We know, from the great Cantor's work, the cardinality of the set of Even numbers = cardinality of the set of Whole numbers. A part = The whole. — TheMadFool

It's more like f(part) = f(whole). The sets aren't equal. They are just have the same cardinality. Like Jefferson and Washington were both presidents.

What's 1 ÷ infinity? If it's 0 then infinity × 0 = 1?? — TheMadFool

Consider that there are lots of systems in math. (Check out an abstract algebra textbook. ) It's possible to extend R with positive and negative infinity, but a few nice features must be sacrificed. I think it's uncontroversial that we can't just add the sideways 8 to ordinary arithmetic.

In this way we can achieve arbitrary precision (infinite) on the value of the sqrt(2). — TheMadFool

That's a fascinating issue right there. Arbitrary precision as infinite...as opposed to the idea of a completed infinity. You wrote N = {0,1,2,...}. So we kinda know what that means, but do 'all' of the elements exist somewhere? What exactly do we mean by those ellipses? Obviously mathematicians can crank out papers and keep the tradition alive. Nuns can smack children's hands with rulers for putting apples in N. Professors can mock confused grad students at a higher level for more complicated 'misunderstandings' (errors in manners).

In other words: social 'games' within social games, wherein we discover what we can and can't get away with saying, without ever necessarily knowing just what we mean. Semantically we're in the fog, whatever that's supposed to mean. 'This is how it's done around here.' That's the spongy bedrock. (That's one vision of Socrates, a guy trying to show people that they don't exactly know what they are talking about, who found himself fatally unpopular, having written no books.)

Thank you for engaging. I've been shut down for maintenance. Check back later!

The quantum mechanics describes reality using quantized elements so it doesn’t lead to singularities or infinities afaik. But the theory of gravity does (general relativity). However, it is expected that spacetime is quantized as well.
On the other hand a simulation could be made in wich aparent infinities would exist because you only need to simulate the brains. More over it is likely we don’t have free will so it’s funny that the Universe may not necessarily be a simulation but a mere playback of brain activity. In my opinion it’s not possible to create or simulate free will other than an illusion

I found this on wikipedia:

While the set of real numbers is uncountable, the set of computable numbers is classically countable and thus almost all real numbers are not computable
— wikipedia — TheMadFool

I have a point that might be of interest to you.

Wiki is correct that if you take the standard rules of math, you can logically deduce the existence of noncomputable numbers. You've quoted an outline of the proof. Cantor's diagonal argument shows that the reals are uncountable; but we can show that there are only countably many Turing machines. So there must be real numbers whose digits can not be cranked out by a Turing machine. QED.

But what is a logical deduction? You start from some axioms, which are strings of symbols. You accept some rules of inference. You mechanically apply the rules of inference to your axioms, and conclude the result. The entire process of deduction could be carried out by a computer. You input the axioms as strings of symbols; you program in the rules of inference; and the computer can determine whether or not that's a valid proof.

So that's the thing. The proof that noncomputable real numbers exist, is itself a computation. In general, proofs are computations. Symbol manipulation according to formal rules.

In other words: A computational intelligence would eventually prove to itself that noncomputable real numbers exist given the rules of standard math.

What do you think about that?

If any of a number of articles (e.g. https://medium.com/cantors-paradise/uncomputable-numbers-ee528830d295) on the Internet are correct, then it is not the case that there does not exist a definable uncomputable real number. I am not versed in all the details, but it does seem to be settled mathematics.

(By 'there exists a definable uncomputable real number' in this context, I mean the ordinary mathematical sense:

There exists a formula F such that

E!x(Fx & x is an uncomputable real number)

is a theorem of set theory.)

Using 'infinity' as a noun in the context of cardinality is incorrect and supposed refutations of set theoretic notions of cardinality by using 'infinity' as a noun are fallacious.

In the context of set theoretic cardinality, there is no object named 'infinity'. Rather, there is the adjective 'is infinite'. So expressions like '1/infinity', et. al are meaningless, and supposed arguments based on such usage are fallacious.

(This does not contradict that there are things such as the extended real number system in which there are points of positive infinity and negative infinity and arithmetic on them, since that is a different context from set theoretic cardinality.)

Moreover, set theory does not assert that there is a set that is equal to a proper subset of itself. Indeed, it is a trivial theorem that there is no set S such that S is equal to a proper subset of S. That does not contradict however that there are sets that have a 1-1 function from the set onto a proper subset of itself. Having a 1-1 function between sets (of which we say 'the sets have a bijection' or 'the sets are equinumerous') is plainly distinct from the fact that the sets are not equal, not identical, not having all the properties of each other.

