was arbitrary we have that $F_{\alpha }\subseteq 𝑸$ so that (1) is satisfied.

Now consider the mapping $f:E_{\alpha }\to F_{\alpha }$ defined by $f(x)=\mathit{𝑝𝑥}+q$ for $x\in E_{\alpha }$. Clearly $F_{\alpha }=\left\{f(x)\mid x\in E_{\alpha }\right\}$ so that $f$ is onto. Consider then $x,y\in E_{\alpha }$ where $x/mo>yso\; that\; we\; have$

$$\begin{array}{ll}\hfill x& /mo>y\hfill & & \hfill & & \hfill \mathit{𝑝𝑥}& /mo>\mathit{𝑝𝑦}\hfill \text{(since }p/mo>0\text{)}& \hfill & & \hfill & \hfill \mathit{𝑝𝑥}+q& /mo>\mathit{𝑝𝑦}+q\hfill & & \hfill & & \hfill f(x)& /mo>f(y).\hfill & & \hfill & \end{array}$$

Hence $f$ is an isomorphism. Thus $F_{\alpha }$ is isomorphic to $E_{\alpha }$ so that clearly it is also isomorphic to $\alpha$ since $E_{\alpha }$ is, thereby showing (2).

Lastly for any $y\in F_{\alpha }$ we have $y=\mathit{𝑝𝑥}+q$ for some $x\in E_{\alpha }$. We then have

$$\begin{array}{ll}\hfill a& \le x/mo>b\hfill & & \hfill & & \hfill \mathit{𝑝𝑎}& \le \mathit{𝑝𝑥}/mo>\mathit{𝑝𝑏}\hfill \text{(since }p/mo>0\text{)}& \hfill & & \hfill & \hfill \mathit{𝑝𝑎}+q& \le \mathit{𝑝𝑥}+q/mo>\mathit{𝑝𝑏}+q\hfill & & \hfill & & \hfill \mathit{𝑝𝑎}+q& \le y/mo>\mathit{𝑝𝑏}+q.\hfill & & \hfill & \end{array}$$

This shows both (3) and the last statement. □

Proof. First consider any $y\in E_{\alpha }\cdot U_{\beta }$ so that there is an $x\in E_{\alpha }$ such that $y\in \mathrm{\Delta }x\cdot U_{\beta }+x$. Since $y\in \mathrm{\Delta }x\cdot U_{\beta }+x$ is an embedding by Theorem 5 it follows that $y\in 𝑸$, which shows (1) since $y$ was arbitrary.

Now since $E_{\alpha }$ is an embedding of $\alpha$ there is an isomorphism $f:\alpha \to E_{\alpha }$. Similarly there is an isomorphism $g:\beta \to U_{\beta }$ since $U_{\beta }$ is an embedding of $\beta$. Consider $\alpha \times \beta$ with lexicographic ordering $\prec$. Then define a mapping $h:\alpha \times \beta \to E_{\alpha }\cdot U_{\beta }$ by

$$h(\delta ,𝜀)=\mathrm{\Delta }f(\delta )\cdot g(𝜀)+f(\delta )$$

for $(\delta ,𝜀)\in \alpha \times \beta$, noting that $\mathrm{\Delta }f(\delta )$ is that guaranteed by Theorem 3 since $f(\delta )\in E_{\alpha }$.

First we claim that $h$ is surjective. So consider any $z\in E_{\alpha }\cdot U_{\beta }$ so that there is an $x\in E_{\alpha }$ such that $z\in \mathrm{\Delta }x\cdot U_{\beta }+x$. By definition then there is a $y\in U_{\beta }$ such that $z=\mathrm{\Delta }x\cdot y+x$. Now let $\delta =f^{-1}(x)$ and $𝜀=g^{-1}(y)$ so that $x=f(\delta )$ and $y=g(𝜀)$, which can be done since $f$ and $g$ are bijections. We then have that

$$h(\delta ,𝜀)=\mathrm{\Delta }f(\delta )\cdot g(𝜀)+f(\delta )=\mathrm{\Delta }x\cdot y+x=z,$$

which shows that $h$ is surjective since $z$ was arbitrary.

