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?
Best Answer
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 $\Bbb R$)