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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!

Akira
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    Very broad hint: for 1) you just need a Dense Linear Order with No Endpoints, since any such is order- isomorphic to $\mathbb{Q}$ by the usual back-and-forth argument. What subsets of the rationals that are dense within it do you know? (Note that being dense within $\mathbb{Q}$ isn't a requirement for being order-isomorphic, either; there are plenty of bounded subsets that are, too...) – Steven Stadnicki Sep 27 '18 at 15:17
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    Hi @StevenStadnicki! from your hints, I think $\mathbb{Q}\setminus {0}$ maybe fine. Is my choice of $\mathbb{Q}\setminus {0}$ correct? – Akira Sep 27 '18 at 15:25
  • Hi @StevenStadnicki ! Could you please confirm my understanding in previous comment? Thank you so much! – Akira Sep 27 '18 at 16:21
  • An example for (2) does not come to me immediately, but I can tell you that much: without the axiom of choice it can not be proved that there is no such set(there may be simple example for such set I can't think on right now) – ℋolo Sep 27 '18 at 16:28
  • Hi @Holo! Does $\mathbb{Q}\setminus {0}$ satisfy (1)? – Akira Sep 27 '18 at 16:30
  • @Akira I can't think about order isomorphism between $\Bbb Q$ to $\Bbb Q\setminus{0}$, but there is easy example: $f(x)=\begin{cases}x&x<0\x+1&x\ge 0\end{cases}$, it is easy to see this is order isomorphism from $\Bbb Q$ to $\Bbb Q\setminus\left([0,1)\cap\Bbb Q\right)$ – ℋolo Sep 27 '18 at 16:45
  • $\mathbb{Q}\setminus{0}$ should work, but the question is whether you have to provide an explicit isomorphism or not. Holo's example is a nice one if so. (There are some very cute explicit order-isomorphisms between $\mathbb{Q}$ and $\mathbb{Q}$ minus one point, but they take more thought) – Steven Stadnicki Sep 27 '18 at 16:59
  • @StevenStadnicki If I remember correctly it was cantor that proved that countable ordered dense chain is order isomorphic to $\Bbb Q$, but I remember the proof use inductive construction, can you give the explicit order isomorphic? – ℋolo Sep 27 '18 at 17:17
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    @holo the cleanest version I can think of of a mapping from $\mathbb{Q}\setminus{0}$ to $\mathbb{Q}$ is to choose a pair of monotonic sequences $r_n, s_n$ of rationals converging to e.g. $\sqrt{2}$ from below and above; map $(-\infty, -1)$ to $(-\infty, r_0)$ and $(s_0, \infty)$ to $(1, \infty)$, then map the intervals $[-1, -1/2), [-1/2, -1/4), \ldots$ to $[r_0, r_1), [r_1, r_2), \ldots$ and likewise map the intervals $(2^{-(n+1)}, 2^{-n}]$ to $(s_{n+1}, s_n]$. – Steven Stadnicki Sep 27 '18 at 18:00
  • @StevenStadnicki oh right! Thanks – ℋolo Sep 27 '18 at 18:33
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    For the second question, see https://mathoverflow.net/questions/131933/is-it-possible-to-construct-an-infinite-subset-of-bbb-r-that-is-not-order-iso – Gerry Myerson Sep 28 '18 at 03:41
  • Thank you @GerryMyerson! I got it. – Akira Sep 28 '18 at 03:44
  • Good, Akira. Maybe you could summarize what you got, and post it as an answer here. – Gerry Myerson Sep 28 '18 at 03:46
  • Hi @StevenStadnicki! Regarding your mapping from $\mathbb{Q}\setminus{0}$ to $\mathbb{Q}$, I have some confusions, please elaborate more. 1. you said a pair of monotonic sequences $r_n, s_n$ of rationals converging to e.g. $\sqrt{2}$ from below and above. But from your description, it seems to me that $r_n, s_n$ each converges to $0$. […] – Akira Sep 28 '18 at 04:08
  • @StevenStadnicki 2. Did you mean that we have an order-isomorphism from each interval in $\mathbb{Q}\setminus{0}$ to each interval in $\mathbb{Q}$. Moreover, these intervals are disjoint, then we take union of them to form an order-isomorphism from .$\mathbb{Q}\setminus{0}$ to $\mathbb{Q}$. […] – Akira Sep 28 '18 at 04:08