We also claim that $h$ is an isomorphism and therefore also injective. So consider any $({\delta }_1,{𝜀}_1)$ and $({\delta }_2,{𝜀}_2)$ in $\alpha \times \beta$ where $({\delta }_1,{𝜀}_1)\prec ({\delta }_2,{𝜀}_2)$. By the definition of lexicographic ordering we have the following:

Case: ${\delta }_1/mo>\; <msub>\; <mrow>\; <mi>\; \delta \; </mi>\; </mrow>\; <mrow>\; <mn>\; 2\; </mn>\; </mrow>\; </msub>.\; Then\; since$ f$is\; an\; isomorphism$ f({\delta }_1)/mo>f({\delta }_2),\; and\; also\; by\; Corollary 4it\; follows\; that$ f({\delta }_1)/mo>f({\delta }_1)+\mathrm{\Delta }f({\delta }_1)\le f({\delta }_2).\; We\; also\; have\; that$$$

$$\mathrm{\Delta }f({\delta }_1)\cdot g({𝜀}_1)+f({\delta }_1)/mo>\mathrm{\Delta }f({\delta }_1)+f({\delta }_1)$$

by Theorem 5 since $g({𝜀}_1)\in U_{\beta }$ and $U_{\beta }$ is a unit embedding. Also since $g({𝜀}_2)\ge 0$ (since $g({𝜀}_2)\in U_{\beta }$) and $\mathrm{\Delta }f({\delta }_2)/mo>0(by\; Theorem 3)\; that$

$$f({\delta }_2)\le \mathrm{\Delta }f({\delta }_2)\cdot g({𝜀}_2)+f({\delta }_2).$$

Combining all this results in

$$h({\delta }_1,{𝜀}_1)=\mathrm{\Delta }f({\delta }_1)\cdot g({𝜀}_1)+f({\delta }_1)/mo>\mathrm{\Delta }f({\delta }_1)+f({\delta }_1)\le f({\delta }_2)=h({\delta }_2,{𝜀}_2).$$

Case: ${\delta }_1={\delta }_2$ and ${𝜀}_1/mo>\; <msub>\; <mrow>\; <mi>\; 𝜀\; </mi>\; </mrow>\; <mrow>\; <mn>\; 2\; </mn>\; </mrow>\; </msub>.\; Then\; obviously$ f({\delta }_1)=f({\delta }_2)$so\; that$ \mathrm{\Delta }f({\delta }_1)=\mathrm{\Delta }f({\delta }_2)$but\; also$ g({𝜀}_1)/mo>g({𝜀}_2)since$ g$is\; an\; isomorphism.\; We\; then\; have$$

$$\begin{array}{ll}\hfill g({𝜀}_1)& /mo>g({𝜀}_2)\hfill & & \hfill & & \hfill \mathrm{\Delta }f({\delta }_1)\cdot g({𝜀}_1)& /mo>\mathrm{\Delta }f({\delta }_1)\cdot g({𝜀}_2)\hfill \text{(since }\mathrm{\Delta }f({\delta }_1)/mo>0\text{)}& \hfill & & \hfill & \hfill \mathrm{\Delta }f({\delta }_1)\cdot g({𝜀}_1)+f({\delta }_1)& /mo>\mathrm{\Delta }f({\delta }_1)\cdot g({𝜀}_2)+f({\delta }_1)\hfill & & \hfill & & \hfill \mathrm{\Delta }f({\delta }_1)\cdot g({𝜀}_1)+f({\delta }_1)& /mo>\mathrm{\Delta }f({\delta }_2)\cdot g({𝜀}_2)+f({\delta }_2)\hfill \text{(since }\mathrm{\Delta }f({\delta }_1)=\mathrm{\Delta }f({\delta }_2)\text{ and }f({\delta }_1)=f({\delta }_2)\text{)}& & \hfill & & \hfill & \hfill h({\delta }_1,{𝜀}_1)& /mo>h({\delta }_2,{𝜀}_2).\hfill & & \hfill & \end{array}$$

Thus in all cases $h({\delta }_1,{𝜀}_1)/mo>h({\delta }_2,{𝜀}_2),\; which\; shows\; that$ h$is\; an\; isomorphism\; since$ ({\delta }_1,{𝜀}_1)$and$ ({\delta }_2,{𝜀}_2)$were\; arbitrary.\; Hence$ E_{\alpha }\cdot U_{\beta }$is\; isomorphic\; to\; the\; lexicographic\; ordering\; of$ \alpha \times \beta $and\; therefore\; also\; to$ \beta \cdot \alpha $by\; Theorem 6.5.8.\; This\; shows\; part\; (2)\; of\; the\; embedding\; definition.$

