This question already has an answer here:

I have recently watched a video by "Undefined Behavior", explaining countable and uncountable infinities, and showing why uncountable infinity is larger than countable infinity. He then stated that a question had been asked if there is some infinity that's in-between the two (larger than countable, but smaller than uncountable), and that it has been proven that this question in unsolvable. However, he did not mention any name of the theorem proving this, and also did not show any proof. What is the proof that this problem is unsolvable?


marked as duplicate by Andrés E. Caicedo, Chappers, hardmath, user223391, max_zorn Jul 11 '18 at 2:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ In short, by our definition of sets, it is impossible to construct a set that has the cardinality strictly between $\mathbb N$ and $\mathbb R$. $\endgroup$ – JungHwan Min Jul 10 '18 at 22:13
  • 1
    $\begingroup$ Your question is not precisely stated. The cardinal number of every countable infinity is $\aleph_0$. The next large cardinal is $\aleph_1$ and it is uncountable. There is no cardinal between them. What is unprovable in standard $ZFC$ is whether the cardinal corresponding to the reals is $\aleph_1$. If the cardinal corresponding to the reals is larger than $\aleph_1$, there is an infinity between the naturals and the reals. All the cardinals greater than $\aleph_0$ are uncountable. $\endgroup$ – Ross Millikan Jul 10 '18 at 22:23
  • 2
    $\begingroup$ @JungHwanMin, I think it is misleading to say "by our definition of sets, it is impossible...", since that is exactly about the continuum hypothesis. One may argue that the independence of CH is an indication that we cannot "construct" (?) any such set, but in my amateur perception of the state of things it is maybe not so clear that that's the meaning. $\endgroup$ – paul garrett Jul 10 '18 at 22:28
  • 1
    $\begingroup$ @JungHwanMin It is impossible (using ZFC) to construct such a set and prove its cardinality is strictly between $\aleph_0$ and $\mathfrak c$. There could be sets we can construct but not compare their cardinalities to $\aleph_0$ or $\mathfrak c$. $\endgroup$ – Robert Israel Jul 10 '18 at 23:04
  • $\begingroup$ @RobertIsrael Indeed, I should have made that more clear. $\endgroup$ – JungHwan Min Jul 10 '18 at 23:06

That person was talking about the continuum hyphothesis. It was proved (by Kurt Gödel and Paul Cohen) that, assuming that set theory is consistent, neither the continuum hyphothesis nor its negation can be proved from the standard set theory axioms (the Zermelo-Fraenkel axioms).


Not the answer you're looking for? Browse other questions tagged or ask your own question.