order isomorphism between two intervals, where both have minimum and maximum elements 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 !
 A: 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.
A: 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.