  • @StevenStadnicki 3. I'm still unable to understand why the union of such intervals generated by $r_n,s_n$ equals to $\Bbb Q$. I think $0$ still does not belong to that union. – Akira Sep 28 '18 at 04:15
  • @StevenStadnicki 4. I'm unable to get how you defined order isomorphism for each interval (if you intended to do do). – Akira Sep 28 '18 at 04:22
  • @Akira The order isomorphism between intervals is just a linear mapping, nearly trivial - mapping [0,1) to [1,5), for instance, is just the mapping $x\mapsto4x+1$. As for the intervals, note that if we start with e.g. $r_0=1$, then $0$ is in the very first interval; we map $(-\infty, -1)$ to $(-\infty, 1)$ and it's certainly the case that $0\in (-\infty, 1)$. – Steven Stadnicki Sep 28 '18 at 15:11
  • The core idea is that the set of intervals $(-\infty, -1), [-1, -1/2), \ldots, (1/4, 1/2], (1/2, 1], (1, \infty)$ cover $\mathbb{Q}-0$, and likewise the intervals $(-\infty, r_0), [r_0, r_1), \ldots, (s_1, s_0], (s_0, \infty)$ cover $\mathbb{Q}$; so we build a piecewise mapping by sending each interval in the first set to the corresponding interval in the second set, preserving the order of intervals. – Steven Stadnicki Sep 28 '18 at 15:16
  • Hi @StevenStadnicki! I have tweaked your method a little bit. Please check if I understand your previous comments correctly! […] – Akira Sep 28 '18 at 15:40
  • @StevenStadnicki I will chose a pair of monotonic sequences of rational numbers $a_n,b_n$ such that $a_n$ is increasing, converges to $0$, and $a_1=-2$, and that $b_n$ is increasing, converges to $\sqrt 2$, and $b_1=-2$. Similarly, I chose $b_n,c_n$ such that $c_n$ is decreasing, converges to $0$, and $c_1=2$, and that $d_n$ is decreasing, converges to $\sqrt 2$, and $d_1=2$. Then we map $(-\infty, -2)$ to $(-\infty, -2)$, $[a_{n},a_{n+1})$ to $[b_{n}, b_{n+1})$, $(c_{n+1},c_n]$ to $(d_{n+1},d_n]$, and $(2,+\infty)$ to $(2,+\infty)$. – Akira Sep 28 '18 at 15:40
  • Hi @StevenStadnicki! I have a picture here to visualize my idea here – Akira Sep 28 '18 at 15:50
  • I have business today but in response to your request in a comment to my comment to an answer I will help you with this tonight (Ontario time) – DanielWainfleet Sep 28 '18 at 18:23
  • Thank you @DanielWainfleet! Best wish for your business! – Akira Sep 29 '18 at 00:12
  • Now that I'm back I see that the "P.S." section of the Answer by bof is about what I would have said for an order-isomorphism from $\Bbb Q$ \ ${0}$ tp $\Bbb Q$. – DanielWainfleet Sep 29 '18 at 02:20
  • Hi @DanielWainfleet! Could you please confirm if my understanding in this comment correct? – Akira Sep 29 '18 at 02:24
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    Yes. You got it. BTW the "back-and-forth" technique is not the only way to prove that theorem of Cantor. I When I had it as homework I found a different proof. I DK what method Cantor used. – DanielWainfleet Sep 29 '18 at 02:56
  • @DanielWainfleet Honestly, Cantor's "back-and-forth" method seems to be quite mechanical to me (Of course, it is since it defines an isomorphism explicitly). I'm curious about your proof too (and I hope that I'm able to learn something from it). Please elaborate ob it if you don't mind! – Akira Sep 29 '18 at 02:59
  • Not sure why you accepted the answer that you did, since it has the wrong guess for your second question. – Andrés E. Caicedo Sep 29 '18 at 15:06
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    Hi @AndrésE.Caicedo! I have changed the accepted answer to @bof's. Thank you so much! – Akira Sep 29 '18 at 15:08
  • Hi @DanielWainfleet Could you please elaborate on your proof which is different from Cantor's one? Please give me some hints on your approach so I can give it a shot! Many thanks for you! – Akira Oct 03 '18 at 05:29

