Exercise 6.5.15

Proof. First, let $A=\{{\alpha }_{\delta }\mid \delta /mo>\gamma \},$ B=\{{\beta }_{\delta }\mid \delta /mo>\alpha \},\; and$ C=\{{\beta }_{{\alpha }_{\delta }}\mid \delta /mo>\gamma \}so\; that$ \alpha =\sup <!--\; FUNCTION\; APPLICATION\; -->A$and\; we\; must\; show\; that$ \sup <!--\; FUNCTION\; APPLICATION\; -->B=\sup <!--\; FUNCTION\; APPLICATION\; -->C$.\; We\; show\; this\; by\; showing\; that$ \sup <!--\; FUNCTION\; APPLICATION\; -->C$has\; the\; least\; upper\; bound\; property\; of$ B$.$$$

So first consider any ${\beta }_{\delta }\in B$ so that $\delta /mo>\alpha .\; It\; then\; follows\; that$ \delta $is\; not\; an\; upper\; bound\; of$ A$so\; that\; there\; is\; a$ \xi /mo>\gamma such\; that$ \delta /mo>\; <msub>\; <mrow>\; <mi>\; \alpha \; </mi>\; </mrow>\; <mrow>\; <mi>\; \xi \; </mi>\; </mrow>\; </msub>.\; From\; this\; we\; have\; that$ {\beta }_{\delta }\le {\beta }_{{\alpha }_{\xi }}\le \sup <!--\; FUNCTION\; APPLICATION\; -->C$since$ \left\{{\beta }_{\delta }\right\}$is\; a\; non-decreasing\; sequence\; and$ {\beta }_{{\alpha }_{\xi }}\in C$.\; Since$ {\beta }_{\delta }$was\; arbitrary\; this\; shows\; that$ \sup <!--\; FUNCTION\; APPLICATION\; -->C$is\; an\; upper\; bound\; of$ B$.$$$

Now consider any $\delta /mo>\sup <!--\; FUNCTION\; APPLICATION\; -->C.\; Then$ \delta $is\; not\; an\; upper\; bound\; of$ C$so\; that\; there\; is\; a$ \xi /mo>\gamma such\; that$ \delta /mo>\; <msub>\; <mrow>\; <mi>\; \beta \; </mi>\; </mrow>\; <mrow>\; <msub>\; <mrow>\; <mi>\; \alpha \; </mi>\; </mrow>\; <mrow>\; <mi>\; \xi \; </mi>\; </mrow>\; </msub>\; </mrow>\; </msub>.\; And\; since$ \xi /mo>\gamma we\; have\; that$ {\alpha }_{\xi }\in A$so\; that$ {\alpha }_{\xi }/mo>\sup <!--\; FUNCTION\; APPLICATION\; -->A=\alpha since$ \alpha $is\; a\; limit\; ordinal.\; Thus$ {\beta }_{{\alpha }_{\xi }}\in B$.\; Since\; we\; have\; that$ \delta /mo>\; <msub>\; <mrow>\; <mi>\; \beta \; </mi>\; </mrow>\; <mrow>\; <msub>\; <mrow>\; <mi>\; \alpha \; </mi>\; </mrow>\; <mrow>\; <mi>\; \xi \; </mi>\; </mrow>\; </msub>\; </mrow>\; </msub>this\; show\; that$ \delta $is\; not\; an\; upper\; bound\; of$ B$.\; Hence\; since$ \delta $was\; arbitrary\; this\; concludes\; the\; proof\; that$ \sup <!--\; FUNCTION\; APPLICATION\; -->B=\sup <!--\; FUNCTION\; APPLICATION\; -->C$by\; the\; least\; upper\; bound\; property\; (Lemma $$$$$$??). □

Proof. That this has the desired property (i.e. that $\omega +\xi =\xi$) was shown in Exercise 6.5.3b.

Now we show that any ordinal $\alpha /mo>\; <msup>\; <mrow>\; <mi>\; \omega \; </mi>\; </mrow>\; <mrow>\; <mn>\; 2\; </mn>\; </mrow>\; </msup>does\; not\; have\; this\; property.\; So\; consider\; any\; such$ \alpha $so\; that\; by\; Lemma$?? there are natural numbers $n$ and $k$ such that $\alpha =\omega \cdot n+k$. Thus we have

$$\omega +\alpha =\omega +\omega \cdot n+k=\omega \cdot 1+\omega \cdot n+k=\omega \cdot (1+n)+k=w\cdot (n+1)+k$$

