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They’re not finitely describable — for any member of that set, no matter how many digits you articulate, you’ve named a rational number. The same is true even

They’re not finitely describable — for any member of that set, no matter how many digits you articulate, you’ve named a rational number. The same is true even for numbers that have finite predicates defining them — because there’s only measure zero finite predicates. Most real numbers are indescribable sequences (ie, not finitely describable), so it’s not surprising we can’t describe them. That’s what happens when you construct something by including all possible infinite sequences — almost all of them are total gibberish. Diagonalization is a consequence of that: we have a schema of descriptions forming our predicate and prove no matter how we index reals by the integers, there’s more stuff out there. To get rid of that, we need to stop completing sequences. Most people aren’t in favor of that finitism/constructive approach now — so I’m not sure why that would change.

xscott 2y ago
Even Cantor's counterexample, where he traverses the diagonal of an enumerated list of numbers, he has a countable number of choices to pick his digits. So we've built a countable set, and he claims it's incomplete by picking one number from a countable set of counterexamples. So now lets union the original set with Cantor's countable set of counterexamples, and get a combined set which is also countable, right?
- zmgsabst 2y ago
  But that schema can be applied to your new set, as well — to create yet more counter examples. Your method didn’t exhaust the real numbers outside the original list. You merely added one of them. For any listable subset of the real numbers, we can provide a counter example. And since we’re proving by contradiction in Cantor’s argument, we assume the original list is exhaustive, ie there’s a bijection between naturals and reals. We show that assumption leads to a contradiction, because there’s at least one real number for which no natural maps to it. Therefore, there can’t be such a bijection. - - - Assume f is a bijection between N and [0,1]. Let an be the nth digit of f(n). Then x = sum(1,inf)[10^-n * (an + 1 % 10)] is a real number in [0, 1] for which there is no f(k) = x. So f is not onto and therefore f is not a bijection. Therefore, no such bijection can exist — by contradiction. - - - You can’t patch that schema up just by adding counter examples to your first choice of bijection — because a counter example is constructable for any bijection attempt. The fundamental problem is that reals include limits — which is what allows the sum to be a real number.
  - xscott 2y ago
    I won't claim to have any proof in hand, but there's still something fishy about it all. I think things like Banach-Tarsky and such ought to be treated as counter examples to show the absurdity of the Reals. At times I've thought maybe the problem with diagonalization is forcing me to do a (countably) infinite number of steps before Cantor gets to take an infinite number of steps. For instance, I can give a trivial counting scheme and claim it will eventually generate every number between zero and one, perhaps by writing a program to write programs that generate digits. It's all very deterministic, and I state exactly what I'm doing up front. So I give the first number (program), and Cantor gives his first digit and says it's not in my list. Then I tell Cantor the countably infinite number of places where his first digit is found, and we select those programs. Then he gives me his second digit, and I select the subset where those are found. So we go back and forth, he keeps saying his number so far is not in my list, and I keep telling him all the places where it is in the list... Feels like stalemate there. There are other places (such as summing the alternating harmonic series) where you have to be very careful about the ordering of operations. If he really has a higher order infinity of numbers not in my list, maybe I should be able to ask him for his list first, no? It's uncountably infinite, but he can't provide even one of them.
    - zmgsabst 2y ago
      You don’t have to do anything but assume such a bijection does exist, as in my comment: its existence leads to a contradiction in the presence of limits — just like I showed. But without that same limiting to build sets, you can’t construct N from the successor function and you can’t discuss the power set of N. So we either need to throw out countable limiting to build sets or accept that such set building rules out bijections between reals and naturals. You think Banach-Taraki is weird (as do most people), but I’d argue a world where you have finite numbers or couldn’t discuss the set of subsets of the naturals is even weirder. > maybe I should be able to ask him for his list first, no? There’s no such list indexable by the naturals, as shown by that contradiction. If you allow indexing by the reals, that request is trivial.
      - xscott 2y ago

xscott 2y ago

Even Cantor's counterexample, where he traverses the diagonal of an enumerated list of numbers, he has a countable number of choices to pick his digits. So we've built a countable set, and he claims it's incomplete by picking one number from a countable set of counterexamples. So now lets union the original set with Cantor's countable set of counterexamples, and get a combined set which is also countable, right?

zmgsabst 2y ago
But that schema can be applied to your new set, as well — to create yet more counter examples. Your method didn’t exhaust the real numbers outside the original list. You merely added one of them. For any listable subset of the real numbers, we can provide a counter example. And since we’re proving by contradiction in Cantor’s argument, we assume the original list is exhaustive, ie there’s a bijection between naturals and reals. We show that assumption leads to a contradiction, because there’s at least one real number for which no natural maps to it. Therefore, there can’t be such a bijection. - - - Assume f is a bijection between N and [0,1]. Let an be the nth digit of f(n). Then x = sum(1,inf)[10^-n * (an + 1 % 10)] is a real number in [0, 1] for which there is no f(k) = x. So f is not onto and therefore f is not a bijection. Therefore, no such bijection can exist — by contradiction. - - - You can’t patch that schema up just by adding counter examples to your first choice of bijection — because a counter example is constructable for any bijection attempt. The fundamental problem is that reals include limits — which is what allows the sum to be a real number.
- xscott 2y ago
  I won't claim to have any proof in hand, but there's still something fishy about it all. I think things like Banach-Tarsky and such ought to be treated as counter examples to show the absurdity of the Reals. At times I've thought maybe the problem with diagonalization is forcing me to do a (countably) infinite number of steps before Cantor gets to take an infinite number of steps. For instance, I can give a trivial counting scheme and claim it will eventually generate every number between zero and one, perhaps by writing a program to write programs that generate digits. It's all very deterministic, and I state exactly what I'm doing up front. So I give the first number (program), and Cantor gives his first digit and says it's not in my list. Then I tell Cantor the countably infinite number of places where his first digit is found, and we select those programs. Then he gives me his second digit, and I select the subset where those are found. So we go back and forth, he keeps saying his number so far is not in my list, and I keep telling him all the places where it is in the list... Feels like stalemate there. There are other places (such as summing the alternating harmonic series) where you have to be very careful about the ordering of operations. If he really has a higher order infinity of numbers not in my list, maybe I should be able to ask him for his list first, no? It's uncountably infinite, but he can't provide even one of them.
  - zmgsabst 2y ago
    You don’t have to do anything but assume such a bijection does exist, as in my comment: its existence leads to a contradiction in the presence of limits — just like I showed. But without that same limiting to build sets, you can’t construct N from the successor function and you can’t discuss the power set of N. So we either need to throw out countable limiting to build sets or accept that such set building rules out bijections between reals and naturals. You think Banach-Taraki is weird (as do most people), but I’d argue a world where you have finite numbers or couldn’t discuss the set of subsets of the naturals is even weirder. > maybe I should be able to ask him for his list first, no? There’s no such list indexable by the naturals, as shown by that contradiction. If you allow indexing by the reals, that request is trivial.
    - xscott 2y ago