12
$\begingroup$

My understanding of logic is really basic, and I ask this question out of curiosity.

Is there an explicit example of a statement whose proof uses the continuum hypothesis and is unprovable in $ZFC + \lnot CH$.

ADDED: Preferably I'm looking for an example outside set theory.

$\endgroup$
  • 12
    $\begingroup$ It might not be what you expect but : CH. $\endgroup$ – Pece Jan 23 '14 at 9:18
  • $\begingroup$ @Pece: Technically, axioms aren't provable. $\endgroup$ – Lucian Dec 5 '15 at 5:24
  • 2
    $\begingroup$ @Lucian: Technically, all axioms are provable, and their proof is a sequence of length $1$. $\endgroup$ – user21820 May 14 '16 at 7:30
  • $\begingroup$ @Lucian: Very technically, $T \vdash φ$ for every axiom $φ$ of $T$. $\endgroup$ – user21820 May 14 '16 at 7:31
16
$\begingroup$

One not-quite-trivial example is

$2^{\aleph_0} < 2^{\aleph_1}$.

This clearly follows from $\mathsf{CH}$ (by Cantor's Theorem), however both $2^{\aleph_0} = 2^{\aleph_1}$ and $2^{\aleph_0} < 2^{\aleph_1}$ are consistent with $\neg \mathsf{CH}$, and so the statement above cannot be proven from $\mathsf{ZFC}+\neg \mathsf{CH}$.


For an example outside of set theory proper, consider the following:

The smallest non-Lebesgue-measurable subset of $\mathbb{R}$ has cardinality $\aleph_1$.

Again, this follows from $\mathsf{CH}$ (since all countable subsets of $\mathbb{R}$ are Lebesgue measurable, but there are non-Lebesgue-measurable sets). However again both the above statement and its negation are consistent with $\neg \mathsf{CH}$


Now to hide virtually all set theory from the statement:

A topological space is called hereditarily separable if every subspace of is separable (so separable metric spaces are hereditarily separable). A regular (T$_3$) hereditarily separable but non-Lindelöf space is called an S-space.

So consider the following statement:

There is an S-space

  • With $\mathsf{CH}$ one may construct $S$-spaces. (The Kunen line is one such example).
  • In 1978 Szentmiklóssy Z. showed that $\mathsf{MA} + \neg \mathsf{CH}$ is consistent with the existence of S-spaces. (Starting in a model of $\mathsf{CH}$, show there is an S-space that cannot be destroyed by ccc forcing, and then force $\mathsf{MA}+\neg\mathsf{CH}$ in the usual manner by iterating all ccc posets of size $< \aleph_2$.)
  • In 1981 S. Todorcevic showed that $\mathsf{PFA}$ implies that S-spaces do not exist. (For those claiming that this result has large cardinal power, the proof may be modified so as to avoid the large cardinals.)
$\endgroup$
  • $\begingroup$ Excellent example. Does the more general statement $\kappa < \nu \rightarrow 2^\kappa < 2^\nu$ follow from $\mathrm{GCH}$? Its certainly not provable in ZFC. $\endgroup$ – goblin Jan 23 '14 at 9:29
  • $\begingroup$ @user18921: yes, since assuming $\mathsf{GCH}$ we know that $2^\kappa = \kappa^+$, and the cardinal successor function is strictly increasing. $\endgroup$ – user642796 Jan 23 '14 at 9:32
  • 1
    $\begingroup$ Thank you; I did not know that. (Although now that you mention it, its utterly obvious from the definitions!!!) $\endgroup$ – goblin Jan 23 '14 at 9:33
  • 2
    $\begingroup$ @user18921: The less obvious part is that the reverse implication doesn't hold (i.e. $\kappa<\lambda\rightarrow 2^\kappa<2^\lambda$ doesn't imply $\sf GCH$). We can, however, easily violate $\sf GCH$ on every regular cardinal, but still keeping the continuum function injective everywhere. $\endgroup$ – Asaf Karagila Jan 23 '14 at 10:44
  • $\begingroup$ @AsafKaragila, thanks. $\endgroup$ – goblin Jan 23 '14 at 10:49
15
$\begingroup$

The following statement is equivalent to $\mathsf{CH}$:

There exists a function $\chi: \mathbb R \rightarrow \mathbb N$ such that for all $x,y,z,t\in\mathbb R$, if $\chi(x) = \chi(y) = \chi(z) = \chi(t)$ and $x+y=z+t$, then at least two of these numbers are equal.

