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Comment by skissane | Hacker News Reader

> So to conclude that there are more reals than naturals, the classical mathematical argument is:

> a) There are more functions N to Bool than naturals.

> b) There are as many reals as functions from N to Bool.

> Now, we of course agree the mistake is in b) not in a).

Given certain foundations, (a) is false. For example, in the Russian constructivist school (as in Andrey Markov Jr), functions only exist if they are computable, and there are only countably many computable functions from N to Bool. More generally, viewing functions as sets, if you sufficiently restrict the axiom schema of separation/specification, then only countably many sets encoding functions N-to-Bool exist, rendering (a) false

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Indeed, what you write is true from an external point of view; just note that within this flavor of constructive mathematics, the set of functions from N to Bool is uncountable again.

There is no paradox: Externally, there is an enumeration of all computable functions N -> Bool, but no such enumeration is computable.

Is it internally uncountable in the strong sense that the system can actually prove the theorem “this set is uncountable”, or only in the weaker sense that it can’t prove the theorem “this set is countable”, but can’t prove its negation either?

If the latter, what happens if you add to it the (admittedly non-constructive) axiom that the set in question is countable?

It should be true in the stronger sense.

Suppose that you've written down a function enumerate, that maps all natural numbers to functions that themselves map all natural numbers to booleans. We then can write down the following program.

    (defn unenumerated (n) (not ((enumerate n) n)))

This function can be recognized as Cantor diagonalization, written out as a program.

If enumerate actually works as advertised, then it can't ever find unenumerated. Because if (enumerate N) is unenumerated, then ((enumerate N) N) will hit infinite recursion and therefore doesn't return a boolean.

This argument is, of course, just Cantor's diagonalization argument. From an enumeration, it produces something that can't be in that enumeration.

> If enumerate actually works as advertised, then it can't ever find unenumerated. Because if (enumerate N) is unenumerated, then ((enumerate N) N) will hit infinite recursion and therefore doesn't return a boolean.

Okay, but if we take the position that only non-halting (for all inputs) programs represent functions over N, if your program “unenumerated” never halts for some N, it doesn’t represent a function, so your argument isn’t a case of “using the enumeration of all functions to produce something which doesn’t belong to the enumeration”

Obviously an enumeration of all computable functions isn’t itself computable. But if we consider Axiom CompFunc “if a function over N is computable then it exists” (so this axiom guarantees the existence of all computable functions, but is silent on whether any non-computable functions exist) and then we consider the additional Axiom CountReals “there exists a function from N to all functions over N”, then are those two axioms mutually inconsistent? I don’t think your argument, as given, directly addresses this question

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.

> 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

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.

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

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 )

Bringing in computability from an external point of view is a mistake here. Markov had no issue with a. He would only disagree with b.

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