A proper subset of $\Bbb Q$ that is order-isomorphic to $\Bbb Q$; An infinite set that is NOT order-isomorphic to any of its proper subset

Question

While it's not hard to define an order isomorphism between $\Bbb N$ and one of its proper subset or between $\Bbb Z$ and one of its proper subset, I'm unable to find sets in below situations:

1. A proper subset of $\Bbb Q$ that is order-isomorphic to $\Bbb Q$. This required subset of $\Bbb Q$ is ordered under the usual ordering $<$.

2. An infinite set that is NOT order-isomorphic to any of its proper subset. This infinite set and all of its subsets have the same ordering.

Please help me find some examples (if exist) for these sets!

Answer 1

This should have been posted as two separate questions.

1. The map $$x\mapsto\begin{cases}\ \ \ \ x\ \ \ \ \text{ if }\ x\lt0\\ x+1\ \text{ if }\ x\ge0\end{cases}$$ is an order-isomorphism from $\mathbb Q$ to $\mathbb Q\setminus[0,1).$

2. A dense subset $S$ of $\mathbb R$ which is not isomorphic to any of its proper subsets can be constructed by transfinite induction with the axiom of choice. In fact $S$ can be constructed so that the only strictly increasing function $f:S\to S$ is the identity function.

P.S. Constructing an order-isomorphism between $\mathbb Q$ and $\mathbb Q\setminus\{0\}$ is more complicated. Here is one way to do it. (Another, more general, way is Cantor's "back-and-forth" method.)

Lemma. Given $a,b,c,d\in\mathbb Q$, $a\lt b$, $c\lt d$, we can define an order-isomorphism $f:\mathbb Q\cap[a,b]\to\mathbb Q\cap[c,d]$.

Proof. Let $f(x)=c\cdot\frac{x-b}{a-b}+d\cdot\frac{x-a}{b-a}$.

Fix sequences $a_1\lt a_2\lt a_3\lt\cdots$ and $b_1\gt b_2\gt b_3\gt\cdots$ such that $\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n=\sqrt2$. (A nice way to do this is to use the continued fraction convergents: $a_1=1$, $b_1=3/2$, $a_2=7/5$, $b_2=17/12$, $a_3=41/29$, etc.)

Define $c_n=-\frac1n$ and $d_n=\frac1n$ so that $c_1\lt c_2\lt c_3\lt\cdots$ and $d_1\gt d_2\gt d_3\gt\cdots$ and $\lim_{n\to\infty}c_n=\lim_{n\to\infty}d_n=0$.

Theorem. There is an order-isomorphism $f:\mathbb Q\to\mathbb Q\setminus\{0\}$ such that $f(a_n)=c_n$ and $f(b_n)=d_n$ for $n=1,2,3,\dots$.

Proof. For $x\in\mathbb Q\cap(-\infty,a_1)$ define $f(x)=x+c_1-a_1$.

For $x\in\mathbb Q\cap[a_1,a_2]$ define $f(x)=c_1\cdot\frac{x-a_2}{a_1-a_2}+c_2\cdot\frac{x-a_1}{a_2-a_1}$.

Further details are left as an exercise for the reader.

Answer 2

1. $(-1,1)\cap \Bbb Q$.
2. I doubt there is any.

Answer 3

In response to a request in the Comments from the Proposer. A proof of a result of Cantor without using a back-and-forth method. It requires much preliminary work on the structure of a linear order, followed by the proof itself, which is short. The main idea is to employ the preliminary result (III). I dk how Cantor proved this theorem..

Theorem. (Cantor). Any countably infinite linear orders $(A,<), (A',<') $ which are order-dense and without end-points are order-isomorphic.

Preliminaries. Let $<$ be a linear order on a countably infinite $A=\{a_n:n\in \Bbb N\}$ with no end-points, and order-dense in itself (i.e. if $x<y$ there exists $z$ with $x<y<z).$ Then

(I). There exists $B\subset A$ which is order-isomorphic to $\Bbb N$ and unbounded above in $A.$ Proof: Let $f(1)=1.$ Recursively, for $n\in \Bbb N$ let $f(n+1)$ be the least $j\in \Bbb N$ such that $a_{f(n)}<a_j. $ Let $B=\{a_{f(n)}:n\in \Bbb N\}.$ To show that $B$ is unbounded above in $A,$ by induction:

(I-i). $a_1=a_{f(1)}<a_{f(2)}$ so $a_1$ is not an upper bound for $B.$

(I-ii). Suppose $n\in \Bbb N$ and no member of $\{a_j:j\leq n\}$ is an upper bound for $B.$ Let $n_0$ be the least $k$ such that $a_{f(k)}\geq \max \{a_j:j\leq n\}.$ Then:

If $a_{n+1}\leq a_{f(n+0)}$ then $a_{n+1}<a_{f(n_0+1}\in B.$

If $a_{n+1}>a_{f(n_0)}$ then $n+1$ is the least $j$ such that $a_j>a_{f(n_0)}$... (because $j\leq n\implies a_j\leq a_{f(n_0)}$)... so by the recursive def'n of $f(n_0+1)$ we have $n+1=f(n_0+1).$ So $a_{n+1}=a_{f(n_0+1)}<a_{f(n_0+2)}\in B.$

(II). There exists $C\subset A$ where $C$ is order-isomorphic to $\Bbb Z$ and $C$ is unbounded above and below in $A.$ Proof: Applying (I) to the reverse-order $<^*$ on $A$ (where $x<^*y$ iff $y<x$), we obtain $B^*\subset A$ with $B^*$ order-isomorphic to the set of negative integers. So with $B$ as in (I), let $C= B\cup B^*$.