This result is due to Erdős. Since it is equivalent to $\mathsf{CH}$, it is not provable in $\mathsf{ZFC+\neg CH}$ assuming it is a consistent theory!

$\endgroup$
  • 1
    $\begingroup$ +1 I think among all answers this is the one best fitting "outside set theory" (it doesn't mention anything about cardinality explcitly) and still looks awfully elementary. $\endgroup$ – Hagen von Eitzen Jan 23 '14 at 10:47
  • $\begingroup$ This is a lovely result (new to me -- not that that means much!) Do you have a pointer to a proof, please? $\endgroup$ – Peter Smith Jan 23 '14 at 12:20
  • 1
    $\begingroup$ Answering my own query, this looks what we need! arxiv.org/pdf/1201.1207v1.pdf $\endgroup$ – Peter Smith Jan 23 '14 at 13:55
6
$\begingroup$

There exists an outer automorphism of the Calkin algebra of a separable [infinite-dimensional] Hilbert space.

Assuming $\sf CH$ we can prove that there exists an outer automorphism of the Calkin algebra [of a separable Hilbert space]. When $\sf CH$ fails it is consistent that all automorphisms are inner. (The first result is due to Christopher-Weaver, the second due to Farah.)

An overview of both results can be found in Farah's paper,

Farah, Ilijas "All automorphisms of the Calkin algebra are inner". Ann. of Math. (2) 173 (2011), no. 2, 619–661. arXiv version.


The global dimension of $\prod_{i=1}^\infty\Bbb C$ is $2$.

This is in fact equivalent to the continuum hypothesis. See more in this MathOverflow thread.


The Kaplansky conjecture [for Banach algebras] fails. There exists a compact Hausdorff space $X$ such that there is a discontinuous homomorphism from $C(X)$ into another Banach algebra.

It was shown that the conjecture fails when one assumes the continuum hypothesis; but it is consistent that it holds, and the continuum hypothesis fails. (See Wikipedia for a bit more, and references.)

$\endgroup$
  • $\begingroup$ It's worth noting that Farah's result is using OCA, which follows from PFA, which is another nice example of what the OP was looking for (think of it in terms of the Baire category theorem if you insist on having examples outside of set theory). $\endgroup$ – Haim Jan 23 '14 at 11:00
3
$\begingroup$

The statement $"\mathfrak{x}_0= \mathfrak{x}_1 "$ for many pairs of cardinal invariants of the continuum $\mathfrak{x}_0,\mathfrak{x}_1$ such that $"\mathfrak{x}_0=\mathfrak{x}_1"$ is not provable in $ZFC$ (for example, $\mathfrak a$ and $\mathfrak d$).

$\endgroup$
  • $\begingroup$ Yeah, I was going to post this (specifically for $\frak b=d$), but then I saw that the OP asked for non-set theoretical applications of $\sf CH$. $\endgroup$ – Asaf Karagila Jan 23 '14 at 10:40
  • $\begingroup$ Oh, c'mon! This is general topology. :) $\endgroup$ – Haim Jan 23 '14 at 10:42
  • $\begingroup$ More like set theoretical topology... :-) $\endgroup$ – Asaf Karagila Jan 23 '14 at 10:44
3
$\begingroup$

There exists a subset $S$ of plane $\mathbb{R}^2$, such that intersection of any horizontal line and $S$ is countable, and intersection of any vertical line with $S$ has countable complement.

It's quite easy to show that if such $S$ exists, then $\aleph_1 = \mathfrak{c}$, so it's not provable in ZFC + $\lnot$ CH. It's a bit harder to show that if CH holds, then such $S$ exists.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy