• dang 18 hours ago |
    Discussed at the time:

    How many real numbers exist? New proof moves closer to an answer - https://news.ycombinator.com/item?id=27845576 - July 2021 (344 comments)

    (Reposts are fine after a year or so! especially for a great article like this. Links to past threads are just to satisfy extra-curious readers)

  • JadeNB 16 hours ago |
    The title:

    > How many real numbers exist? New proof moves closer to an answer

    and the subhed:

    > For 50 years, mathematicians have believed that the total number of real numbers is unknowable. A new proof suggests otherwise.

    both suggest a mystery that isn't there. Not that there isn't a mystery—there is!—but I think it's unreasonable to describe it as the title and subhed do.

    As the rest of the article, which is at Quanta's usual excellent expository standard, explains, there is absolutely no debate about the number of real numbers; it is 2^{\aleph_0}. You don't have to know what that is, but it is important to know that it is a definite, precise answer. You can ask whether it's the same as \aleph_1, and that's something we don't know. But that doesn't mean that we don't know the answer to how many reals there are, only that we don't know whether some other potential answer is also the answer.

    • dullcrisp 16 hours ago |
      I had the same initial reaction, because in my mind 2^{ℵ₀} is a more fundamental concept than ℵ₁.

      But the notation, and probably the theory, suggest otherwise, in which case saying there are 2^{ℵ₀} reals might be a bit like saying that there are |ℝ| reals. But I feel like I’m being pedantic regardless.

      • JadeNB 15 hours ago |
        > I had the same initial reaction, because in my mind 2^{ℵ₀} is a more fundamental concept than ℵ₁.

        > But the notation, and probably the theory, suggest otherwise, in which case saying there are 2^{ℵ₀} reals might be a bit like saying that there are |ℝ| reals.

        That's a good point, but, interestingly, and as you doubtless know, this article points out a rigorous sense in which it's the other way around: we know very explicitly at least two sets whose size we know is 2^{\aleph_0}, whereas we don't know any whose size we know is \aleph_1!

        • dullcrisp 14 hours ago |
          I believe that it’s defined as the size of ω₁, the set of wellorderings of the natural numbers. I don’t think there’s any real sense in which that’s less of a concrete set than the power set of the natural numbers, even if it’s harder to picture.
          • JadeNB 3 hours ago |
            > I believe that it’s defined as the size of ω₁, the set of wellorderings of the natural numbers. I don’t think there’s any real sense in which that’s less of a concrete set than the power set of the natural numbers, even if it’s harder to picture.

            The definition with which I am familiar is that \aleph_1 is the least cardinal that is strictly greater than \aleph_0, but, of course, one can use any of an equivalent collection of properties as the definition.

            Nonetheless, at least any definition in line with usual mathematical practice should probably not provably require that \aleph_1 is at least 2^{\aleph_0}, which I claim that your definition does. Let me refer to positive integers, and non-negative integers, to avoid having to commit to what the natural numbers are. I define a function from the power set of the positive integers to the set of total orderings of the non-negative integers as follows. For every power set A, write B for the complement of A in the positive integers, and let <_A be the order that restricts to the usual order on A and on B (though not on their union), and that declares 0 to be greater than every element of A and less than every element of B. Since the usual order on the positive integers is a well order, so is <_A. Clearly, A may be recovered from <_A (as the set of elements preceding 0), so the function is injective.

            • dullcrisp 39 minutes ago |
              Sorry, classes of well-orderings of subsets of the natural numbers up to order isomorphism: https://en.wikipedia.org/wiki/Hartogs_number. And yes it’s equivalent to your definition due to ordinal magic as far as I can tell (or maybe because cardinals are by definition well ordered, I’m clearly not an expert).
    • billfruit 9 hours ago |
      I find this style of writing from Quanta very jarring and detracting from the usefulness of the article.

      A factual head line is very much sufficient to get the readers interest.

  • ajkjk 15 hours ago |
    It's a trick! Despite the name, the real numbers aren't real at all. Most of them don't exist...

    Now the computable reals, on the other hand... maybe.

    • skissane 14 hours ago |
      Or the definable reals. Definable meaning you can “name” it uniquely, by writing down some proposition which is true of that real and false for every other. Since there are only a countable number of formulas, there are only countably many definable reals. Definable reals are a superset of computable reals - Chaitin’s constant (for a given computational formalism) is definable but not computable.

      There is not a single set of definable reals, since different formal languages produce different sets (e.g. there are reals definable in second order logic but not first order logic). But, given any formal language composed of finite sentences from a finite alphabet, the set of sentences in the language is countable so the corresponding set of definable reals is too.

