Well ordered sets

Question

$(X,\le)$ is totally ordered. How do you prove that if every non empty countable subset of $X$ is well ordered then $(X,\le)$ is well ordered?

Answer 1

Now you have the correct conditions, the proof goes like this:

Let $Y\subseteq X$ be non-empty. If it is not well ordered then it is uncountable. Let $y_1\in Y$, we know $y_1$ is not the minimal element of $Y$, hence by totality there is $y_2\neq y_1\in Y$ such that $y_2\leq y_1$.

Repeat this process inductively and you have an infinite chain of distinct elements

$$y_1\geq y_2\geq y_3\geq\ldots$$ This is a countable set and so it is well ordered, however it is clear that there is no smallest element. A contradiction.

Hence $Y$ is well-ordered.

Answer 2

The below answers an older revision of this question; the current revision (4) is answered here.

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This can't be done, for the following reason:

Exercise: Every finite totally ordered set is well-ordered.

Any finite subset of a totally ordered set is also totally ordered, hence well-ordered.

Therefore, the truth of your conjecture would imply that every totally ordered is well-ordered. But consider $\Bbb Q$.