(III). $A= \{J_n:n\in \Bbb N\}$ such that

(III-i). $a_1\in J_1.$

(III-ii). Each $J_n$ is order-isomorphic to $\Bbb Z $ and is unbounded above and below in $A.$

(III-iii). $J_n\subset J_{n+1}$ for all $n.$ And whenever $x,y$ are consecutive members of $J_n$ with $x<y,$ there is exactly one $z \in J_{n+1}$ \ $J_n$ such that $x<z<y$... (Note: $x,y$ are consecutive members of $J_n$ iff no member of $J_n$ is between $x$ and $y$).

Proof: Let $J_1=\{a_1\}\cup C$ where $C$ is as in (II). Recursively define $J_{n+1}$ \ $J_n$ as follows: For any consecutive $x,y$ in $J_n$ with $x<y$ let $m$ be the least $j$ such that $x<a_j<y,$ and let $a_m$ be the unique member of $J_{n+1}$ \ $J_n$ that lies between $x$ and $y$. Then (III-i),(III-ii),(III-iii) are satisfied.

It remains to show, by induction, that the set $J=\cup_{n\in \Bbb N}J_n$ is equal to $A$. As follows, by induction:

(III-i'). $a_1\in J_1\subset J.$

(III-ii'). Suppose $\{a_j:j\leq n\}\subset J.$ Let $n_0$ be the least (or any ) $k$ such that $\{a_j:j\leq n\}\subset J_k.$

If $a_{n+1}\in J_{n_0}$ then $a_{n+1} \in J.$

If $a_{n+1}\not \in J_{n_0}$ then there are consecutive $x,y$ in $J_{n_0}$ with $x<a_{n+1}<y,$ but no member of $J_{n_0},$ and a fortiori no member of $\{a_j:j\leq n\}$ lies between $x$ and $y$, therefore $n+1$ is the least $j $ such that $x<a_j<y.$ By the recursive def'n of $J_{n_0+1}$ \ $J_{n_0},$ therefore $a_{n+1}$ is the unique member of $J_{n_0+1}$ \ $J_{n+0} $ between $x$ and $y.$ So $a_{n+1}\in J_{n_0+1}\subset J.$

$\bullet$. After all this preliminary work, to finally prove the Theorem: Let $A=\cup_{n\in \Bbb N}J_n$ and $A'=\cup_{n\in \Bbb N}J'_n$ as in (III). Since $J_1$ and $J'_1$ are each order-isomorphic to $\Bbb Z,$ let $f$ map $J_1$ order-isomorphically onto $J'_1.$

Inductively, define $f$ on each $J_n$ as follows: Suppose $f$ maps $J_n$ order-isomorphically onto $J'_n.$ If $x,y$ are any consecutive members of $J_n$ and $z$ is the unique member of $J_{n+1} \setminus J_n$ between $x$ and $y$ then let $z'$ be the unique member of $J'_{n+1} \setminus J'_n$ between $f(x)$ and $f(y).$ Now let $f(z)=z'.$ Then $f$ maps $J_{n+1}$ onto $J'_{n+1}$ order-isomorphically.

Since $J_n\subset J_{n+1}$ for all $n$ and since $A=\cup_{n\in \Bbb N}J_n,$ if $x,y\in A$ with $x<y$, then $\{x,y\}\subset J_n$ for some $n$, and $f$ is strictly order-preserving on $J_n$, so $f(x)<'f(y)$.

Remark: This theorem can be used to prove that the real interval $(0,1)$ is uncountable. Suppose not. Then there is an order-isomorphism $f:(0,1)\to \Bbb Q\cap (0,1),$ which we can extend to an order-isomorphism $f:[0,1]\to [0,1]\cap \Bbb Q$ by letting $f(0)=0$ and $f(1)=1.$ Using the fact that there is a rational beteen any two reals, we can readily show that $f$ is continuous when considered as a function from $[0,1]$ into $[0,1].$ And a basic result of analysis (the Intermediate Value Property) then implies that $[0,1]=\{f(x): 0\leq x\leq 1\}.$ But $\{f(x):0\leq x\leq 1\}=\Bbb Q \cap[0,1],$ a contradiction.

Another way is that if $(0,1)$ were countable there would be an order-isomorphism $g:(0,1)\cap \Bbb Q\to (0,1),$ but then $\{g(q):q\in \Bbb Q\cap (0,1/\sqrt 2\,)\}$ would be a non-empty subset of $(0,1)$ with no $lub .$