3 ms·

Meanwhile all I've ever wanted for Christmas was a notion of size that admits (A ⊆ B) ∧ (A ≠ B) ⇒ size(A) < size(B).

• pfdietz 2y ago
  That's a notion that doesn't work for infinite sets.
  • dataflow 2y ago
    Could you expand on why it can't work?
    • heyitsguay 2y ago
      So if N is the natural numbers {1, 2, 3, ...} and E is the even numbers {2, 4, 6, ...}, by the definition you propose, we must have size(E) < size(N), right? But now let's divide each element of E by 2 to produce a set D. Now that set D is {1, 2, 3, ...} aka N the natural numbers. But we just applied a function to each element of E, so how can D have a different size than E? Each element of E maps to exactly one element of D: 2 -> 1, 4 -> 2, 6 -> 3, etc. So does size(D) = size(E)? Or does size(D) = size(N)?
      • dataflow 2y ago
        Thanks. How I would've imagined this could work (if it could be made to work) is to somehow account for the notion that your "infinity" just got halved (yes I get that infinity is not part of the set, but let me wave my hands here while we're breaking our axioms), and everything "after" it is no longer in the set. Which would imply that "{1, 2, 3, ...}" is no longer a sufficient description for the set of natural numbers; you'd probably need extra information (maybe a "scale factor" for the infinity or something). I imagine you're right that this leads to a contradiction somewhere, but I (obviously) haven't thought it through. I just would love to see someone try to break enough axioms to make it work and see what comes out of it, or show that it contradicts either itself or something we see in the real world if we do that.
        • dullcrisp 2y ago
          You can have the partial order A ≤ B iff A ⊆ B. But if you want size({1}) = size({2}), size({2}) = size({3}), etc., then you’ll find that size({1,2,3,…}) = size({2,3,4,…}) and there’s really nothing you can do about it. But yes, the ordinals might be more to your liking. If you equip your sets with more structure you can say more things about them.
      • jhghikvhu 2y ago
        > But now let's divide each element of E by 2 to produce a set D. You are assuming that this doesn't change the size and certainly that's how the normal notion of size works. But the question is whether we can create any order relationship on the sets with the desired properties. The properties he mentioned defines a partial order and partial orders can be extended to total orders (given axiom of choice). So it is in fact possible.
        • pfdietz 2y ago
          A partial order of the elements of some set can be extended to a total order. But a partial order of all sets? How is this function "size" even defined? Functions don't have a domain of all sets in conventional set theory.
• bubblyworld 2y ago
  There's natural density, which is a notion of size for subsets of the naturals that can differentiate between subsets of equal cardinality. There are also ordinals, which are much finer than cardinals but have the disadvantage that they only apply to well-ordered sets, and ordinal arithmetic works very different to the naturals. And then a final one - exotic models of the real numbers like the surreals contain infinite quantities of different sizes that can be compared, divided, added just as you like and the order relation works intuitively. The disadvantage here is that they are generally harder to define and reason about than the ordinary reals. The reason people like them is that you they contain infinitesmals too, which you can use to formalise an alternative foundation for calculus. (although strictly speaking this last one isn't about sets and sizes, it's in a similar cluster of ideas about infinite arithmetic)
  • dataflow 2y ago
    Lovely, thank you!
• munchler 2y ago
  Couldn't you write this more simply as A ⊂ B ⇒ size(A) < size(B)?
  • dataflow 2y ago
    Yeah I knew someone was gonna ask. The trouble is a lot of people use ⊂ to mean ⊆. So I would've gotten a comment correcting me on that instead.