Comment by btilly

Comment by btilly 5 days ago

It's just a straight up liar's paradox. If enumerate is a function that works as advertised, then unenumerated is as well. If enumerate tries to list unenumerated, then enumerate can't work as advertised.

For the argument that I gave to work, you need what you might call Axiom ComposeFunc, you may always compose a new function by taking an existing function and doing something that is known to be valid with it. Obviously this axiom is true about the computable universe. But, more than that, it is also true about any generalization of "function" that behaves like most would expect a function to have.

Now it is true that your Axiom CompFunc and Axiom CountReals do not necessarily contradict each other. But CompFunc, ComposeFunc and CountReals do contradict each other, and the contradiction can be built by following Cantor's diagonalization argument.

skissane 5 days ago

> But, more than that, it is also true about any generalization of "function" that behaves like most would expect a function to have.

It isn’t true in NFU though, correct? At least not in the general case. Because Cantor’s argument fails in NFU

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btilly 5 days ago

If I understand what I just Googled correctly (definitely not guaranteed), the reason why Cantor's argument can fail in NFU is that NFU does not necessarily allow you to build a function that returns things in X, out of a function that returns functions returning things in X.
So it has non-computable functions, but also has a type system that tries to avoid allowing self-reference. And that type system gets in the way of Cantor's argument.
I clearly hadn't thought of this possibility. It is a constraint of a kind that doesn't show up in the computable universe.
And so, going back, if the the Russian constructivist school does indeed center on computability, then my first answer to you was correct.

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- skissane 5 days ago
  
  Right, I think we are in agreement - a pure Russian constructivist approach which only permits computable functions cannot prove the reals are countable. However, I still am sceptical it can prove they are uncountable-if you limit yourself to computable constructions, you can’t actually computably construct a Cantor diagonal, so his argument fails just like it does in NFU.
  The (un)countability of the reals is known to be independent of NFU-it is consistent both with the reals being countable and them being uncountable. There are two different axioms which it is standard to add to NFU to decide this-AxCount≤ which implies the reals are countable and AxCount≥ which implies the reals are uncountable.
  I guess I was suggesting that in the same way, an additional axiom could be added to computable set theory which renders the reals countable or uncountable. If an additional axiom asserting the countability of the reals involves the existence of a function from the naturals to functions over the naturals, that would obviously be introducing an uncomputable function-but for that to produce an inconsistency, it would need to enable Cantor’s argument-and, given your “ComposeFunc” in the computable universe is already restricted to only operating over computable functions, it is reasonable to limit its application to computable functions in an extension, which would mean the addition of this uncomputable function would still not permit Cantor’s argument
  
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  gylterud 5 days ago
  
  There is nothing uncomputable with the cantor diagonalisation. The Russians gladly accept it, I assure you. Here is a Haskell implementation:
  diag :: (nat -> (nat -> Bool)) -> (nat -> Bool)
  diag f n = not (f n n)
  The following argument is also constructive. Just like in classical mathematics, crustructive mathematics proves the negation of A by assuming A and deriving a contradiction. (But you cannot prove A by assuming it’s negation and deriving a contraction):
  Assume a surjectiom f :: nat -> Bool. Then there is x such that f x = diag f. Since these two functions are equal, they take equal values when we evaluate them in any point. In particular f x x = diag f x, but since diag f x = not (f x x), by definition, we have that f x x = not (f x x). This is a contradiction since not does not have fixed points.
  ( I made nat a type variable here since this works for any type / set )
  
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