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I remember at a certain point early in my mathematical studies learning that the Axiom of Choice is equivalent to the following statement on Cartesian products:

If $\{ X_i \}_{i \in I}$ is any collection of nonempty sets indexed by an index set $I$, then $\prod_{i \in I} X_i$ is nonempty.

To me, this settled the question of whether to use Axiom of Choice in practical contexts (although it's still interesting to consider systems of math where it doesn't hold, and the interdependence of various other theorems/results/lemmas/axioms on $AC$).

My first question is:

Question 1--Is there any similarly fundamental lemma or theorem which depends on the continuum hypothesis or its negation? That is, are there any basic facts in set theory, topology, measure theory, etc. which are (a) "self-evident" and (b) equivalent to $CH$ or $\lnot CH$?

I would also be interested in hearing if such a statement existed for $GCH$ or its negation $\lnot GCH$, although to me $GCH$ seems "less likely" to be true than $CH$ just because it makes a much broader statement over the class of all cardinals, whereas $CH$ is a relatively narrow statement about the relationship of two cardinals $2^{\aleph_0}$ and $\aleph_1$.

Currently, the two "simplest" results (that I know of) in this vein that would directly depend on $CH$ or $\lnot CH$ are:

  1. Wetzel's problem

  2. Whether or not $\Bbb{R}^\omega$ is normal in the box topology

But neither of these seem intuitively true or false, much less so essential that we had better accept them one way or another if we want to get any serious math done in the related field.

I'm aware that attempts have been made to resolve $CH$ one way or another (e.g. Freiling's axiom of symmetry) that are basically trying to reduce $CH$ to such an obviously true/false statement of general set theory/topology/measure theory. So I have a follow-up:

Question 2--What seem to be the obstacles to finding such a resolution of $CH$ or $\lnot CH$? That is, why is it so difficult to make concrete and testable statements (i.e. not trivial things like "There exists an element of $2^{2^{\aleph_0}}$ which is neither countable nor of size $\mathfrak{c}$") dependent on $CH$'s truth or falsity? And, should this difficulty be taken as evidence one way or the other for $CH$? Slash, is it actually considered evidence one way or the other for $CH$?

For instance: every Borel set is either of size $\aleph_0$ (if countable) or of size $2^{\aleph_0}$ (if uncountable). Is our difficulty in constructing a set of intermediate cardinality (as opposed to the ease with which we can construct a non-measurable set) evidence that no such intermediate-cardinality set exists?

I'll also mention that I take the "Platonic view" of $CH$. That is, I believe that despite the existence of models of set theory where either $CH$ or $\lnot CH$ holds, the statement

"If $S = 2^\Bbb{N}$ is the set of all subsets of $\Bbb{N}$, then for $A \subset S$ any subset of $S$, either $A$ is countable, or there exists a 1-1 correspondence between $A$ and $S$"

has a canonical and demonstrable true/false answer.

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  • $\begingroup$ For 1), I'd say no; for 2), I'd say see Solutions to the Continuum Hypothesis. The two questions are similar enough that I've voted to close. $\endgroup$ – Matt F. Aug 7 at 14:19
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    $\begingroup$ Matt, I actually pored over a number of related questions before asking this to make sure my questions weren’t answered there, including the one you suggested. If you’re not sure how my questions are distinct, I’m happy to edit for clarification. $\endgroup$ – Rivers McForge Aug 7 at 15:43
  • $\begingroup$ amazon.com/Hypothese-Du-Continu-W-Sierpinski/dp/B000JUOD6S) $\endgroup$ – bof Aug 8 at 1:48
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    $\begingroup$ It's not at all evident to me that $\prod_{i\in I} X_i$ is non-empty because there is no basic process by which to define an element when $I$ is uncountable (unless you assume that $I$ is well-ordered which assumes AC in the first place). Countable Choice and even Dependent Choice, however, are totally obvious. Some evidence for the truth of CH may be that it is implied by AD+DC. CH is also equivalent to the statement that every free ultrafilter on $\omega$ induces the same ordering on $\omega^\omega$. This seems unlikely, but is not a problem under AD because free ultrafilters do not exist. $\endgroup$ – D.S. Lipham Aug 8 at 17:35
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    $\begingroup$ @RiversMcForge That element of the product is a choice function. $\endgroup$ – D.S. Lipham Aug 8 at 19:07
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Concerning your first question, there is a simple, if not "self-evident", order-theoretic statement equivalent to $CH$ admitting a generalization equivalent to $GCH$:

  • If $L$ is a linear ordering of size $2^{\omega}$, then $L$ embeds every cardinal less than $2^{\omega}$ or $L^*$ ($L$ reversed) embeds every cardinal less than $2^{\omega}$.