Lastly consider any $z\in E_{\alpha }\cdot U_{\beta }$ so that there is an $x\in E_{\alpha }$ such that $z\in \mathrm{\Delta }x\cdot U_{\beta }+x$. Then since $U_{\beta }/mo>1we\; have\; that$ z/mo>\mathrm{\Delta }x+x\le bby\; Theorems 5and3.\; Also\; since$ 0\le U_{\beta }$it\; follows\; from\; Theorem 5that$ a\le x\le z$since$ a\le E_{\alpha }$.\; Since$ z$was\; arbitrary\; this\; shows\; that$ a\le E_{\alpha }\cdot U_{\beta }/mo>b,\; which\; shows\; (3).\; This\; completes\; the\; proof.\; \square $$$

Proof. First consider any $n$ and $m$ in $\omega$ where $n\ne m$. We can assume without loss of generality $n/mo>m.\; Then$ f(n)/mo>f(m)since$ f$is\; an\; isomorphism\; and,\; moreover,\; it\; follows\; from\; Corollary 4that$ \mathrm{\Delta }f(n)+f(n)\le f(m)$.\; Hence\; by\; Theorem 5we\; have\; that$$

$$A_n=\mathrm{\Delta }f(n)\cdot U_{{\alpha }_n}+f(n)/mo>\mathrm{\Delta }f(n)+f(n)\le f(m)\le \mathrm{\Delta }f(m)\cdot U_{{\alpha }_m}+f(m)=A_m$$

since $U_{{\alpha }_n}$ and $U_{{\alpha }_m}$ are unit embeddings. Hence all the $A_n$ are disjoint and moreover $A_0/mo>A_1/mo>A_2/mo>\dots .\; Also\; clearly\; by\; Theorem 5each$ A_n$is\; isomorphic\; to$ {\alpha }_n$since$ U_{{\alpha }_n}$is.$

Now for $p,q\in 𝑸$ let $[p,q)=\{x\in 𝑸\mid q\le x/mo>q\}.\; We\; then\; claim\; that$ E_{\alpha }\cap [a,f(n+1))$for\; any$ n\in \omega $is\; isomorphic\; to$ {\alpha }_n$,\; which\; we\; shall\; show\; by\; induction\; on$ n$.\; For$ n=0$we\; clearly\; have\; that$ a\le A_0\le \mathrm{\Delta }f(0)+f(0)\le f(1)\le A_m$for\; any$ m\ge 1$.\; From\; this\; it\; follows\; that$ E_{\alpha }\cap [a,f(n+1))=E_{\alpha }\cap [a,f(1))=A_0$,\; which\; is\; isomorphic\; to$ {\alpha }_0={\alpha }_n$by\; what\; was\; shown\; above.$

Now suppose that $E_{\alpha }\cap [a,f(n+1))$ is isomorphic to ${\alpha }_n$. We clearly have $E_{\alpha }\cap [a,f(n+2))=\left[E_{\alpha }\cap [a,f(n+1))\right]\cup \left[E_{\alpha }\cap [f(n+1),f(n+2))\right]$ and that $E_{\alpha }\cap [f(n+1),f(n+2))=A_{n+1}$, which is isomorphic to ${\alpha }_{n+1}$ by what was shown above. Then $E_{\alpha }\cap [a,f(n+2))$ is the sum of $E_{\alpha }\cap \mathit{𝑐𝑙𝑜𝑝𝑎},f(n+1)$ and $E_{\alpha }\cap [f(n+1),f(n+2))=A_{n+1}$ so that by Theorem 6.5.3, the induction hypothesis, and the given property of $\left\{{\alpha }_n\right\}$ it is isomorphic to ${\alpha }_n+{\alpha }_{n+1}={\alpha }_{n+1}$. This completes the inductive proof.

