Exercise 24.12

Proof of $(a)$. We first note order-preserving maps are injective. Letting $f:A\to B$ be such a map, if $a_1,a_2\in A$, then one is larger than the other by the comparability property of order relations, so one of $f(a_1),f(a_2)$ is larger than the other, hence unequal.

Now suppose $[a,c)$ has order type $[0,1)$, and let $f:[0,1)\to [a,c)$ be the order isomorphism. We claim

$$g(x)=f\{[f^{-1}(b)]x\},h(x)=f\{f^{-1}(b)+[1-f^{-1}(b)]x\}$$

define order isomorphisms $g:[0,1)\to [a,b)$ and $h:[0,1)\to [b,c)$. They are order-preserving since if $x,y\in [0,1)$,

$$\begin{array}{llll}\hfill x<y& ⟹[f^{-1}(b)]x<[f^{-1}(b)]y⟹g(x)=f\{[f^{-1}(b)]x\}<f\{[f^{-1}(b)]y\}=g(y)& \hfill & \\ \hfill x<y& ⟹f^{-1}(b)+[1-f^{-1}(b)]x<f^{-1}(b)+[1-f^{-1}(b)]y& \hfill & \\ \hfill & ⟹h(x)=f\{f^{-1}(b)+[1-f^{-1}(b)]x\}<f\{f^{-1}(b)+[1-f^{-1}(b)]y\}=h(y)& \hfill & \end{array}$$

where the first implications are due to our linear transformations being strictly monotonic increasing, and the second since $f$ is order-preserving. This also implies injectivity by the above. It remains to show surjectivity. Let $z\in [a,b),z^{\prime }\in [b,c)$. Then, $0\le f^{-1}(z)<f^{-1}(b)$ and $f^{-1}(b)\le f^{-1}(z^{\prime })<1$, and so

$$g\left[\frac{f^{-1}(z)}{f^{-1}(b)}\right]=z,h\left[\frac{f^{-1}(z^{\prime })-f^{-1}(b)}{1-f^{-1}(b)}\right]=z^{\prime }.$$

Conversely, suppose $[a,b)$ and $[b,c)$ have order type $[0,1)$, and let $g:[0,1)\to [a,b)$, $h:[0,1)\to [b,c)$ be the order isomorphisms. We claim

$$f(x)=\{\begin{array}{cc}g(2x)\hfill & \text{if}0\le x<1∕2\hfill \\ h(2x-1)\hfill & \text{if}1∕2\le x<1\hfill \end{array}$$

is an order isomorphism. It preserves orders since $g,h$ preserve orders on their respective domains, and since if $x<1∕2\le y$, applying $f$ gives

$$f(x)=g(2x)<b\le h(2y-1)=f(y).$$

This also shows injectivity by the above. $f$ is surjective since if $z\in [a,c)$,

$$z<b⟹f[g^{-1}(z)∕2]=z,z\ge b⟹f\{[h^{-1}(z)+1]∕2\}=z.$$

□

Proof of $(b)$. Suppose $[x_0,b)$ has order type $[0,1)$. For any $i\in {\mathbb{Z}}_{\text{+}}$, by $(a)$, $[x_i,b)$ has order type $[0,1)$; applying $(a)$ again gives that $[x_i,x_{i+1})$ has order type $[0,1)$.

Now suppose every $[x_i,x_{i+1})$ has order type $[0,1)$. If $f_i:[0,1)\to [x_i,x_{i+1})$ are order isomorphisms, first define

$$f:[0,\infty )\to [x_0,b),x\mapsto f_i(x-i)\text{if}x\in [i,i+1),$$

which is well-defined since any $x\in [0,\infty )$ is in some set of the form $[i,i+1)$. We claim $f$ is an order isomorphism. If $x,y\in [0,\infty )$, then $x\in [i,i+1),y\in [j,j+1)$ for some $i,j$. Suppose $x<y$. Then,

