What is the most "concrete-feeling" equivalent formulation of the Continuum Hypothesis that you can think of?

Question

There are many equivalent formulations of the Continuum Hypothesis, but I think the most standard one is that

there is no infinite cardinality lying strictly between the cardinality of the natural numbers and the cardinality of the real numbers.

But, imagining trying to explain Continuum Hypothesis in, say, a popular-mathematics context, I can imagine people thinking, "Well, all this 'looking for different sizes of infinity in terms of one-to-one mappings' stuff sounds really cool, but it's ultimately just mathematicians engaging in abstract word-logic games." And, as I understand it, a significant question in the philosophy of mathematics is whether such people would actually be right! But nevertheless, we could still at least ask: what is the least 'abstract-word-games'-feeling, most concrete-feeling, equivalent statement to the Continuum Hypothesis?

Here's my own best attempt:

(I take $\mathbb{N}$ not to include $0$.)

Recall Zeno's paradox: the sequence $(a_n)_{n \in \mathbb{N}}$ defined recursively by $a_1=0$ and $a_{n+1}=\frac{1}{2}(a_n+1)$ is a strictly increasing infinite sequence, and yet still has a finite upper bound.

A Generalised Zeno Paradox is a partition of $[0,1)$ of the form $$ \Big\{ [ q_n , q_n+2^{-n} ) \ \Big| \ n \in \mathbb{N} \Big\} $$ for some sequence of rational numbers $(q_n)_{n \in \mathbb{N}}$.

We say that two Generalised Zeno Paradoxes $$ \Big\{ [ q_n , q_n+2^{-n} ) \ \Big| \ n \in \mathbb{N} \Big\} \quad \text{and} \quad \Big\{ [ \tilde{q}_n , \tilde{q}_n+2^{-n} ) \ \Big| \ n \in \mathbb{N} \Big\} $$ are qualitatively equivalent if there exists a strictly increasing bijection between the set $\{q_n | n \in \mathbb{N}\}$ and the set $\{\tilde{q}_n | n \in \mathbb{N}\}$.

A Qualitatively Generalised Zeno Paradox (QGZP) is a qualitative-equivalence class of Generalised Zeno Paradoxes.

An infinite-bit string is an $\mathbb{N}$-indexed family of $0$s and $1$s.

Continuum Hypothesis: There exists a set of pairings-of-a-QGZP-with-an-infinite-bit-string in which

• every possible infinite-bit string has at least one QGZP paired to it, and
• every QGZP is paired to at most one infinite-bit string.

Are there other equivalent formulations that feel at most as "abstract" as (or perhaps, even less "abstract" than) this one?

Answer 1

Since we can force $CH$ and $\neg CH$ between transitive models of ZFC, we know that $CH$ cannot be equivalent to a statement that is absolute between transitive models, including arithmetic statements which I think most people would consider the most concrete. This already implies that any answer to this question has to be somewhat abstract. For this reason, I don't think we can provide an answer that would convince average person that we are not dealing with abstract logic games.

That said, I think we can convince the average mathematician that $CH$ is relevant to their field by providing concrete not-seemingly-logic-related statements. Here is an example of algebraic flavor: The projective dimension of $\mathbb{R}(x,y,z)$ as a module over $\mathbb{R}[x,y,z]$ is equal to $2$ if $CH$ holds and is not equal to $2$ (indeed, is equal to $3$) if $\neg CH$ holds.

If you allow an average person to understand what it means to color each point in $\mathbb{R}^2$ using countably many colors, here is a concrete statement of combinatorial flavor equivalent to $CH$:

The plane $\mathbb{R}^2$ can be colored with countably many colors in such a way that no right triangle is monochromatic, i.e. its vertices are of the same color.

This is proven in

P. Erdős, P. Komjáth; Countable decompositions of R² and R³, Discrete Comput. Geom.5(1990), no.4, 325–331

with recent corrections in

B. Bursics, P. Komjáth; A Coloring of the Plane Without Monochromatic Right Triangles, Studia Sci. Math. Hungar.60 (2023), no.1, 91–95.

Answer 2

Here are a few of my favorite characterizations of the continuum hypothesis:

• Sierpiński (1951) proved that CH is equivalent to the assertion that there is a partition of the plane into two sets $\mathbb{R}^2=S_1\sqcup S_2$ such that every horizontal line meets $S_1$ only countably and every vertical line meets $S_2$ only countably. (The one direction is easy to see by taking $S_1$ as the graph of a well-order in order type $\omega_1$.)
• More impressively, and more concretely from the perspective of your question, CH is equivalent to the existence of a partition of space $\mathbb{R}^3=S_1\sqcup S_2\sqcup S_3$ such that every line parallel to axis $i$ meets $S_i$ only finitely. (See further generalizations in Erdős, Paul; Jackson, Steve; Mauldin, R. Daniel, On partitions of lines and space, Fundam. Math. 145, No. 2, 101-119 (1994). ZBL0809.04004.)
• CH is equivalent to the negation of the assertion that every function $x\mapsto A_x$ from reals to countable sets of reals must admit some pair of distinct reals $x,y$ with $x\notin A_y$ and $y\notin A_x$. This is the axiom of symmetry of Freiling. Freiling, Chris, Axioms of symmetry: Throwing darts at the real number line, J. Symb. Log. 51, 190-200 (1986). ZBL0619.03035.

And let me add another new favorite of mine:

• In current work of myself and Ben De Bondt, we have proved that CH is equivalent to the assertion that the size of the smallest dictionary $\Delta\subset\omega^\omega$ for which the codebreaker has no winning strategy to win infinite Wordle using dictionary $\Delta$ is the same as the size of the smallest dictionary $\Delta$ for which the absurdist has a winning strategy in infinite Absurdle.

