2
$\begingroup$

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.

$\endgroup$
11
  • 3
    $\begingroup$ 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.) $\endgroup$
    – Dave L. Renfro
    Apr 22, 2016 at 17:44
  • 2
    $\begingroup$ @Matt Dyer: akkarin is asking whether there exists such a set on the real line, not whether there exists such a set somewhere. $\endgroup$
    – Dave L. Renfro
    Apr 22, 2016 at 17:45
  • 2
    $\begingroup$ 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. $\endgroup$
    – Dave L. Renfro
    Apr 22, 2016 at 17:51
  • 1
    $\begingroup$ @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. $\endgroup$
    – Dave L. Renfro
    Apr 22, 2016 at 17:52
  • 2
    $\begingroup$ 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. $\endgroup$
    – Noah Schweber
    Apr 22, 2016 at 18:52

1 Answer 1

Reset to default
9
$\begingroup$

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 $\Bbb Q$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .