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I've also been convinced that any comparative might have issues. Let's take the set of all primes. On the one hand, I can trivially demonstrate they are less n

I've also been convinced that any comparative might have issues. Let's take the set of all primes. On the one hand, I can trivially demonstrate they are less numerous than the set of integers On the other hand, Cantor's exercises for showing larger cardinality can be exercised equally on the set of prime numbers At least as far as I've been able to see. This shows a contradiction that can be easily accommodated for with multi-dimensional or an otherwise richer descriptor set. Any refutation that starts like "assume there exist a function f that generates all the primes as a sequence of integers" I reject. The whole point is that I'm not assuming that - maybe a proof that no such function can exist for the set of primes is an undiscovered property. If you decide to be extremely disagreeable and stubbornly throw out any speculative, at least I'm left with only with the contradiction. They seem to behave way closer to the irrationals then anything else. I think a more useful exercise is whether the peano successor function can be defined. For the irrationals, I think the answer is "not a chance". For the primes the intuition is "maybe" but when you start stating why, nearly every argument you give (statistically you can do it - with enough information you can do it, it is a discrete number and so on) you can defend for the irrationals with the same reasons.

hinkley 4y ago
Refresh my memory. Cantor's argument is that you can number the primes in order, right? So as long as primes are infinite (we believe they are but they just get more sparse), you can map every single whole number to a distinct prime number.
- kristopolous 4y ago
  Sure. You can do that with the irrationals as well. Whenever you have a new irrational, I'll increment my counter and assign it thusly. You show that there's irrationals between two assigned ones that were missed and I will continue to increment the counter accordingly and assign it to the missed irrationals. Then we will have our mapping to the whole numbers like you claimed. The point is that rule is insufficient to demonstrate the distinction we wish for. There's other ones such as the diagonalization or the column proof, you cannot do these directly with the primes, only with this middle step of assigning them a cardinality If you can only do them if you abstract your representation and assume you can assign them like that which I've showed, then you can do for any other set and it makes it no longer valid. It's only about what proofs can be demonstrated and what proofs cannot directly. Cantor's work is, in this way, more about the nature of the "generators" of the sequences then the sequences themselves and categorizing them like algorithmic complexity. The act of mapping is contingent on the sequences peano successor being algebraic. When you assign them whole number cardinality, you apply another "generator" that normalizes the thing you're inspecting and removes it from inspection. There's a distinction here that's not appreciated. But I've honestly lost hope
karpierz 4y ago
Here's a simple function from positive integers to primes. Let f : Z+ -> Z+ Let f(0) = 2 Let f(n) = p, where p is not divisible by f(n') for all positive integers n' such that n' < n, and p <= p' for all positive integers p' such that p' is not divisible by f(n'') for all positive integers n'' such that n'' < n.
- kristopolous 4y ago
  Sure. You just used jargon and formality to define it. That's not an actual algebraic definition in the same way you can express a peano successor function for say, all natural numbers. My claim is that's not an equivalent class of functions. You can define the irrationals in an almost identical way to that which you proposed You could claim that one can come to an agreement on the Nth prime but we cannot come to an agreement on the Nth irrational. But I say sure you can. You can number both of them off however you please. For every prime, we take a square root and get an irrational. There's a really simple one. Of course there's lots of holes but the point is the behaviors are far closer then they are to the rationals. I bet the farm that when Riemann Zeta is solved it will show just this connection. I can only see it in a blurry fuzzy way. But I don't think it's an illusion
  - karpierz 4y ago
    > That's not an actual algebraic definition in the same way you can express a peano successor function for say, all natural numbers. I guess it's not clear to me how you define an "actual algebraic definition" of a function. Can you provide a formal definition for this? Like some rules that I could use to determine whether a function is algebraically defined. I think it'll clear up whatever misunderstanding I seem to have.

hinkley 4y ago

Refresh my memory. Cantor's argument is that you can number the primes in order, right? So as long as primes are infinite (we believe they are but they just get more sparse), you can map every single whole number to a distinct prime number.

kristopolous 4y ago
Sure. You can do that with the irrationals as well. Whenever you have a new irrational, I'll increment my counter and assign it thusly. You show that there's irrationals between two assigned ones that were missed and I will continue to increment the counter accordingly and assign it to the missed irrationals. Then we will have our mapping to the whole numbers like you claimed. The point is that rule is insufficient to demonstrate the distinction we wish for. There's other ones such as the diagonalization or the column proof, you cannot do these directly with the primes, only with this middle step of assigning them a cardinality If you can only do them if you abstract your representation and assume you can assign them like that which I've showed, then you can do for any other set and it makes it no longer valid. It's only about what proofs can be demonstrated and what proofs cannot directly. Cantor's work is, in this way, more about the nature of the "generators" of the sequences then the sequences themselves and categorizing them like algorithmic complexity. The act of mapping is contingent on the sequences peano successor being algebraic. When you assign them whole number cardinality, you apply another "generator" that normalizes the thing you're inspecting and removes it from inspection. There's a distinction here that's not appreciated. But I've honestly lost hope

karpierz 4y ago

Here's a simple function from positive integers to primes. Let f : Z+ -> Z+ Let f(0) = 2 Let f(n) = p, where p is not divisible by f(n') for all positive integers n' such that n' < n, and p <= p' for all positive integers p' such that p' is not divisible by f(n'') for all positive integers n'' such that n'' < n.

kristopolous 4y ago
Sure. You just used jargon and formality to define it. That's not an actual algebraic definition in the same way you can express a peano successor function for say, all natural numbers. My claim is that's not an equivalent class of functions. You can define the irrationals in an almost identical way to that which you proposed You could claim that one can come to an agreement on the Nth prime but we cannot come to an agreement on the Nth irrational. But I say sure you can. You can number both of them off however you please. For every prime, we take a square root and get an irrational. There's a really simple one. Of course there's lots of holes but the point is the behaviors are far closer then they are to the rationals. I bet the farm that when Riemann Zeta is solved it will show just this connection. I can only see it in a blurry fuzzy way. But I don't think it's an illusion
- karpierz 4y ago
  > That's not an actual algebraic definition in the same way you can express a peano successor function for say, all natural numbers. I guess it's not clear to me how you define an "actual algebraic definition" of a function. Can you provide a formal definition for this? Like some rules that I could use to determine whether a function is algebraically defined. I think it'll clear up whatever misunderstanding I seem to have.