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Forgive me if this question is unclear or misguided, I have almost no knowledge in this area. It is well known (though not well understood) that the Continuum Hypothesis is independent from ZFC. The way I interpret this is that we can assume CH is true or false, and this will not cause any contradiction anywhere else in the mathematics that follows. In this vein, assuming $\neg$CH, we can say the there is a set $S$ such that $|\mathbb{N}| < |S| < |\mathcal{P}(\mathbb{N})|$, but what is this statement even meaningful? Can we discover any new mathematics by declaring the existence of such a set, and is there a way of writing down such a set as, say, a subset of $\mathbb{R}$? If not, why don't we just accept CH and move on?

If this question doesn't meet any guidelines or is unclear, please let me know of a way to amend it.

Nico
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  • I am not aware of the specific implications of adding CH or its negation to ZFC, but in general adding axioms to ZFC can be very fruitful in some areas of mathematics. See the following for a mind bending read: https://en.wikipedia.org/wiki/Large_cardinal – Valborg Mar 26 '18 at 04:32
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    Yes, of course there are many consequences of CH that are not provable without extra assumptions, and there are many consequences of its negation (or of strengthenings of its negation). – Andrés E. Caicedo Mar 26 '18 at 04:34
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    "It is well known (though not well understood) that the Continuum Hypothesis is independent from ZFC." Um, it's very well understood, and the independence of CH is a standard topic in early graduate set theory classes. "assuming CH, we can say the there is a set S such that $\vert\mathbb{N}\vert<\vert S\vert<\vert\mathcal{P}(\mathbb{N})\vert$" Did you mean "assuming $\neg$CH"? – Noah Schweber Mar 26 '18 at 04:43
  • As to whether there can be a "concrete" failure of CH (= "simply definable" set of reals with intermediate cardinality), the answer is: it depends what you mean by "concrete." No such set can be Borel, but consistently with ZFC there could be a coanalytic set which has intermediate cardinality. The problem is that although we can indeed write down a reasonably concrete definition of a set which consistently could have intermediate cardinality, we can't decide in ZFC alone whether it does in fact have intermediate cardinality. – Noah Schweber Mar 26 '18 at 04:45
  • This seems to be a mixture of "What is CH exactly?" and "What are some consequences of CH and its negations?", both of which have been asked before. – Asaf Karagila Mar 26 '18 at 07:24
  • CH and GCH are instant results of Occam's razor. – William Elliot Mar 26 '18 at 07:31
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    @WilliamElliot: That's a naive statement. Both introduce complicated objects and both have compelling axioms which contradict them. – Asaf Karagila Mar 26 '18 at 08:05
  • Also relevant to my first comment are these threads: https://math.stackexchange.com/q/499560/622 https://math.stackexchange.com/q/189471/622 https://math.stackexchange.com/q/2622162/622 https://math.stackexchange.com/q/494099/622 https://math.stackexchange.com/q/2445262/622 https://math.stackexchange.com/q/675400/622 https://math.stackexchange.com/q/2198404/622 and there are probably a handful of these out there. – Asaf Karagila Mar 26 '18 at 14:28
  • CH implies that there is a well-ordering of $\mathbb R$ in which each element has only countably many predecessors. Is there a way of writing down such a well-ordering? If not, why don't we just accept $\neg$CH and move on? (To avoid misunderstanding: I'm not advocating $\neg$CH; I'm just pointing out that the "just move on" suggestion in the question can work against CH as well as in favor of CH.) – Andreas Blass Apr 13 '18 at 02:30

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Let me start with a fun consequence of denying or accepting CH.

In 1962, John Wetzel asked the following question. Say a family $F$ of analytic functions on some common domain $D$ is pointwise countable if for each $z\in D$, the set of values $\{f(z)|f\in F\}$ is countable. Does it follow that $F$ is countable?

Pál Erdős answered the question very soon afterwards: pointwise countability implies countability if and only if the continuum hypothesis is false. Erdős's paper is available at online. It also appears in the book Proofs from THE BOOK by Aigner and Ziegler.

Osofsky proved a result in homological algebra using GCH. There's an entry in mathoverflow about it. (I only know about this because I was a grad student when she proved it, and Prof. Osofsky pulled me into her office to tell me about it, since I was in mathematical logic.)

The answer to this question gives some more applications of CH in "everyday mathematics".

Now, as to your question, "why don't we just accept CH and move on?", here's one answer. The two results I mentioned stand out because, for the most part, CH and GCH don't seem to have many consequences outside of axiomatic and descriptive set theory (and higher recursion theory). That is, if you're a working mathematician in algebraic geometry, or PDEs, or number theory, or lots of other branches, you just don't care that much about the value of $2^{\aleph_\alpha}$, or even of $2^{\aleph_0}$. Indeed, Osofsky was excited to tell me about her result because it was the first time in her career she had ever encountered a use for GCH. Likewise, a good part of what makes the Erdős result cool, is that one doesn't typically find applications of CH in complex analysis. In contrast, the axiom of choice pervades so much of "modern" mathematics (at least 100 years old by now), that most mathematicians accept it regardless of philosophical issues.

