Do you mean the underlying math we need for probability theory does not allow us to construct or derive the part of number theory that Gödel’s theorem needs? That seems wrong somehow. (Going by Wikipedia, it means that we don’t know a non-empty set of probabilities has a least upper bound. That seems like it would cause problems sooner or later.) Where would I go to find out why this holds true?
You cannot get number theory out of the first-order theory of the reals; how would you write a first-order predicate to tell you whether or not a real number was an integer?
I don’t see what least upper bounds have to do with anything. Again, Gödel’s theorem deals with systems that can model number theory (i.e. the integers, the whole numbers); it has nothing to do with the real numbers.
The Wikipedia article and this shorter account both say that some form of Gödel’s incompleteness theorem applies to second-order logic. I asked about the limits of the first-order approach to the reals because it looks like we’d need to use that if we want to stop the theorem from applying.
That approach still seems odd, but I can sort of see how you could do probability that way. I’ll edit the OP to reflect my real question as soon as I feel up to it.
Do you mean the underlying math we need for probability theory does not allow us to construct or derive the part of number theory that Gödel’s theorem needs?
Essentially. The specifics are explained in more detail in the comments above.
Going by Wikipedia, it means that we don’t know a non-empty set of probabilities has a least upper bound.
That is not true. What Wikipedia article are you referring to?
Second-order logic says that, “if the domain is the set of all real numbers,” then “one needs second-order logic to assert the least-upper-bound property for sets of real numbers, which states that every bounded, nonempty set of real numbers has a supremum.”
Do you mean to say we can somehow prove this in first-order logic for numbers between 0 and 1, but we can’t extend it to the real number line as a whole?
It’s not about what you can prove, it’s what you can state. The first-order theory of the reals doesn’t even have the concepts to state such a thing. If your base theory is the reals, then sets of reals are a second-order notion.
No. I meant that we can prove that some specific sets of numbers have least upper bounds. What we cannot prove is that every bounded, nonempty set has a least upper bound.
Do you mean the underlying math we need for probability theory does not allow us to construct or derive the part of number theory that Gödel’s theorem needs? That seems wrong somehow. (Going by Wikipedia, it means that we don’t know a non-empty set of probabilities has a least upper bound. That seems like it would cause problems sooner or later.) Where would I go to find out why this holds true?
You cannot get number theory out of the first-order theory of the reals; how would you write a first-order predicate to tell you whether or not a real number was an integer?
I don’t see what least upper bounds have to do with anything. Again, Gödel’s theorem deals with systems that can model number theory (i.e. the integers, the whole numbers); it has nothing to do with the real numbers.
The Wikipedia article and this shorter account both say that some form of Gödel’s incompleteness theorem applies to second-order logic. I asked about the limits of the first-order approach to the reals because it looks like we’d need to use that if we want to stop the theorem from applying.
That approach still seems odd, but I can sort of see how you could do probability that way. I’ll edit the OP to reflect my real question as soon as I feel up to it.
Essentially. The specifics are explained in more detail in the comments above.
That is not true. What Wikipedia article are you referring to?
Second-order logic says that, “if the domain is the set of all real numbers,” then “one needs second-order logic to assert the least-upper-bound property for sets of real numbers, which states that every bounded, nonempty set of real numbers has a supremum.”
Do you mean to say we can somehow prove this in first-order logic for numbers between 0 and 1, but we can’t extend it to the real number line as a whole?
It’s not about what you can prove, it’s what you can state. The first-order theory of the reals doesn’t even have the concepts to state such a thing. If your base theory is the reals, then sets of reals are a second-order notion.
No. I meant that we can prove that some specific sets of numbers have least upper bounds. What we cannot prove is that every bounded, nonempty set has a least upper bound.