Order preserving bijection from a countable subset of R to a subset of N Let A be a countable subset of $\Bbb R$ which is well-ordered with respect
to the usual ordering on $\Bbb R$ (where ‘well-ordered’ means that every
nonempty subset has a minimum element in it). Then A has an order
preserving bijection with a subset of $\Bbb N$.
It is given that the above statement is FALSE.
But I think it is true. The following is my reasoning.
If A is empty then the statement is true. Because I can give a order preserving bijection from empty set to empty set.
Let A be a non empty set. Since A is well ordered I can arrange the elements of A as $ \{x_1, x_2,x_3,....,x_n\} $ in such a way that $x_1<x_2<....<x_n.$ . So I can give a bijective function $f$ from {1,2,3,...n} to $\{x_1,x_2,...,x_n\}$ defined by $f(i)=x_i$ which is order preserving.
Similarly I can do it for infinite set too.
Somehow I am not satisfied with my reasoning for the empty set. I want to know
whether the given statement is true or false. If it is false why? 
 A: Try $$A=\{-\tfrac1n\mid n\in\Bbb N\}\cup \{42\}$$
which is an inifnite well-ordered set with a largest element - a property $\Bbb N$ does not have.
Or even try
$$A=\{k-\tfrac1n\mid n,k\in\Bbb N\}$$
(and there are still weirder countable well-ordered subsets of $\Bbb R$)
A: The statement is False. To prove that, we just need to provide an example of well-ordered subset of R that doesn't have order-preserving bijection with a subset of N.
Any set A of this type $((-1/(n)^k)UL)$
where $n$ and $k$
belong to $N$
and $L$ is any proper subset of $(0UN)$ except empty set.
This is generalization of the answer above. Awesome example which I didn't see clearly before & I thought was (-n) (because I got a test tomorrow and I'm maxed out).
Also any set A of type $0U1/(n)^k$ where $n$ and $k$ belong to $N$.
Both of the sets have an infimum which makes them well-ordered and they are countable subsets of R. But also, they've a supremum, a property which N does not have.
Why this works is because in order-preserving bijection
$x1<x2$  implies
$f(x1)<f(x2)$
And so if a certain countably infinite set has a supremum and has an order-preserving bijection with $N$, then it would imply $N$ also has a supremum, which is a contradiction.
