# Is there any well-ordered uncountable set of real numbers under the original ordering?

I know that the usual ordering of $\mathbb R$ is not a well-ordering but is there an uncountable $S\subset \mathbb R$ such that S is well-ordered by $<_\mathbb R$?

Intuitively I'd say there is no such set but intuitively I'd also say there is no well-ordered uncountable set at all, which is obviously wrong. I still struggle to grasp the idea of an uncountable, well-ordered set.

• FYI, a frequently asked question on the Indiana University (Bloomington) Ph.D. real variables qualifying exams back in the late 1960s to the mid 1970s was to prove that any well-ordered subset of the reals is countable. (Sometimes the question was whether an uncountable well-ordered subset of the reals exists, and you were to prove your answer.) Apr 22, 2016 at 17:44
• @Matt Dyer: akkarin is asking whether there exists such a set on the real line, not whether there exists such a set somewhere. Apr 22, 2016 at 17:45
• Hint: You can get a pairwise disjoint collection of open intervals by picking one open interval lying between each of the points and the next point of the well ordered set. Apr 22, 2016 at 17:51
• @Matt Dyer: True, but the ordering you get may not have the numbers being in the same order as they appear on the number line, which is what akkarin wants. Apr 22, 2016 at 17:52
• Why the vote to close? It's possible this is a duplicate (I vaguely remember the same question being asked earlier), but it's certainly not off-topic. Apr 22, 2016 at 18:52

There can't be. If $S\subseteq \Bbb R$ is well-ordered by the usual ordering, for every element $s_{\alpha}\in S$ that has an immediate successor $s_{\alpha+1}\in S$ (every element of $S$ except the greatest element if there is one), the set of rationals $Q_{\alpha}$ between the element and its successor is nonempty: $(s_{\alpha}, s_{\alpha+1}) \cap \Bbb Q \ne \emptyset$, and the $Q_{\alpha}$ are disjoint. If $S$ were uncountable, then $\bigcup_{\alpha < length(S)} Q_{\alpha}$ would also be uncountable — impossible, as it's a subset of $\Bbb Q$.