Answer 3

Theorem (Erdős [1964]). The following assertion is equivalent to $\mathsf{CH}$: There exists an uncountable family $ℱ$ of entire analytic functions such that for each $z ∈ ℂ$, the set of values $\{ f(z) \mid f ∈ ℱ \}$ is countable.

Answer 4

A cloud around $x$ in $\Bbb R^2$ is a set $C$ such that for all lines $\ell$ through $x$, $C\cap\ell$ is finite. A cloud is a cloud around some point.

Theorem: (Komjáth, Schmerl) the continuum hypothesis holds if and only if three clouds cover the plane.

More generally Komjáth proved that for every finite $n$, $\mathfrak c\leq\aleph_n$ implies that $\Bbb R^2$ is covered by $n+2$ clouds, while Schmerl showed the converse in this paper.

Answer 5

Morayne showed the following in this paper:

Theorem. The continuum hypothesis is equivalent to the existence of a surjection $F : \mathbb{R} \to \mathbb{R}^2$ such that for every $x \in \mathbb{R}$, at least one of the coordinate functions of $F$ is differentiable at $x$.

One thing to note is that $F$ is of course badly discontinuous. Furthermore, Morayne showed that such a function cannot have either of its coordinate functions be measurable.

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In some sense I don't actually find this characterization to be that concrete, but I think that intuition comes from being exposed to a lot of set theory. The collection of arbitrary functions from $\mathbb{R}$ to $\mathbb{R}^2$ is actually quite wild and beyond this, pointwise differentiability is actually a pretty subtle condition, despite how early people are exposed to it in mathematics.

In any case, I find the characterization amusing enough that I couldn't resist sharing it.

Answer 6

The continuum hypothesis is equivalent to the assertion that for any set $X\subseteq\mathbb R$ either White or Black has a winning strategy in the following infinitely long game $G(X)$.

At the $n^\text{th}$ move, first White chooses a set $W_n\subseteq X$, and then Black chooses a set $B_n\in\{W_n,X\setminus W_n\}$. White wins if $\bigcap_{n\in\mathbb N}B_n\ne\varnothing$; Black wins if $\bigcap_{n\in\mathbb N}B_n=\varnothing$.

The game $G(X)$ is a win for White iff $|X|=2^{\aleph_0}$, a win for Black iff $|X|\le\aleph_0$, undetermined iff $\aleph_0\lt|X|\lt2^{\aleph_0}$.

I don't know the reference but I think this is due to Mycielski and Solovay.

Answer 7

This is a Hilbert-hotel variant I came up with that I'm putting in a philosophy of math book I'm currently writing.

In the infinite binary tree shown, imagine that it is a hotel, and infinitely many guests arrive. The hotel can accommodate up to $|\mathbb{R}|$-many guests. The individual $\aleph_0$-many branch points of the tree are the cheap hotel rooms, while the $|\mathbb{R}|$-many "leaves" (i.e., branch tips) of the tree are expensive water-front properties. CH is true if and only if, for every given infinite group of at most $|\mathbb{R}|$-many guests, the hotel manager can either assign all guests to occupy every cheap branch-point property, or else assign all guests to occupy every expensive water-front (branch-tip) property. CH is false if and only if there is a group of guests that is too large to be accommodated by the cheap properties and too small to occupy all of the expensive properties.

Additionally: Note that, unlike many other known equivalents of CH, this one is quite easy and direct to prove equivalent to CH, i.e., it is almost a direct paraphrase, once one realizes that there are $|\mathbb{R}|$-many "leaves" on the tree, or, equivalently, there are $|\mathbb{R}|$-many maximal paths through the infinite binary tree (which are in natural one-to-one correspondence with the power set of the set of all positive integers). Just because an easily grasped statement is equivalent to CH doesn't mean at all that it is obviously equivalent to CH. Arguably, if one really wants to understand CH via one of its equivalents, then one ought also to understand why the given equivalent is in fact equivalent. (I refer you to the hilarious joke about the Axiom of Choice, the Well-Ordering Principle, and Zorn's Lemma.) That's why I particularly like this reformulation.

Answer 8

My favorite equivalence is that CH fails if and only if whenever we color $\mathbb{R}$ with countably many colors, we can find monochromatic (and distinct) $x$, $y$, $u$, and $v$ such that $x+y = u + v$. I am not sure who gets the credit for this one, but it is included in the book Problems and Theorems in Classical Set Theory by Komjath and Totik.

Answer 9

The global dimension of the ring $\prod_{n = 1}^\infty \mathbb{F}_2$ is $k+1$ if and only if $2^{\aleph_0} = \aleph_k$. Note that the given ring, in another guise, is the power set of the set $\mathbb{Z}_{>0}$ of all positive integers under the ring operations $\oplus$ and $\cap$, which itself has cardinality $2^{\aleph_0}$. This is a great equivalence, as it also provides two simultaneous ring-theoretic interpretations of the cardinal $2^{\aleph_0}$, whether or not CH is true.

https://www.ams.org/journals/tran/1968-132-01/S0002-9947-1968-0224606-4/S0002-9947-1968-0224606-4.pdf

What the heck is the Continuum Hypothesis doing in Weibel's Homological Algebra?

Answer 10

This is probably a somewhat peculiar perspective, but for me personally the most concrete formulation of the continuums hypothesis is:

There is a cofinal chain in the Turing degrees.