For people who are working in a branch where CH or GCH matters, no one hypothesis has emerged as "the best" new axiom. That is, $2^{\aleph_\alpha}=F(\alpha)$ is consistent with ZFC for a plethora of $F$'s. (See Easton's theorem.) Why should we choose this $F$ over that $F$? You can imagine aesthetic, philosophical, or pragmatic arguments, but a consensus has not coalesced around GCH or an alternative. Anyway, there's no problem with stating a theorem as "GCH implies foobar".

Michael Weiss
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    Forcing axioms are also very good reasons to reject CH (and therefore GCH). MA+not CH, and even more so, PFA and other extensions thereof are pretty good reasons to reject CH from a philosophical, utilitarian, and even aesthetic points of view. – Asaf Karagila Mar 26 '18 at 14:39
  • First, thank you for all the links in your other comment. Second, do you disagree with my "a consensus has not coalesced"? If not, how could I improve my answer to address your concerns? – Michael Weiss Mar 26 '18 at 14:47
  • No, I am not disagreeing with that. Some people would tell you that CH is true, others will you it's false. Some people will tell you that it's not even a "valid question". – Asaf Karagila Mar 26 '18 at 14:48
  • Some conjectures have first been proven assuming CH and later proven from ZFC. The earlier work was valuable in that it showed that the conjecture was consistent with CH, justifying the later effort of trying to prove it in ZFC. (I think this happenef in the search for a Dowker space.) – DanielWainfleet Mar 27 '18 at 03:21
  • Hmm, I thought the existence of a Dowker space was proved in ZFC alone; later attempts to construct a Dowker space with "small" cardinality used CH at first (maybe other set-theory axioms?) – Michael Weiss Mar 27 '18 at 05:37
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I saw a nice example years ago. Define a function $S: [0,1]\to $countable subsets of $[0,1]$. Is it possible to find two numbers $x,y$ such that $x\not \in S(y), y\not \in S(x)?$ It seems obvious that you can because countable sets are so small, so you should be able to take $x,y$ "at random" and succeed. However, if CH is true you cannot. Under CH you can well order the reals in $[0,1]$ such that each real has countably many predecessors. Let $S(x)$ be the set of predecessors of $x$. Then for any two different reals, one precedes the other in the order. Say $x$ precedes $y$, then $x \in S(y)$. This fails if CH is false because there will be reals with uncountably many predecessors in the well order.

Ross Millikan
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As for "writing down" an $S$, I don't know how to do it as a literal subset of $\mathbb R$, but here's something we can do:

The ordinal $\omega_1$ is by definition the first uncountable ordinal. It will be smaller than $|\mathbb R|$ if and only if CH fails. This means that we can use the set of all countable ordinals as $S$.

Transfinite ordinals may be a bit too abstract for comfort, but when all the ones we're interested in have the same cardinality, we can represent each of them more concretely as an equivalence class of well-ordering relations on $\mathbb N$. This allows us to write, $$ S = \{ x \in \mathcal P(\mathcal P(\mathbb N\times \mathbb N)) \mid \cdots \} $$ where "$\cdots$" is a concrete formula that quantifies only over $\mathcal P(\mathbb N\times\mathbb N)$, $\mathcal P(\mathbb N)$, and $\mathbb N$. (Technically, this has the arguable advantage over using $\omega_1$ itself that all sets involved in defining $S$ have small rank).

ZFC now proves $$ \neg{\rm CH} \leftrightarrow |\mathbb N|<|S|<|\mathbb R| $$

hmakholm left over Monica
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  • A typo: don't you mean that $\neg\text{CH}\leftrightarrow$ etc.? Also, I'm a bit puzzled by your formula for $S$. Where are you taking equivalence classes? Or do you mean $S$ to contain one representative from each equivalence class? – Michael Weiss Apr 12 '18 at 23:00
  • @MichaelWeiss: Right about the missing negation; fixed. As for the expression for $S$: Each well-ordering relation is an element of $\mathcal P(\mathbb R^2)$, so each equivalence class is an element of $\mathcal P(\mathcal P(\mathbb R^2))$. The "$\cdots$" is a formula that expresses: "$x$ is a set of well-ordering relations, and any two orders in $x$ are isomorphic, and every order on $\mathbb N$ that is isomorphic to an element of $x$ is itself an element of $x$". – hmakholm left over Monica Apr 12 '18 at 23:43
  • Oh, of course. Somehow my eyes/brain didn't take in the second $\mathcal{P}$. – Michael Weiss Apr 13 '18 at 15:59