Regarding sets with size distinct from a power set of $\mathbb{N}$

Question

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.

Answer 1

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".

Answer 2

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.

Answer 3

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