We also show that $E_{\alpha }\cap [a,f(n+1))$ is an initial segment of $E_{\alpha }$ for any $n\in \omega$. So consider any such $n$, any $x\in E_{\alpha }\cap [a,f(n+1))$, and any $y\in E_{\alpha }$ where $y/mo>x.\; Since$ a\le E_{\alpha }$clearly$ a\le y$.\; We\; also\; have$ y/mo>x/mo>f(n+1)(since$ x\in [a,f(n+1))$).\; Hence$ y\in [a,f(n+1))$so\; that\; also$ y\in E_{\alpha }\cap [a,f(n+1))$,\; which\; shows\; that$ E_{\alpha }\cap [a,f(n+1))$is\; an\; initial\; segment\; of$ E_{\alpha }$by\; definition.$$

Now we claim that $E_{\alpha }$ is a well-ordered set. So consider any nonempty subset $B$ of $E_{\alpha }$. Then there is some $x\in B$ and since $x\in E_{\alpha }$ there is an $n\in \omega$ such that $x\in A_n$. Then clearly $x\in B\cap [a,f(n+1)]$ so that $B\cap [a,f(n+1)]$ is a nonempty subset of $E_{\alpha }\cap [a,f(n+1))$. It was shown above that $E_{\alpha }\cap [a,f(n+1))$ is isomorphic to ${\alpha }_n$ so that it is a well-ordered set. Hence $B\cap [a,f(n+1)]$ has a least element $y$. We claim that this is the least element of $B$, so consider any $z\in B$. If $z/mo>f(n+1)then\; clearly$ z\in B\cap [a,f(n+1))$so\; that\; obviously$ y\le z$since$ y$is\; the\; least\; element\; of$ B\cap [a,f(n+1))$.\; On\; the\; other\; hand\; if$ z\ge f(n+1)$then$ y/mo>f(n+1)\le z(since$ y\in [a,f(n+1))$)\; so\; that\; again$ y\le z$.\; Since$ z$was\; arbitrary\; this\; shows\; that$ y$is\; in\; fact\; the\; least\; element\; of$ B$.\; Since$ B$was\; an\; arbitrary\; subset\; of$ E_{\alpha }$this\; shows\; that$ E_{\alpha }$is\; well-ordered.$$

Since $E_{\alpha }$ is a well-ordered set it is isomorphic to some ordinal $\gamma$ by Theorem 6.3.1. We then claim that $\gamma =\alpha$. Letting $C=\left\{{\alpha }_n\mid n\in \omega \right\}$, we show this by showing that $\gamma$ is the least upper bound of $C$, which shows that $\gamma =\alpha$ by the least upper bound property (Lemma 2) since $\alpha =\sup <!--\; FUNCTION\; APPLICATION\; -->C$ by definition. So first consider any ${\alpha }_n\in C$. It was shown above that ${\alpha }_n$ is isomorphic to $E_{\alpha }\cap [a,f(n+1))$, and this was shown to be an initial segment of $E_{\alpha }$, which itself is isomorphic to $\gamma$. Thus it follows that that ${\alpha }_n/mo>\gamma so\; that$ {\alpha }_n\le \gamma $is\; true.\; Since$ {\alpha }_n$was\; arbitrary\; this\; shows\; that$ \gamma $is\; an\; upper\; bound\; of$ C$.$

Now consider any ordinal $\delta /mo>\gamma so\; that$ \delta \in \gamma $.\; Let$ g$be\; the\; isomorphism\; from$ \gamma $to$ E_{\alpha }$since\; it\; has\; been\; shown\; that\; they\; are\; isomorphic.\; Then\; since$ g(\delta )\in E_{\alpha }$so\; that\; there\; is\; an$ n\in \omega $such\; that$ g(\delta )\in A_n$.\; Then\; also$ g(\delta )\in E_{\alpha }\cap [a,f(n+1))$.\; Since$ E_{\alpha }\cap [a,f(n+1))$is\; an\; initial\; segment\; of$ E_{\alpha }$(just\; shown\; above)\; it\; follows\; that$ g^{-1}[E_{\alpha }\cap [a,f(n+1))]$is\; an\; initial\; segment\; of$ \gamma $.\; Since$ \gamma $is\; an\; ordinal\; it\; follows\; from\; Lemma 1that$ g^{-1}[E_{\alpha }\cap [a,f(n+1))]$is\; an\; ordinal.\; Since$ g^{-1}\upharpoonright E_{\alpha }\cap [a,f(n+1))$is\; an\; isomorphism\; (since$ g$is)\; and$ E_{\alpha }\cap [a,f(n+1))$is\; isomorphic\; to$ {\alpha }_n$(shown\; above),\; it\; follows\; that$ g^{-1}[E_{\alpha }\cap [a,f(n+1))]$is\; in\; fact$ {\alpha }_n$!\; Then,\; since$ g(\delta )\in E_{\alpha }\cap [a,f(n+1))$we\; have\; that$ \delta \in g^{-1}[E_{\alpha }\cap [a,f(n+1))]={\alpha }_n$so\; that$ \delta /mo>\; <msub>\; <mrow>\; <mi>\; \alpha \; </mi>\; </mrow>\; <mrow>\; <mi>\; n\; </mi>\; </mrow>\; </msub>.\; Hence$ \delta $is\; not\; an\; upper\; bound\; of$ C$.\; Since$ \delta $was\; arbitrary\; this\; shows\; that$ \gamma $is\; in\; fact\; the\; least\; upper\; bound\; of$ C$so\; that$ \gamma =\alpha $.$$

