Prove: Let $\displaystyle X \subset \bold{N} $. Then $\displaystyle X $ is at most countable. Let $\displaystyle A_{0} := X $, and let $\displaystyle a_{0} := \min(A_{0}) $. Then $\displaystyle A_{n++} := A_{n} - \{a_{n} \} $ and $\displaystyle a_{n++} := \min(A_{n++}) $. So basically after removing the first smallest element, $\displaystyle a_0 $, then $\displaystyle a_1 $ is the second smallest element, etc.. Is this on the right track? Could I maybe let $\displaystyle a_{0} := \max(A_0) $, and $\displaystyle A_{n++} := A_{n}- \{a_{n}\} $ and $\displaystyle a_{n++} := \max(A_{n++}) $ (e.g. take away the maximum elements)?