We then have that

$$\begin{array}{llllll}\hfill \alpha & =\omega \cdot n+k=\left(\omega \cdot n+k\right)+0& \hfill & & \hfill & \\ \hfill & /mo>\left(\omega \cdot n+k\right)+\omega \hfill \text{(by Lemma 6.5.4a since }0/mo>\omega \text{)}& \hfill & & \hfill & \hfill & =\omega \cdot n+(k+\omega )& \hfill \text{(associativity of addition)}& & \hfill & & \hfill \\ \hfill & =\omega \cdot n+\omega =\omega \cdot n+\omega \cdot 1& \hfill & & \hfill & \\ \hfill & =\omega \cdot (n+1)& \hfill \text{(distributive law)}& & \hfill & & \hfill \\ \hfill & =\omega \cdot (n+1)+0& \hfill & & \hfill & \\ \hfill & \le \omega \cdot (n+1)+k& \hfill \text{(by Lemma 6.5.4 since }0\le k\text{)}& & \hfill & & \hfill \\ \hfill & =\omega +\alpha & \hfill & & \hfill & \end{array}$$

so that clearly $\omega +\alpha \ne \alpha$. Since $\alpha /mo>\; <msup>\; <mrow>\; <mi>\; \omega \; </mi>\; </mrow>\; <mrow>\; <mn>\; 2\; </mn>\; </mrow>\; </msup>was\; arbitrary\; this\; shows\; our\; result.\; \square$

Proof. First we have

$$\begin{array}{lll}\hfill \omega \cdot \xi =\omega \cdot {\omega }^{\omega }={\omega }^1\cdot {\omega }^{\omega }={\omega }^{1+\omega }={\omega }^{\omega }=\xi & & \hfill \end{array}$$

where we have used Exercise 6.5.13a. Thus $\xi ={\omega }^2$ has the desired property.

Now consider any $0/mo>\alpha /mo>\; <msup>\; <mrow>\; <mi>\; \omega \; </mi>\; </mrow>\; <mrow>\; <mi>\; \omega \; </mi>\; </mrow>\; </msup>.\; Since\; by\; Definition 6.5.9c\; we\; have\; that$ {\omega }^{\omega }=\underset{n/mo>\omega }{\sup <!--\; FUNCTION\; APPLICATION\; -->}it\; follows\; from\; the\; least\; upper\; bound\; property\; (Lemma $$??) that there is an $n/mo>\omega such\; that$ \alpha /mo>\; <msup>\; <mrow>\; <mi>\; \omega \; </mi>\; </mrow>\; <mrow>\; <mi>\; n\; </mi>\; </mrow>\; </msup>since$ \alpha $is\; not\; an\; upper\; bound\; of$ \{{\omega }^n\mid n/mo>\omega \}.\; From\; this\; is\; follows\; that\; the\; set$ A=\{k\in \omega \mid \alpha /mo>\; <msup>\; <mrow>\; <mi>\; \omega \; </mi>\; </mrow>\; <mrow>\; <mi>\; k\; </mi>\; </mrow>\; </msup>\}is\; not\; empty.\; Since\; this\; is\; a\; set\; of\; natural\; numbers\; (which\; is\; well-ordered)\; it\; has\; a\; least\; element$ m$.\; Note\; also\; that\; it\; has\; to\; be\; that$ m/mo>0since\; were\; it\; the\; case\; that$ m=0$then\; we\; would\; have$ \alpha /mo>{\omega }^m={\omega }^0=1,\; which\; implies\; that$ \alpha =0$,\; which\; contradicts\; our\; initial\; supposition\; that$ 0/mo>\alpha .\; Thus$ m\ge 1$so\; that$ m-1$is\; still\; a\; natural\; number.$$$$$$$

Since $m$ is the least element of $A$ it follows that $\alpha \ge {\omega }^{m-1}$ since otherwise $m-1$ would be the least element of $A$. Hence we have

$${\omega }^{m-1}\le \alpha /mo>{\omega }^m.$$

It then follows from Exercise 6.5.8b that

$$\begin{array}{llllll}\hfill \omega \cdot {\omega }^{m-1}& \le \omega \cdot \alpha & \hfill & & \hfill & \\ \hfill {\omega }^1\cdot {\omega }^{m-1}& \le \omega \cdot \alpha & \hfill & & \hfill & \\ \hfill {\omega }^{1+m-1}& \le \omega \cdot \alpha & \hfill \text{(by Exercise 6.5.13a)}& & \hfill & & \hfill \\ \hfill {\omega }^m& \le \omega \cdot \alpha .& \hfill & & \hfill & \end{array}$$