Suppose the following is the complete list of computable irrational numbers between e and pi — TheMadFool

Right from the start of your argument, the merely ostensive (and not specified by actual mathematical description) list you gave is either actually not defined or it's finite. It ends with a certain value, yet doesn't specify how to squeeze in a countably infinite number of values in between the first and the last.

There are proofs that there exist definable uncomputable reals, but you haven't given such a proof.

it is not the case that there does not exist a definable uncomputable real number. — GrandMinnow

That's a lot of negations, but for clarity, there does exist a definable yet noncomputable real, namely Chaitin's Omega. I haven't mentioned it because I didn't want to further complicate the conversation.

https://en.wikipedia.org/wiki/Chaitin%27s_constant

Suppose the following is the complete list of computable irrational numbers between e and pi
— TheMadFool
Right from the start of your argument, the merely ostensive (and not specified by actual mathematical description) list you gave is either actually not definable or it's finite. — GrandMinnow

No that's not right. There are a countably infinity of Turing machines hence a countable infinity of computable numbers, hence a bijection between the natural numbers and the noncomputable numbers.

As I indicated earlier, there is no computable bijection; but there is a bijection.

Definability is a much more subtle concept and is best left out of the discussion. For one thing, first-order definability is not itself formally first-order definable. There are models in which everything is definable. Joel David Hamkins has written on this topic. It's not relevant here and would take us far beyond the topic.

The negative statement was deliberately chosen. Of course, it is equivalent to saying that there does exist a definable uncomputable real.

The negative statement was deliberately chosen. — GrandMinnow

I only meant to add clarity. It's a small point. The OP has disappeared and I was hoping he'd return long enough to comment on my observation that the proof of the existence of a noncomputable real is itself computable. That is, proofs are computations, hence a computational intelligence would be able to prove the existence of noncomputable phenomena, given the standard axioms and inference rules of math.

the merely ostensive (and not specified by actual mathematical description) list you gave is either actually not definable or it's finite. — GrandMinnow

I believe your assertion that there is no definable list of noncomputable reals may well be true but I'm not sure that the proof would be at all elementary. Do you happen to have such a proof? There's surely no computable list, since that would solve the Halting problem. But why not a definable list? I don't know if what you said is true, but I do know that it's not trivial either way. Definability is very murky as Hamkins points out. See his brilliant response in this thread.

https://mathoverflow.net/questions/44102/is-the-analysis-as-taught-in-universities-in-fact-the-analysis-of-definable-numb

your assertion that there is no definable list of noncomputable reals — fishfry

I didn't make that assertion. I only commented on the poster's particular argument. (I later edited 'definable' to 'defined', but in either case, my comment pertained only to the poster's own argument.)

Anyway, there is no list (no enumeration) of the the set of uncomputable reals, since the set of uncomputable reals is uncountable.

/

That thread you linked to includes an argument that uses 'least undefinable ordinal' to throw shade on the "naive" notion of definability. But one would not claim that 'definabie' itself is a predicate in the theory. I only mentioned a certain syntactical fact - regarding formulas of a certain form. I wouldn't say in set theory itself, about set theory, that there exists a definable something or other. To even speak of that "something' is to speak of an object that exists per a set theoretic model, but indeed, as we well know, set theory (if it is consistent, which I take as a "background" assumption) does not prove the existence of a model of set theory. (Though, I'm not expert enough to defend against possible other complications in the matter.)

I changed this post greatly:

So to make the argument work that there are only countably many definable real numbers, maybe something like this in a set theoretic meta-theory for set theory:

Let 'Rx' be the set theory formula 'x is a real number'. Let M be any model of set theory such that any subset S of the universe of M satisfies 's is countable' if and only if S is countable. Let D (the set of definable reals) be the subset of the universe of M by D = {d | there exists a formula F of set theory such that (E!x(Fx & Rx) is a theorem of set theory & d satisfies Fx and d satisfies Rx). Then D is countable.