$$\begin{array}{llll}\hfill i\ne j& ⟹f(x)=f_i(x-i)<x_{i+1}\le x_j\le f_j(y-j)=f(y),& \hfill & \\ \hfill i=j& ⟹f(x)=f_i(x-i)<f_i(y-i)=f(y),& \hfill & \end{array}$$

since the $f_i$ are order-preserving; this also implies injectivity by the above. To show surjectivity, we first know $f$ maps onto ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_i[x_i,x_{i+1})$ by definition. So let $z\in [x_0,b)$. Since $b$ is the least upper bound of the $\{x_i\}$, $z$ is not an upper bound, so there exists $i$ such that $z\in [x_i,x_{i+1})$. But then, since the $f_i$ are bijective as well, $f(f_i^{-1}(z)+i)=z$.

Now let $g:[0,1)\to [0,\infty )$ be defined as $x\mapsto x∕(1-x)$; this is an order isomorphism since it has inverse $x∕(1+x)$, and since it is strictly monotonic increasing. Thus, $f\circ g:[0,1)\to [x_0,b)$ is a bijection, and preserves orientation since $f,g$ do. □

Proof of $(c)$. Let $a>a_0$. We proceed by transfinite induction. $S_{\mathrm{\Omega }}$ is a well-ordered set, and so if we let $J$ be the set of $a\in S_{\mathrm{\Omega }}$ such that the claim holds, it suffices to show that for every $a\in J$, $S_a\subseteq J⟹a\in J$.

We first show that either $a$ has an immediate predecessor or there exists a sequence $\{a_i\}\subseteq S_a$ such that $a=\sup <!--\; FUNCTION\; APPLICATION\; -->\{a_i\}$. Suppose $a$ does not have an immediate predecessor. Then, we have the section $S_a=\{b_i\}$, which is countable by definition of $S_{\mathrm{\Omega }}$. $(b_1,a]\ne \varnothing$ since $a$ has no immediate predecessor, and so let $a_1\in (b_1,a]$. We construct the $a_i$ inductively as follows: if we have $a_n$, let $a_{n+1}\in (\sup <!--\; FUNCTION\; APPLICATION\; -->\{a_n,b_{n+1}\},a]$, which is nonempty as above. We then get a sequence of elements $a_1<a_2<\cdots <a$. But since $a_n>b_n$ for all $n$ by construction, we see that $a\ge \sup <!--\; FUNCTION\; APPLICATION\; -->\{a_i\}$. Moreover, if $a>\sup <!--\; FUNCTION\; APPLICATION\; -->\{a_i\}$, then $\sup <!--\; FUNCTION\; APPLICATION\; -->\{a_i\}=b_k$ for some $k$, for $S_a$ contains all elements less than $a$, and hence $\sup <!--\; FUNCTION\; APPLICATION\; -->\{a_i\}<a_k$, contradicting that $\sup <!--\; FUNCTION\; APPLICATION\; -->\{a_i\}$ is an upper bound.

Now suppose $S_a\subseteq J$. If $a$ has an immediate predecessor $a-1$, then $[a_0\times 0,a\times 1)=[a_0\times 0,(a-1)\times 1)\cup [a\times 0,a\times 1)$ has order type $[0,1)$ by $(a)$, for we have the order isomorphism $[a\times 0,a\times 1)\to [0,1)$ defined by $a\times x\mapsto x$, which is trivially bijective and order-preserving since $S_{\mathrm{\Omega }}\times [0,1)$ was constructed with the dictionary order. On the other hand, if $a$ does not have an immediate predecessor, then there exists a sequence $\{a_i\}\subseteq S_a$ such that $a=\sup <!--\; FUNCTION\; APPLICATION\; -->\{a_i\}$, and so the claim follows by $(b)$. □