Putting this together, we have that

$$\alpha /mo>{\omega }^m\le \omega \cdot \alpha$$

so that clearly $\omega \cdot \alpha \ne \alpha$. Since $1/mo>\alpha /mo>\; <msup>\; <mrow>\; <mi>\; \omega \; </mi>\; </mrow>\; <mrow>\; <mi>\; \omega \; </mi>\; </mrow>\; </msup>was\; arbitrary,\; this\; shows\; that$ {\omega }^{\omega }$is\; the\; least\; such\; ordinal\; \square$

Proof. To show that $𝜀$ has the required property we must first show that the sequence defined by ${\{n\omega }^{\}}for$ n/mo>\omega is\; non-decreasing\; even\; though\; this\; is\; somewhat\; obvious.\; We\; show\; this\; by\; induction.\; For$ n=0$we\; have$$

$${}^n\omega {=}^0\omega =1/mo>\omega ={\omega }^1={\omega }^{{}^0\omega }{=}^1\omega {=}^{n+1}\omega$$

Now suppose that ${}^n\omega {/mo>n+1}^{\omega }so\; that\; we\; have$

$$\begin{array}{llllll}\hfill {}^{n+1}\omega & ={\omega }^{{}^n\omega }& \hfill & & \hfill & \\ \hfill & /mo>{\omega }^{{}^{n+1}\omega }\hfill \text{(by Exercise 6.5.14b and the induction hypothesis)}& & \hfill & & \hfill & \hfill & {=}^{n+2}\omega ,& \hfill & & \hfill & \end{array}$$

which completes the induction.

Returning to the main problem, we then have

$$\begin{array}{ll}\hfill {\omega }^{𝜀}& =\underset{\alpha /mo>𝜀}{\sup <!--\; FUNCTION\; APPLICATION\; -->}\hfill \text{(by Definition 6.5.9c since }𝜀\text{ is clearly a limit ordinal)}& & \hfill & & \hfill & \hfill & =\underset{n/mo>\omega }{\sup <!--\; FUNCTION\; APPLICATION\; -->}\hfill \text{(by Lemma }\text{1<!-- tex4ht:ref: lem:ord:props:sups -->}\text{ since }{\{n\omega }^{\}}\text{ is non-decreasing)}& \hfill & & \hfill & \hfill & =\underset{\hfill & }{\sup <!--\; FUNCTION\; APPLICATION\; -->{}^{n/mo>\omega }n+1\omega {=}^{\omega }\omega =𝜀}\end{array}$$

so that $𝜀$ does have the desired property.

Now we must show that it is the least such ordinal that has this property. So consider any $\alpha /mo>𝜀then\; since$ 𝜀=\underset{$ n/mo>\omega such\; that$ \alpha {/mo>n}^{\omega }by\; the\; least\; upper\; bound\; property\; (Lemma $$}{\sup <!--\; FUNCTION\; APPLICATION\; -->{}^{n/mo>\omega }n\omega }$$??). It then follows that the set $A=\{k\in \omega \mid \alpha {/mo>k}^{\omega }\}is\; not\; empty.\; Since\; this\; is\; a\; set\; of\; natural\; numbers\; it\; follows\; that\; it\; has\; a\; least\; element$ m$.$

Case: $m=0$. Then $\alpha {/mo>m}^{\omega }so\; that\; it\; must\; be\; that$ \alpha =0$.\; But\; then\; we\; have$

$${\omega }^{\alpha }={\omega }^0=1\ne 0=\alpha$$

so that $\alpha$ does not have the property.

Case: $m/mo>0.\; Then$ m\ge 1$so\; that$ m-1$is\; still\; a\; natural\; number.\; It\; then\; follows\; that$ \alpha {\ge }^{m-1}\omega $since\; otherwise$ m-1$would\; be\; the\; least\; element\; of$ A$.\; Thus\; we\; have$

$${}^{m-1}\omega \le \alpha {/mo>m}^{\omega }$$

It then follows from Exercise 6.5.14b that

$$\begin{array}{llll}\hfill {\omega }^{{}^{m-1}\omega }& \le {\omega }^{\alpha }& \hfill & \\ \hfill {}^m\omega & \le {\omega }^{\alpha }.& \hfill & \end{array}$$

Thus we have

$$\alpha {/mo>m}^{\omega }$$

so that clearly $\alpha$ does not have the property since ${\omega }^{\alpha }\ne \alpha$. Since $\alpha$ was arbitrary and the cases exhaustive this shows that $𝜀$ is indeed the least such ordinal. □