I just posted a note on PhilPapers, which is also forthcoming in Philosophy of Science. (It is a bit weird for a completely mathematical paper to show up in Philosophy of Science, but it is what it is…)
The main feature of this note is that it describes a question that I could not solve. But this question is so simple that it can be explained to any mathematics undergrad or even high school student. So here we go…
Cantor tells us that infinite sets have different sizes.
Click for quick review of Cantor…
According to Cantor, the size of set \(A\) being equal to the size of set \(B\) means there is a perfect correspondence between them. For example, we have \(0\mapsto 0, 1\mapsto 2, 2\mapsto 4, \dots, n\mapsto 2n, \dots\), so the set of natural numbers \(\{0, 1, 2, 3, \dots\}\) and the set of even numbers \(\{0, 2, 4, \dots\}\) have the same size. Sets that have the same size as the natural numbers are called “countable”, and they are the smallest among the infinite sets. On the other hand, the set of real numbers is larger than countable sets, since Cantor discovered that for any countable list of real numbers we can always find yet another real number that is not in the list.
Cantor’s theory has a interesting feature: although the even numbers are only “half of” the natural numbers, in Cantor’s theory these sets have the same size!
Consequently, the possible number of sizes is quite small. For example, any infinite subset of a set of size \(\aleph_7\) (the 8th level of infinity, with the countable sets at the 1st level \(\aleph_0\)) must be at one of these 8 levels — although there are \(2^{\aleph_7}\)-many subsets, a huge number in comparison to 8! In other words, we are classifying these \(2^{\aleph_7}\)-many subsets into merely 8 baskets.
What if we want to say that the set of even numbers has a smaller size than the set of natural numbers? We would need an alternative definition of size. The part-whole principle says if set \(A\) is strictly contained in set \(B\), then the size of \(A\) is strictly less than the size of \(B\).
The question I could not solve is the following: suppose we are given an infinite set \(\kappa\), and we assign a size for its subsets, while respecting the part-whole principle. That is, we are classifying \(2^\kappa\)-many subsets into a number of baskets, according to their size. What is the minimum number of baskets that we need?
Clearly we would need many more baskets than Cantor (e.g. the natural numbers and the even numbers were in the same basket, now we need two different ones), and this number will grow as \(\kappa\) grows. But how many?
In my note I determined that this number is at least \(\text{ded } \kappa\) — a quantity associated with \(\kappa\) that is greater than \(\kappa\), an improvement over previous results.
The quantity \(\text{ded } \kappa\) is of independent interest, since it relates to another bizarre feature of the infinite. Now let me briefly tell you what it is.
Think of a line segment, a rope, a banana, or what have you. If you cut it once in the middle, you get two segments. If you cut it 2 times, you get 3 segments. If you cut it \(n\)-times, you get \(n+1\)-many segments. Sounds quite trivial, right?
However, the situation is a bit different when you can cut infinitely many times. Think of your line segment as the interval \([0, 1]\) of real numbers. Now let us make cuts at all the rational numbers. How many segments do we get? Observe that any two irrational numbers belong to different segments, since between them there is a rational number where we made a cut, by the density of rationals. From Cantor, we know that there are countably many rational numbers but uncountably many irrational numbers. So this is indeed a very bizarre situation: with countably many cuts, we can produce uncountably many pieces! This forms a sharp discontinuity with the finite case, where the number of pieces \(n+1\) is effectively the same as the number of cuts \(n\).
One mathematician who identified this phenomenon was Richard Dedekind, who actually defined real numbers as the collection of cuts that you can make on rational numbers. What if we make even more than countably many cuts?
Consider an infinite number \(\kappa\) and let us make \(\kappa\)-many cuts. The maximal number of pieces you can get is exactly \(\text{ded }\kappa\), named after Dedekind. Just as we observed, you can cut countably many times and get uncountably many pieces, it is generally true that \(\text{ded }\kappa>\kappa\).
Cantor and Dedekind, generated by ChatGPT Image 2
Finally let me tell you why I can’t solve my problem.
We have \(2^\kappa\)-many sets to be classified, and it is known that it is possible to classify them into \(2^\kappa\)-many baskets while respecting the part-whole principle. Now that I showed one needs at least \(\text{ded } \kappa\)-many baskets, if \(\text{ded }\kappa=2^\kappa\), the question would be solved. But the problem is, while \(\text{ded }\kappa\le 2^\kappa\), whether they are equal for all \(\kappa\) is independent of the fundamental assumptions of mathematics, known as \(\mathsf{ZFC}\). In other words, there are some mathematical universes where \(\text{ded }\kappa< 2^\kappa\) at some \(\kappa\). Can we make the assignment with less than \(2^\kappa\)-many baskets in these situations?
I tried to settle this question for a bit but it appears beyond me. I hope it will be answered one day by some mathematician or perhaps, some AI! I can see several possibilities of how this turns out…
An argument (perhaps relatively simple but hitherto evades us) shows that we always need at least \(2^\kappa\)-many baskets.
A \(\mathsf{ZFC}\) construction (perhaps somewhat involved) shows that it is always possible to use merely \(\text{ded }\kappa\)-many baskets.
A forcing construction shows that there is a \(\mathsf{ZFC}\)-universe where we can use less than \(2^\kappa\)-many baskets.
