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(ZFC) Let X ⊆ R is a well ordered form the standart order in R. Prove that X is either finite or countable.

My attempt: Since X is well ordered then if L is a chain where every countable subset of X is well-ordered, L is well ordered as well. In other words: then L contains a finite descending sequence a0 > a1 > a2 > . . . with a minimal element an, hence X is finite;

If I manage to find a function which maps every element of my set into a unique ordinal I think that this is satisfying. Still I know that well-ordered set is order-isomorphic to a unique ordinal. However, I do not how to make use of it.

I hope that someone will help me. Thanks!

Tiki
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1 Answers1

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I might be missing something, but it seems to me that since the well-ordering comes from $\mathbb{R}$, we can use the countability of the rationals:

For each $a_i$ (with $i \in I$, your indexing set), we can assign a rational $q_i \in (a_i, a_{i+1} )$. This $q_i$ is only mapped to by $a_i$, so we have an injection, $I \to \mathbb{Q}$. Thus, $|I| \le | \mathbb{Q} |$, so it is either countable or finite.

Daniel P.
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  • I do not think that you have missed anything but I have a question. Do I have to justify my choice for Q being the index set using the axiom of Choice? – Tiki Jan 12 '21 at 19:41