Parts (1) and (3) of the definition of an embedding are trivial to show by the same arguments as those used in the proof of Theorem 6. Hence $E_{\alpha }$ is an embedding of $\alpha$ in $[a,b)$. □

Proof. Clearly $U_{\omega }\subseteq 𝑸$ so that (1) is satisfied. To show (2) let $g:\omega \to U_{\alpha }$ be defined by $g(n)=f(n)=1-1∕(n+1)$, which is clearly onto based on the definition of $U_{\alpha }$. We also claim that $g$ is an isomorphism. To this end consider any $n,m\in \omega$ where $n/mo>m.\; We\; then\; have$

$$\begin{array}{ll}\hfill n& /mo>m\hfill & & \hfill & & \hfill n+1& /mo>m+1\hfill & & \hfill & & \hfill \frac{n+1}{m+1}& /mo>1\hfill \text{(since }m+1\ge 1/mo>0\text{ since }m\ge 0\text{)}& \hfill & & \hfill & \hfill \frac{1}{m+1}& /mo>\frac{1}{n+1}\hfill \text{(since }n+1\ge 1/mo>0\text{ since }n\ge 0\text{)}& \hfill & & \hfill & \hfill -\frac{1}{m+1}& /mo>-\frac{1}{n+1}\hfill & & \hfill & & \hfill 1-\frac{1}{m+1}& /mo>1-\frac{1}{n+1}\hfill & & \hfill & & \hfill g(m)& /mo>g(n)\hfill & & \hfill & & \hfill g(n)& /mo>g(m).\hfill & & \hfill & \end{array}$$

Hence $g$ is an isomorphism so that $U_{\omega }$ is indeed isomorphic to $\omega$, which shows (2).

Lastly consider any $x\in U_{\omega }$ so that $x=f(n)=1-1∕(n+1)$ for some $n\in \omega$. Then we have

$$\begin{array}{ll}\hfill n& \ge 0/mo>-1\hfill & & \hfill & & \hfill n+1& \ge 1/mo>0\hfill & & \hfill & & \hfill 1& \ge \frac{1}{n+1}/mo>0\hfill \text{(since }n+1\ge 1/mo>0\text{)}& \hfill & & \hfill & \hfill -1& \le -\frac{1}{n+1}/mo>0\hfill & & \hfill & & \hfill 0& \le 1-\frac{1}{n+1}/mo>1\hfill & & \hfill & & \hfill 0& \le x/mo>1.\hfill & & \hfill & \end{array}$$

Since $x$ was arbitrary this shows that $0\le U_{\omega }/mo>1so\; that\; (3)\; holds\; and$ U_{\omega }$is\; a\; unit\; embedding.$

Now let $E_1=\left\{1\right\}$. Clearly this is an embedding of the ordinal 1. Moreover we have $0\le U_{\omega }/mo>1\le E_1/mo>2so\; that$ U_{\omega }$and$ E_1$are\; disjoint.\; Then\; clearly$ U_{\omega }\cup E_1$is\; the\; sum\; of$ U_{\omega }$and$ E_1$so\; that\; it\; is\; isomorphic\; to$ \omega +1$by\; Theorem 6.5.3\; and\; thus\; an\; embedding\; of$ \omega +1$since\; it\; is\; trivial\; to\; show\; that$ 0\le U_{\omega }\cup E_1/mo>2.\; \square $$