      The definable reals are all the reals we can talk about individually. (Well, there are some definable reals we can’t in practice talk about individually-e.g. those for which the shortest sentence uniquely identifying them is too long to fit in this physical universe.) Undefinable reals are an uncountable amorphous mass we can only discuss collectively, we can’t refer to any of them as individuals.

      • dullcrisp 13 hours ago |
        For any concrete definition of definable reals and any well-order of the reals, I can refer to the first real that’s undefinable in that system.
        • skissane 13 hours ago |
          If we don’t limit ourselves to formal systems, we can speak about definable in natural language. In which case you just uttered a paradox.

          Now, you can object to “natural language definability” as lacking “concreteness” since maybe it can’t be formalised, but then the claim in your comment can’t be formalised either…

          Actually, are we sure natural language can’t be formalised? Obviously any formalisation of natural language has to be paraconsistent (maybe even dialetheic) rather than consistent, but granted that…

          • dullcrisp 13 hours ago |
            I think you’re right, which I guess might prove that it’s impossible to refer to a well-ordering of the real numbers. I guess that’s what they mean when they say non-constructible, huh?

            But still my intuition is that for any consistent formal system I should be able to find a more powerful system that can express strictly more things, even if my argument above was silly. But I don’t think we can talk about the set of things that can be defined in natural language in any meaningful way.

            • chriswarbo 9 hours ago |
              > But still my intuition is that for any consistent formal system I should be able to find a more powerful system that can express strictly more things

              The problem with defining more and more powerful systems is that we have an arbitrary selection of axioms to choose from. For example, if we start with the Peano axioms and define a Goedel sentence (i.e. a specific instance of "this sentence cannot be proven"), then we can choose to extend that system either by adding that sentence as an axiom, or adding its negation as an axiom. Both would result in a consistent system. Then we can define a Goedel sentence in that system, and make a more powerful system by making that or its negation an axiom. And so on.

              This process of choosing new axioms does make our formal system strictly more powerful (it can prove/refute sentences which were unprovable in the weaker systems, and their consequences); but it's not very useful or interesting, since we're just arbitrarily choosing whether those extra sentences should be provable or not.

              This is related to the fact that any particular formal system can only compute a few digits of Chaitin's constant (since we can only ever solve the halting problem for a subset of all possible programs). We can consistently extend our chosen system by specifying subsequent digits as axioms, but that doesn't really tell us anything; we're just asserting it, for no particular reason one way or another.

              • dullcrisp an hour ago |
                But doesn’t that still call into question the notion of “definability” as a property of a number? If I invent an axiomatic system called “ZFC + Chaitin's constant is c” does that make Chaitin’s constant any more definable than it was before?
        • EGreg 12 hours ago |
          Have you heard of https://en.wikipedia.org/wiki/Berry_paradox ?

          It seems to run into the same sort of thing that led to ZF axioms, namely that you can have a language that is a bit too "loose" and therefore leads to contradictions. I think Kurt Godel showed that any non-trivial language is going to have something like that.

          • dullcrisp 12 hours ago |
            Thanks, I have now!
        • Dylan16807 10 hours ago |
          Can you coherently refer to the "first" real outside some set, just in general?

          Like, what is the first positive real?

          If you have to involve infinitesimals or something, I think it's perfectly fine for you to be able to take the definable reals and then extend them in that way. it doesn't invalidate the idea of a "definable" real.

        • Someone 10 hours ago |
          Can you? There’s no “first real larger than 2”, for example, so if I define the definable reals as {0, 1, 2}, what is that first real?

          In general, you also can’t use Cantor’s diagonal argument on the list of definable reals because there are numbers in there that aren’t computable.

      • ajkjk 13 hours ago |
        I don't know a ton about this (just heard about it in something I read somewhere?), but I prefer the computables--I feel like the idea of a computable number is the formalized version of the idea that (non-integer) numbers in nature do not have definite values, but instead have processes by which you can overturn more digits with effort, and thus they can only be said to be "pretty equal" but never totally equal.

        If a number can be defined in a language but you can't tell me if it's different from another number that I have in finite time, then it is not very interesting to me.

        • skissane 12 hours ago |
          > I prefer the computables--I feel like the idea of a computable number is the formalized version of the idea that (non-integer) numbers in nature do not have definite values, but instead have processes by which you can overturn more digits with effort, and thus they can only be said to be "pretty equal" but never totally equal.

