order isomorphism between two intervals, where both have minimum and maximum elements

Question

Note to viewers of this question: I made two attempts to construct an isomorphism between [0,1) and a larger interval. My original posting had some typos that made the two answers I received hard to follow. So, I took another my typo-less definitions of the two alleged isomorphisms, and I realize that the answer I received from James is still valid: Neither of my alleged isomorphisms work. I will explain why I agree with the given answer in my comment.

--- original question ----

I am looking at a problem which asks if there is an order isomorphism between the two sets A and D (both subsets of the rationals, Q), as defined below:

A = { $1 - \frac{1}{n+1}$ }

D = { $1 - \frac{1}{n+1}$ } $\cup$ { $2 - \frac{1}{n+1}$ }

Note: $n \epsilon \Bbb N$

I answered yes because I did not realize that A does not include 1, whereas the set D does. When n=0, then the following expression in the definition of D:

$2 - \frac{1}{n+1}$

becomes

$2 - \frac{1}{0+1}$

which is 1. So 1 is part of set D. This means A lacks a maximum element while D has one. So, no order isomorphism here.

Then I got to thinking (and that's when trouble usually starts for me ;^): what if we defined A and D as follows:

A = { $1 - \frac{1}{n+1}$ } $ \cup$ {1}

D = { $1 - \frac{1}{n+1}$ } $\cup$ { $2 - \frac{1}{n+1}$ } $ \cup $ {2}

And then we defined a bijection (at least I think it would be a bijection) 'g' as follows:

g(n) =

$1 - \frac{1}{\frac{n}{2} +1}$ -- if n is even and $n \gt 0$

$1 - \frac{1}{\frac{n-1}{2} +1}$ -- if n is odd

$n \epsilon \Bbb N$

With the revised A and D both now have maximum elements. And I believe my bijection from A to D will hit all points on D and is 1-1. But I am hoping wiser minds than mine might weigh in. Thanks in advance !

Answer 1

Independent of you fixing the old $A$ and $D$, your new $A$ and $D$ are not order-isomorphic:

$A$ is order-isomorphic to $\omega+1$ (i.e. you have something that looks like the natural numbers, with something stuck on top) where as $D$ is order isomorphic to $\omega + \omega + 1$ i.e. you have two copies of the natural numbers, one bigger than the other, and an element on top.

You can even see more. As both $A$ and $D$ have a least element, $0$, any order isomorphism between $A$ and $D$ sends $0$ to $0$. Then, inductively, everything of the form $1 - \frac{1}{n}$ in $A$ has to be sent to $1 - \frac{1}{n}$ in $D$ (these are both the least elements of what is left). Then after this, all you have left in $A$ is the top element, however you have a lot more things left in $D$, hence, an isomorphism can't exist (you can however embed $A$ as an initial segment of $D$ by mapping $1$ to $1$).

The $g$ you have written is not $1-1$ as it sends $2$ and $3$ to the same thing (supposing you mean what I think you mean). It's possible you were trying to send "even" $1-\frac{1}{2n}\mapsto 1 - \frac{1}{n}$ and "odd" $1-\frac{1}{2n + 1} \mapsto 2 - \frac{1}{n}$ but this isn't order preserving.

Answer 2

As it stands, $g$ is not defined, because you haven't specified what $x$ and $n$ have to do with each other.

Neither $A$ nor $D$ have a maximum element--$1$ is less than infinitely-many elements of $D$--and adding one in will not make them order isomorphic. The kicker, here, is that $1$ is a limit element of $D$, while $A$ has no such element--all of the elements of $A$ are either the initial element or an immediate successor.