Proof of $(d)$. Let $a\times b,a^{\prime }\times b^{\prime }$ be two points in $L$; suppose without loss of generality that $a\times b<a^{\prime }\times b^{\prime }$. By $(c)$, $[a_0\times 0,a\times 1)$ and $[a_0\times 0,a^{\prime }\times 1)$ have order type $[0,1)$; by $9a)$, this implies $[a_0\times 0,a\times b)$ and $[a_0\times 0,a^{\prime }\times b^{\prime })$ have order type $[0,1)$. Hence, by $(a)$, $Y=[a\times b,a^{\prime }\times b^{\prime })$ has order type $[0,1)$. Let $f:[0,1)\to Y$ be the order isomorphism. We claim $f$ is continuous. First, since $Y$ is an interval, it is convex, and so by Theorem $16.4$ the order topology on $Y$ is the same as the subspace topology on $Y$ inherited from $L$. Then, for any basis set $A=(c\times d,c^{\prime }\times d^{\prime })\subseteq Y$, $f^{-1}(A)=(f^{-1}(c\times d),f^{-1}(c^{\prime }\times d^{\prime }))$ since $f$ is an order isomorphism, and moreover this preimage is open. Also, for any basis set $B=[a\times b,c^{\prime }\times d^{\prime })\subseteq Y$, $f^{-1}(B)=[f^{-1}(a\times b),f^{-1}(c^{\prime }\times d^{\prime }))$, which is again open. Thus, $f$ is continuous. Finally, if we define

$$F(x)=\{\begin{array}{cc}f(x)\hfill & \text{if}x\in [0,1)\hfill \\ a^{\prime }\times b^{\prime }\hfill & \text{if}x=1\hfill \end{array}$$

we have a continuous path $F:[0,1]\to [a\times b,a^{\prime }\times b^{\prime }]$ by the pasting lemma (Theorem $18.3$), and so $L$ is path connected. □

Proof of $(e)$. Let $a\times b$ be a point in $L$. Since $S_{\mathrm{\Omega }}\times [0,1)$ does not have a maximal element, there is some $a^{\prime }\times b^{\prime }>a\times b$. Now by $(c)$, there exists an order isomorphism $f:[0,1)\to [a_0\times 0,a^{\prime }\times b^{\prime })$. Restricting $f$ to $(0,1)$, we get another order isomorphism $f:(0,1)\to (a_0\times 0,a^{\prime }\times b^{\prime })$. The set $[a_0\times 0,a^{\prime }\times b^{\prime })$ is open in $S_{\mathrm{\Omega }}\times [0,1)$, and so $(a_0\times 0,a^{\prime }\times b^{\prime })$ is open in $L$, and is a neighborhood of $a\times b$.

We claim $(a_0\times 0,a^{\prime }\times b^{\prime })$ is homeomorphic to $(0,1)$. We already have a bijection that is continuous by the same argument as in $(d)$, and so it suffices to show $f$ is open as well. But if $(x,y)\subseteq (0,1)$ is a basis set, then $f(x,y)=(f(x),f(y))$ since $f$ is an order isomorphism, and moreover open since the topology on $(a_0\times 0,a^{\prime }\times b^{\prime })$ is the order topology. □

Proof of $(f)$. Suppose $L$ could be imbedded in ${ℝ}^n$; then, every subspace of ${ℝ}^n$ has a countable basis by Example $30.1$, and since $L$ is homeomorphic with such a subspace, it also has a countable basis. Now, since $X=(S_{\mathrm{\Omega }}\times \{0\})\setminus \{a_0\times 0\}$ is a convex subset of $L$, the subspace topology on $X$ is the same as the order topology by Theorem $16.4$. Thus, the intersection of the countable basis for $L$ with $X$ forms a countable basis by Theorem 30.2. This implies that there is a countable subset $Y$ of $X$ that is dense in $X$ by Theorem $30.3$. By Theorem $10.3$, though, this subset $Y$ has an upper bound $x$ in $X$. Thus, $\varnothing \ne (x,\mathrm{\Omega })\subseteq X\setminus Y$, and so the closure of $Y$ is not all of $X$, a contradiction. □