          Consider an irrational number, where the fastest possible algorithm (given a standard Turing machine) to compute its nth digit requires (10^(10^100))*(n+1) operations. Now, by the standard definition, such a number is computable – but in practice, we will never be able to compute even its first digit. In practice, its first digit is just as unknowable to us as the first uncomputable digit of (any instantiation of) Chaitin's constant is.

          "Computable" by the standard definition is unphysical, because it includes computations that require unphysically large amounts of time and space. Now, many uncomputable (by the standard definition) numbers can be computed by a hypercomputer (such as a Turing machine with an oracle, or a computer which can execute supertasks) – hypercomputers are unphysical, but "standard" computers (in the mathematical sense) are unphysical too. Why denounce one form of unphysicality but not another? It would seem to be more consistent to either embrace the unphysicality of mathematics and imbibe the psychedelic broth of the Cantorian paradise, or else reject unphysicality entirely and embrace ultrafinitism. The "infinity is verboten but unphysically large finitudes are A-okay" approach of (non-ultrafinitist) construcitivsm/intuitionism seems rather arbitrary.

          > If a number can be defined in a language but you can't tell me if it's different from another number that I have in finite time, then it is not very interesting to me.

          If a number can be defined in a language but I can't tell you whether it's different from another number that you have in a physical amount of time (let's say less than the age of the universe), is it interesting to you?

          • tromp 10 hours ago |
            > Consider an irrational number, where the fastest possible algorithm (given a standard Turing machine) to compute its nth digit requires (10^(10^100))*(n+1) operations

            There can be no such number. You can always make a Turing machine output initial digits faster by adding some more states to do so.

            • skissane 9 hours ago |
              Yes, if you knew ahead of time that the first digit of some such number happened to be seven, you could just hardcode your program to start with "IF n = 0 THEN RETURN 7". But that's cheating...

              Let's denote the nth prime by P(n). Now, consider the real number Q, where 0.0 < Q < 1.0, and the nth decimal digit (after the decimal point) of Q, Q(n), is given by Q(n) = P(10^(10^100) + n) mod 10. Q, as defined, is computable, but (unless we discover some mathematical "shortcut" for computing the last digit of a prime number), I doubt we are ever going to know what even its first digit is (well, we know it must be 1, 3, 7 or 9–but which of those four?). And even if some such "shortcut" were discovered, I'm sure someone could cook up another such number for which there is no known "shortcut".

              How about this: let N(n) be any FNP-complete [FNP is to function problems what NP is to decision problems] computable endofunction of the natural numbers (there are many to choose from, pick any), then define:

              Q(n) = P(N(10^(10^100) + n)) mod 10

              You actually think we'll ever know the first digit of the real number so defined?

          • ajkjk 4 hours ago |
            Well, I am glad to learn an even better notion for the concept I'm talking about! I'm just saying, I think "definable" is too loose, because it presupposes that I can even talk about real numbers in a definite way. "Computable" feels closer because it is like an idealized version of the notion, but I agree, it's too loose also.

            The reason I've latched onto it is that I read about the idea that in constructive mathematics all functions are continuous, basically because you can't tell that they're not continuous (because to tell that would imply you could inspect infinite digits of function's inputs and tell that they lead to discontinuities in the outputs). I quite like that idea because it rhymes with the intuition that nature doesn't care about later digits of numbers either, and that properties of the real numbers like being uncountable, dense, etc cannot show up in any physical system and therefore aren't "real".

            But I'm definitely looking for a better way of understanding this, so curious if you can suggest anything.

      • bandrami 12 hours ago |
        > Since there are only a countable number of formulas, there are only countably many definable reals

        It always seemed like it should matter that the constructable formula strings include strings that define multiple numbers (e.g. "three; also pi squared") which at least seems like it should be aleph one rather than aleph nought, but smarter people than me assure me it doesn't work that way for (reasons)

        • EGreg 12 hours ago |
        • NooneAtAll3 12 hours ago |
          that redundancy will only decrease the number of definables, no?

          and countable infinity is already the smallest one, so you can't get smaller

    • danbruc 9 hours ago |
      What do you mean when you say a number does or does not exist?
  • neuroelectron 14 hours ago |
    tl;dr: Godel's incompleteness theorem. When you try to create completely self-consistent axioms, it just creates an infinite hierarchy of further unprofitable axioms. Numbers don't really exist, so that was your first mistake.

    tl;dr^2: Mentally move the hyphen one word to the right.

    • bubblyworld 12 hours ago |
      This is a somewhat different situation. Gödel's theorem gives you independent statements but they are highly contrived (containing expressions encoding provability and consistency in an ad-hoc fashion).

