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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 num

by JadeNB 2y 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 2y 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 2y 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 2y 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 2y 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 2y ago
        Sorry, classes of well-orderings of subsets of the natural numbers up to order isomorphism: https://en.wikipedia.org/wiki/Hartogs_number 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 2y 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.

by JadeNB 2y ago

dullcrisp 2y 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 2y 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 2y 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 2y 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 2y ago
      Sorry, classes of well-orderings of subsets of the natural numbers up to order isomorphism: https://en.wikipedia.org/wiki/Hartogs_number 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 2y 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.