3 Answers3

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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.

bof
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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 $lub .$

  • As an amusing digression, I wonder whether anyone who takes the time to read all this notices that at no point do I say that $m\ne n \implies a_m\ne a_n.$ – DanielWainfleet Oct 03 '18 at 20:18
  • Preliminary result (I) can also be obtained, perhaps more simply, using the Axiom of Choice (AC), but I chose to derive it, and all the rest, without AC. – DanielWainfleet Oct 03 '18 at 20:38
  • Thank you @Daniel! I will read your proof as carefully as possible. I will soon inform you about my result. – Akira Oct 03 '18 at 23:55
  • It also implies that if $U$ is a non-empty open subset of $\Bbb R$ and if $A$ is a countable dense subset of $U$ then $A$ is order-isomorphic and reverse-order-isomorphic to $\Bbb Q$.... ("Reverse-order" means that $x<y\implies f(x)>f(y)$),..... which has further implications. But the back & forth proof is, I think, a lot simpler, and can also be phrased so as to avoid AC. – DanielWainfleet Oct 04 '18 at 03:02
  • Hi @DanielWainfleet! After three days of working hard, I have presented a detailed proof here. Please have a check on it! Many thanks for your dedicated help! Honestly, I only look at your lemmas and don't dare to look at the proofs, so that I can try to give them a shot by myself. – Akira Oct 07 '18 at 04:05
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  1. $(-1,1)\cap \Bbb Q$.
  2. I doubt there is any.
Akira
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    For 2., see the first A in the link in the comment ( to the Q) by Gerry Myerson for an example of an uncountable subset of $\Bbb R$ that satisfies 2.... For 1., we have a theorem of Cantor: For $i\in {0,1}$ let $A_i $ be a countably infinite set and let $<_i$ be a linear order on $A_i$ with no end-points, and order-dense (i.e. if $x<_iy$ there exists $z$ with $x<_iz<_iy)$... Then $(A_0,<_0)$ is order-isomorphic to $(A_1,<_1).$.. We may take $A_0=\Bbb Q$ and $A_1=\Bbb Q$ \ ${0}$ with the usual order on each of them – DanielWainfleet Sep 28 '18 at 07:19
  • Or we may take $A_0=\Bbb Q$ and $A_1=(-1,1)\cap \Bbb Q,$ as in your A. – DanielWainfleet Sep 28 '18 at 07:23
  • Could you please give me an explicit order isomorphism between $\Bbb Q$ and $\Bbb Q \setminus {0}$ without relying on that theorem? – Akira Sep 28 '18 at 07:56
  • In @StevenStadnicki's comment: the cleanest version I can think of of a mapping from $\mathbb{Q}\setminus{0}$ to $\mathbb{Q}$ is to choose a pair of monotonic sequences $r_n, s_n$ of rationals converging to e.g. $\sqrt{2}$ from below and above; map $(-\infty, -1)$ to $(-\infty, r_0)$ and $(s_0, \infty)$ to $(1, \infty)$, then map the intervals $[-1, -1/2), [-1/2, -1/4), \ldots$ to $[r_0, r_1), [r_1, r_2), \ldots$ and likewise map the intervals $(2^{-(n+1)}, 2^{-n}]$ to $(s_{n+1}, s_n]$. But I'm unable to understand this example. – Akira Sep 28 '18 at 07:59