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Assume the continuum hypothesis is false, and add that as an axiom to ZF set theory. How many cardinalities are between the rationals and the reals in this case? Only one? Infinitely many? Countably many? Uncountably many? What are the possibilities to the quantity of counterexamples?

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I think the answers to this question answer most of this. –  Arthur Fischer Nov 18 '12 at 17:24
As many as you want; see the question to which Arthur linked. –  Brian M. Scott Nov 18 '12 at 17:25
The technical answer is "all hell breaks loose" ... –  Peter Smith Nov 18 '12 at 17:25
@BrianM.Scott can one give an example for such a set assuming that CH is false? –  Seyhmus Güngören Nov 18 '12 at 17:29
@Seyhmus: Sure: $\omega_1$, the set of all countable ordinals. –  Brian M. Scott Nov 18 '12 at 17:31

1 Answer 1

The continuum hypothesis merely says $2^{\aleph_0}=\aleph_1$. Its negation, if so, says $2^{\aleph_0}\neq\aleph_1$. However assuming the axiom of choice the continuum can be well-ordered and therefore this translates to $2^{\aleph_0}>\aleph_1$.

The negation of CH does not tell us the value of the continuum, rather it tells us what it is not. It is not $\aleph_1$. Much like saying that if $x\in\mathbb R$ and $x\neq 0$ does not provide us with any actual information, except that $x$ is not zero.

If ZF(C) is consistent then it is consistent that the continuum is almost any cardinal, with the exception of cardinals which can be expressed as a countable union of smaller cardinals.

It could be that $2^{\aleph_0}=\aleph_2$, or it could be that the continuum is actually much larger. It could be that there are exactly $2^{\aleph_0}$ cardinals between $\aleph_0$ and $2^{\aleph_0}$, or it could be that there are much less (clearly there cannot be more).

If one does not assume the axiom of choice then there are two non-equivalent ways to formulate the continuum hypothesis. You could say $2^{\aleph_0}=\aleph_1$, or you could say that every uncountable set of real numbers is equipotent with the real numbers. It is important to make this distinction because it exists. However the answer as to how many cardinals are between the rationals and the reals in those cases can be even harder to answer, because there can be so many pathologies, that one cannot even begin to fathom them.

For extra reading points:

  1. A problem with Cantor's continuum hypothesis
  2. Implications of continuum hypothesis and consistency of ZFC
  3. Relationship between Continuum Hypothesis and Special Aleph Hypothesis under ZF
  4. Can forcing push the continuum above a weakly inacessible cardinal?
  5. Can the cardinality of continuum exceed all aleph numbers in ZF?
  6. Cardinality of sets of subsets of $\mathbb{N}$
  7. bound on the cardinality of the continuum? I hope not
  8. is there a cardinality between the rational and the irrationals?
  9. Can the real numbers be forced to have arbitrary cardinality?
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Thanks Guys. I wonder why people ever thought that the continuum hypothesis was true. It is well known it's independent (of ZF). I mean, for finite sets, you have many numbers in between others, many cardinality classes of subsets of a set on n elements. Well n, actually if you exclude the empty set. Whose to say that a better continuum hypothesis would be that there are uncountably many cardinalities strictly between the naturals and the power set of the naturals? –  John Smith Nov 20 '12 at 13:04
@John: People think differently at different times. You could ask the same question about Aristotle's physics or how the scientists of the king decreed that the black death hit Europe because of some astrological thing... –  Asaf Karagila Nov 20 '12 at 13:11
@John: Furthermore, if you look at it... open sets of real numbers are empty or have size continuum. Closed sets are countable or have size continuum. This continues on. There seemed to be a good evidence that the continuum hypothesis is true. At least at first glance. –  Asaf Karagila Nov 20 '12 at 13:42

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