This statement can be read in general terms as follows: In order to arrange $2^{\omega}$ points in a line one cannot bypass a smaller cardinal (in the sense that it will appear directly or reversed).

The following generalization is equivalent to $GCH$:

  • For every cardinal $\lambda$, if $L$ is a linear ordering of size $\lambda$, then $L$ embeds every cardinal less than $\lambda$ or $L^*$ embeds every cardinal less than $\lambda$.

It says, in general terms, that for every cardinal $\lambda$, in order to arrange $\lambda$ points in a line one cannot bypass a smaller cardinal.

It is important to remark that the regularity thus stated is trivially true in the finite realm (for finite $\lambda$), contrary to $GCH$ (there are, in general, many numbers between $n$ and $2^n$). Therefore, it is, at least, a more uniform statement generalizing a fact of the finite realm for which, presumably, we have a more reliable "intuition". This feature is present in the usual axioms of set theory.

Addendum:

I have just recalled the paper below which seems to be relevant to your question. It connects $CH$ with a more or less concrete problem of machine learning:

  • Shai Ben-David, Pavel Hrubeš, Shay Moran, Amir Shpilka and Amir Yehudayoff, Learnability can be undecidable, Nat. Mach. Intell. 1 (2019) 44–48, doi:10.1038/s42256-018-0002-3.

($CH$ is equivalent to a version of learnability)

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    $\begingroup$ See also the freely-accessible paper: William Taylor, Learnability Can Be Independent of ZFC Axioms: Explanations and Implications arxiv.org/abs/1909.08410 $\endgroup$ – David Roberts Sep 4 at 6:11
  • $\begingroup$ Thanks for your help. $\endgroup$ – Rodrigo Freire Sep 4 at 11:47
  • $\begingroup$ @rodrigo In your answer, when you say $2^{\omega}$, you meant $2^{\aleph_0}$, right? Under ordinal exponentiation, $2^{\omega} = \sup_n 2^n = \omega$. $\endgroup$ – Rivers McForge Oct 30 at 21:24
  • $\begingroup$ Yes, I mean cardinal exponentiation, that is, $2^{\aleph_0}$. $\endgroup$ – Rodrigo Freire Oct 30 at 21:56
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As it is stated in the comments, one reference is Sierpinski's book, Hypothese Du Continu, though it is not in English.

Another reference is Propositions Equivalent to the Continuum Hypothesis.

See also Some propositions equivalent to the continuum hypothesis and The continuum hypothesis (CH) and its equivalent.

You may be also interested in Eliminating the Continuum Hypothesis.

Let me also state one equivalent of CH. I have taken it from Interactions between (set theory, model theory) and (algebraic geometry, algebraic number theory ,...):

Let $R$ be a ring and $D(R)$ its unbounded derived category. Let $D^c(R)$ be the full subcategory of compact objects (in the explicit example below it is spanned by bounded complexes of f.g. projective modules). We say that $D(R)$ satisfies Adams representability if any cohomological functor $D^c(R)^{op}\rightarrow Ab$, i.e. additive and taking exact triangles to exact sequences, is isomorphic to the restriction of a representable functor in $D(R)$ (in particular it extends to the whole $D(R)$), and any natural transformation between restrictions of representable functors $D^c(R)^{op}\rightarrow Ab$ is induced by a morphism in $D(R)$ between the representatives.

Let $\mathbb C\langle x,y\rangle$ be the ring of noncommutative polinomials on two variables. The statement '$D(\mathbb C\langle x,y\rangle)$ satisfies Adams representability' is equivalent to the continuum hypothesis.

For another interesting equivalent of CH see: Reductions between certain incidence problems and the continuum hypothesis.

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