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?
Counterexamples to the continuum hypothesis
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2@Seyhmus: Sure: $\omega_1$, the set of all countable ordinals. – 2012-11-18
1 Answers
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:
- A problem with Cantor's continuum hypothesis
- Implications of continuum hypothesis and consistency of ZFC
- Relationship between Continuum Hypothesis and Special Aleph Hypothesis under ZF
- Can forcing push the continuum above a weakly inacessible cardinal?
- Can the cardinality of continuum exceed all aleph numbers in ZF?
- Cardinality of sets of subsets of $\mathbb{N}$
- bound on the cardinality of the continuum? I hope not
- is there a cardinality between the rational and the irrationals?
- Can the real numbers be forced to have arbitrary cardinality?
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0@AsafKaragila If the continuum hypothesis is false, are there fields of arbitrary cardinality between Q and R? – 2017-08-04