Before I read this paper, I thought that the continuum hypothesis probably did not have a truth value, and that ZFC + CH and ZFC + ¬CH are systems that equally well capture what it means to be a set. Now I think that the continuum hypothesis is actually false, and that ZFC + CH would not be an adequate representation of our intuitive notion of sets.
The argument in the paper is not a formal proof of ¬CH. Instead, it’s a refutation of the following form: “The continuum hypothesis is equivalent in ZFC to this statement which is obviously false. So the continuum hypothesis must be false.”
So what is this obviously false statement? It is a claim that an event that happens with 100% probability is impossible. More exactly, for any function that assigns to each real number a countable set of real numbers, the claim is that it’s impossible for there exist two real numbers x and y such that x is not in the countable set assigned to y and y is not in the countable set assigned to x.
Why is this obviously false? Well, imagine throwing darts at the real line, so as to randomly select a real each time. If we throw one dart, there is a probability of zero that we would hit a natural number. Why? Because the cardinality of the naturals is less than the cardinality of the reals. By the same token, there is a probability of zero that our dart hits a rational number (as the rationals are also only countably large).
Now, if we throw a second dart, the probability that we hit a real number that’s some rational distance away from the first real number is also zero. Again, this is just because we’re considering a countable set of candidates from an uncountable set of possibilities. So by the same token, if we assign to the first dart ANY countable set of real numbers, then there is probability zero that the second dart lands within that countable set. And if we also assign a countable set of real numbers to the second dart, there is also probability zero that the first dart happened to fall within the countable set assigned to the second.
So, to summarize in more general terms: for any function f that maps each real number to a countable set of real numbers, a random choice of x and y will give us that x ∉ f(y) and y ∉ f(x) with 100% probability. Now, the statement that is equivalent to the falsity of the continuum hypothesis (i.e. the existence of cardinalities between |ω| and |𝒫(ω)|) is that it’s possible to find two real numbers x and y that are not in each others’ countable sets.
Notice how weak this is! By randomly choosing real numbers, we will end up with two that are not in each others’ countable sets with probability 100%. But we’re not even saying that the probability 100% event will always happen! We’re just saying that it’s POSSIBLE that it happens! I.e. imagine that by some miraculous coincidence it happens that the first two real numbers we choose are within each others’ countable sets (technically, only one of them needs to be in the other’s countable set). After all, probability-zero events do happen. Not a problem! Just pick two new real numbers! And if this fails, pick again!
Pretty cool right? Well, it gets even cooler. We just saw that from the statement “for any function f that maps reals to countable sets of reals, there exists x and y such that x ∉ f(y) and y ∉ f(x)”, you can prove that |𝒫(ω)| > |ω1| (where the cardinality of ω1 is by definition the next cardinality after ω). We can now consider THREE reals x, y, and z. Consider mappings from two real numbers to a countable set of reals. We can now state that for any such mapping, none of the three reals is in the countable set assigned to the others. And this entails that we can prove that |𝒫(ω)| > |ω2|! In other words, we can prove that there are at least TWO cardinalities in between the reals and the naturals!
And for ANY n, if we take the statement that there are n reals such that none is inside the countable set mapped to by the others, then we can use this to prove that |𝒫(ω)| > |ωn-1|. Notice: whatever n we choose, there’s still probability 100% that the n reals don’t “overlap” in this way! So for any n, we’re again only claiming that it’s possible for an event with 100% probability to happen. And if we accept this, then this entails that there’s not just a single cardinality between |ω| and |𝒫(ω)|. There’s definitely at least two. And at least three. And so on for ANY finite value of n. Which means that there are infinitely many cardinalities between the naturals and the reals!
Ok, this is a good point to stop reading if you don’t want to know the technical details of how EXACTLY to prove the result and are satisfied to just know what the result says. But if not, then read on! (The proof is actually really short and simple, though it relies on some background knowledge about ordinals.)
I’ll prove the result for n = 2. Suppose that |𝒫(ω)| = ℵ1 = |ω1|. Then by the axiom of choice, there’s some well-ordering of the reals of order type ω1. Call this well-ordering <. We define f(x) to be {y | y ≤ x}. Notice that since ω1 is the FIRST uncountable ordinal, f(x) is always a countable set, no matter what x is! Then, since for any two reals, x ≤ y or y ≤ x, this entails that for any two reals, x ∈ f(y) or y ∈ f(x).
And we’re done! We’ve shown that if the continuum hypothesis is true, then there exists a function f such that for any two reals, x ∈ f(y) or y ∈ f(x). So if we think that there is no such function, then we must deny the continuum hypothesis.