      This is talking about the cardinalities of very basic sets, and it was (is? haha) extremely unclear to what extent statements about them are independent, if at all. I would personally be shocked if you can prove independence of CH directly from the incompleteness theorem, though I would love to be wrong! It required a lot more delicate machinery to be developed (Cohen's forcing method).

      Numbers certainly exist in a useful sense - they model real properties of the universe. But I agree that higher set theories may be moving away from that...

      • neuroelectron 10 hours ago |
        Lots of heuristics are useful.
        • bubblyworld 9 hours ago |
          I suggest asking yourself what you think you mean by "exists".
  • dataflow 13 hours ago |
    Meanwhile all I've ever wanted for Christmas was a notion of size that admits (A ⊆ B) ∧ (A ≠ B) ⇒ size(A) < size(B).
    • pfdietz 13 hours ago |
      That's a notion that doesn't work for infinite sets.
      • dataflow 13 hours ago |
        Could you expand on why it can't work?
        • heyitsguay 13 hours 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 13 hours 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 12 hours 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 12 hours 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.

    • bubblyworld 13 hours 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 13 hours ago |
        Lovely, thank you!
  • qnleigh 11 hours ago |
    I wish I understood how mathematicians conceptualize the notion of 'right' and 'wrong' axioms in light of Gödel. I would've thought that the incompleteness theorems would be taken as evidence against Platonism, but clearly this didn't happen. What do people think is the source of 'ground truth'? Is it more than aesthetic appeal? Are there any articles or books that convey it?

    I've talked to a friend with a doctorate in algebraic geometry about this several times, but all I've really gotten out of her is that if I studied a lot more math, I would gradually get the same sense.

    (For reference, I have about as much mathematics background as a US student who minored in maths.)

    • n4r9 10 hours ago |
      > the incompleteness theorems would be taken as evidence against Platonism

      I wonder if this is a map/territory thing. Incompleteness theorems apply to models (axiomatisations) of mathematics, e.g. ZFC set theory. Platonism applies to the actual act of doing mathematics, i.e. persuading other mathematicians that certain statements about abstract entities are true or false.

      • dr_dshiv 9 hours ago |
        Platonism is about the reality of math behind the notation and beyond the human mind —

        Like that spheres are real even though there is no perfect sphere in the material world.

        This is a common opinion among mathematicians and physicists even though it seems unscientific (as it poses reality beyond matter)

        • n4r9 8 hours ago |
          Right. According to Platonism, a mathematician publishing a result about spherical geometry is in fact exploring an objective reality beyond human minds. Contrast that with, say, intuitionism, according to which they are simply describing the mental activity which arises as a result of considering an object of shared understanding known as a "sphere".
    • vlz 10 hours ago |
      This comment from a previous thread

      https://news.ycombinator.com/item?id=27846084

      links to a two-part paper called "Believing the axioms", part 1:

      https://www.cs.umd.edu/~gasarch/BLOGPAPERS/belaxioms1.pdf

      I have only taken a short look and am not qualified to summarize, but it seems this would have an answer if there is one.

    • skissane 9 hours ago |
      > I would've thought that the incompleteness theorems would be taken as evidence against Platonism, but clearly this didn't happen

      I don't think they really are evidence against it. The first incompleteness theorem says (to put it simply) there are truths about the natural numbers you can't prove, and (if we equate proof with knowledge) can't know. I don't know why a Platonist would find that objectionable. I mean, naïve materialism would imply there are lots of facts about the material world we are never going to be able to know (e.g. the particular arrangement of rocks on a lifeless planet in a distant galaxy right now, or as close to right now as relativity permits). If unknowable truths isn't evidence against materialism, why would it be evidence against Platonism?

      Really, Gödel's theorems were a much bigger problem for formalism than Platonism. Formalists wanted to identify mathematical truth with provability, and Gödel shattered that dream. Platonists never dreamt the dream, so its destruction didn't discourage them.

  • trehalose 5 hours ago |
    > But Cantor discovered that natural numbers can’t be put into one-to-one correspondence with the continuum of real numbers. For instance, try to pair 1 with 1.00000… and 2 with 1.00001…, and you’ll have skipped over infinitely many real numbers (like 1.000000001…). You can’t possibly count them all; their cardinality is greater than that of the natural numbers.

    I take issue with this paragraph. It makes it sound like the infinitude of reals between 1 and 1.00001 is proof there are more reals than naturals. One might think this same fact proves that there are more rationals than naturals...