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If I understand the continuum hypothesis (and its undecidability) correctly, then one can safely assert the existence of some set $A$ whose cardinality is between that of the integers and that of the real numbers, and this assertion will not cause a contradiction within ZFC set theory.

Have such sets ever been useful within mathematics or physics? I'll leave "useful" up to your own interpretation, but here are some examples of what I mean:

  1. The assumption that there exists a type of number whose square is $-1$ led to complex analysis, which has been useful in both pure math and applied physics.

  2. The rejection of Euclid's 5th postulate led to non-Euclidean geometry, which was also useful in both math and physics.

Some starting criteria for "useful" might be:

  • Launching a new field of math
  • Shedding new light on existing fields of math
  • Helping prove something that was already an open question
  • Playing any role in theoretical physics or some other field
WillG
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3 Answers3

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The assumption that there exists a type of number whose square is −1 led to complex analysis, which has been useful in both pure math and applied physics.

The rejection of Euclid's 5th postulate led to non-Euclidean geometry, which was also useful in both math and physics.

In both cases this was not an assumption, but a definition.

  • You define $\Bbb C$ as the quotient ring $\Bbb R[X]/{\left<X^2+1\right>}$ and observe that in this structure, $X$ is an element with square $-1$.

  • Similarly, non-Euclidean geometry is not assumed, it is observed in frameworks defined within the ZFC axioms.

An example of a set with cardinality between $\omega$ and $\mathfrak{c}$, by definition, cannot be constructed within ZFC. You can always assume its existence in a non-constructive manner, but then there is very little you can actually do with it. Or you could possibly construct it if you forgot about ZFC and worked in a suitable set of axioms, but then again it would not be useful to the ZFC community.

Community
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Arnaud Mortier
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    It looks to me like you are describing how $\mathbb{C}$ and non-Euclidean geometries are taught these day, not how they were discovered (or invented, if you prefer). I'm pretty sure the question is more concerned with the latter. – JonathanZ Jun 07 '18 at 19:14
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    @JonathanZ The point is not how these are defined or taught, but the very fact that they are constructed by hand and observed, not assumed as the OP was stating. So the comparison doesn't hold. – Arnaud Mortier Jun 07 '18 at 19:19
  • Check out Girolamo Cardano and his use of complex numbers to solve the causus irreducibilis cubic in 1545. ["Complex And Unpredictable Cardano", by Artur Ekert, Section 6 has a nice write up.] He doesn't define or construct them, he just starts using expressions like $5 + \sqrt{-15}$ and manipulating them like any other number containing a square root. That's assuming. – JonathanZ Jun 07 '18 at 20:14
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    @Jonathan true, even though I doubt that this was the OP's point of view. But eventually we were able to build such things. In the case of the CH it is hopeless in conventional settings and we know that already. – Arnaud Mortier Jun 07 '18 at 21:40
  • In the same way that you can build worlds where Euclid's postulate fails, you can build worlds where the continuum hypothesis fails... – Maxime Ramzi Jun 24 '18 at 08:52
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Let $P(x)$ denote the power-set of a set $x.$

Let $L[x]$ be the class of sets that are constructible from $x.$ Then $L[x] $ satisfies ZF.

An exercise from K. Kunen's text:

Let $M$ be a countable transitive model for (enough of) ZFC. We may obtain two Forcing extensions $M(1$) and $M(2)$ with a set $x\in M(1)\cap M(2)$ such that $$(i)\quad x=(P(\omega))^{M(1)}=(P(\omega))^{M(2)}$$ $$(ii)\quad (|x|^+)^{M(1)}<(|x|)^{M(2)}. $$

Now $(L[x])^{M(1)}=(L[x])^{M(2)}.$ Denote this set by $S.$ Then $S$ satisfies ZF. But we obtain a contradiction from (i) and (ii) if we assume there exists some $W\in S$ such that $S$ satisfies "$W$ is a well-order of $x$ "

So from the use of to different models for ZFC + ($\neg$ CH) we obtain a model for ZF+ ($\Bbb R$ cannot be well-ordered) and hence show that the Axiom of Choice is not a theorem of ZF (unless $1=0$ is).

DanielWainfleet
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If the continuum hypothesis is false, there is a canonical "counterexample": the set of all countable ordinals, commonly denoted nowadays by the symbon $\omega_1.$

Note that it's not the existence or the uncountability of $\omega_1$ that's undecidable; those are not in question. What's undecidable is whether $\omega_1$ is smaller than the continuum. If the continuum hypothesis is false (and the axiom of choice holds) then $\omega_1$ is smaller than the continuum.

And $\omega_1,$ the set of countable ordinals, is definitely useful in mathematics, regardless of the status of the continuum hypothesis (although in a sense it is even more useful if the continuum hypothesis is true). If I remember right, Cantor's interest in transfinite cardinals was motivated by his study of countable ordinals, which he found useful in his study of trigonometric series.

bof
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