Let $(X,⪯$) be totally ordered set, prove that if every non empty countable subset of $X$ is well ordered then X is well ordered.
It does seem obvious that any subset should have a minimum but I am not sure of how to prove it. Any help would be appreciated, and sorry I put up the same problem earlier but forgot to include the 'countable' subset.
Best Answer
If $X$ were not well-ordered, you'd have a nonempty $Y\subseteq X$ with no smallest element. As $Y$ is nonempty, pick some $y_0\in Y$. As $y_0$ isn't minimal in $Y$, pick some $y_1\prec y_0$ in $Y$. As $y_1$ isn't minimal in $Y$, pick some $y_2\prec y_1$ in $Y$. Continue in this fashion, obtaining an infinite sequence $y_0\succ y_1\succ\dots\succ y_n\succ y_{n+1}\succ\dots$. The set $\{y_n:n\in\mathbb N\}$ is a countable subset of $Y$ with no smallest element.