So to make the countability argument work, maybe something like this in a set theoretic meta-theory for set theory: — GrandMinnow

Duty calls for other stuff this evening, will get back to you later tonight or tomorrow to continue this interesting convo. But who on earth ever suggested that the set of noncomputable reals was anything other than uncountable? Surely not me and not the OP as far as I could tell. The claims (mine anyway) are:

* The set of computable reals is countable.
* The set of computable reals is NOT computably countable; and
* The question of whether the set of computable reals is definable is Hamkins-murky and way beyond my pay grade. Though if I understand Hamkins correctly, there are models of the reals in which everything is definable. That's why it's murky.

I don't suggest that you suggested that the set of noncomputable reals is countable.

Meanwhile, I changed my post above to pertain to definability rather than computability, It's not a problem to show in set theory that the set of computable reals is countable, but there are issues in showing the countability of the set of definable reals, as mentioned in the article you linked to. I tried to sketch a sense in which the notion of definability can be used without such contradictions as 'the least undefinable ordinal' that was mentioned in that article.

No that's not right.There are a countably infinity of Turing machines hence a countable infinity of computable numbers, hence a bijection between the natural numbers and the noncomputable numbers. — fishfry

You mean there is a bijection between the naturals and the computable reals. And I didn't claim otherwise. I only pointed out the incoherence of a particular poster's argument.

There are a countably infinity of Turing machines hence a countable infinity of computable numbers, hence a bijection between the natural numbers and the noncomputable numbers. — fishfry

Sorry. Am I missing something here?

I think he made a typo and actually meant 'between the natural numbers and the computable numbers'.

I think he made a typo and actually meant 'between the natural numbers and the computable numbers'. — GrandMinnow

Yes sorry typo.

Okay. In preparation for this response I read the introductory material in Hamkins's paper, Pointwise Definable Models in Set Theory. This is his formal paper on the material he discussed informally in the Mathoverflow link I posted earlier. Hamkins is a particularly lucid and engaging writer and this material is accessible even to nonspecialists.

Hamkins shows that (assuming set theory is consistent) there is a model of set theory -- uncountably many models, in fact -- in which everything is definable. Every element, every set, every collection of sets, etc. In particular, every real number in this model is definable. Of course if set theory is not consistent it has no models at all, so it's necessary to assume consistency to get the discussion off the ground.

Now such a model must necessarily be countable, because there are after all only countably many possible definitions or predicates. That's easy to see, there are only countably many finite strings over a countable alphabet.

But, in such a model, the real numbers are still uncountable! That's because Cantor's theorem is a theorem of ZF, and is therefore true in any model of ZF. Cantor's theorem says that the reals, being bijectively equivalent to the powerset of the natural numbers, are uncountable.

So what's going on? The trick is the famous Lowenheim-Skolem theorem, which says that if a collection of axioms has an infinite model, it has models of all cardinalities. In particular there are countable models of set theory. Yet in those models, the real numbers are still uncountable. How does that work?

What does it mean for a set to be uncountable? It means there is no bijection between the set and the natural numbers.

Conceptually here is what's going on. Suppose that we have some countable set that is a model of set theory. Remember that a bijection is just a special kind of function; and that a function is just a particular kind of set (namely a set of ordered pairs in which each first element only appears once). So what we can do is go into the model, and remove all the bijections between the natural numbers and our set of interest.

If we are careful to remove bijections in such a way that the remaining elements still form a model of set theory, we'll have a countable universe that contains a set that is uncountable. In other words there are no bijections between the set and the natural numbers, because we've carefully removed them. But the universe is still a countable set. Of course the "carefulness" involved in throwing out the bijections while still satisfying the axioms of set theory can be made rigorous.

As seen from the outside, our uncountable set is actually countable. But as seen from inside the model, it's uncountable, because we've removed the bijections. This shows that the notion of countability is a "relative" property. It depends on whether you're looking at it from inside or from outside a given model.

See https://en.wikipedia.org/wiki/Absoluteness

and in particular, https://en.wikipedia.org/wiki/Absoluteness#Failure_of_absoluteness_for_countability

So in a model where everything is definable, that model is must necessarily be countable as seen from the outside. But it does contain the real numbers (which exist in any model of set theory); and by Cantor's theorem, the real numbers are uncountable (as seen from within the model) yet all real numbers are definable.

I mention all this to provide context and hopefully some clarity to my following comments.

That thread you linked to includes an argument that uses 'least undefinable ordinal' to throw shade on the "naive" notion of definability. — GrandMinnow

By "throwing shade" you mean "flat out falsifying," in the same way that Russell's paradox "throws shade" on Frege's notion of unrestricted set formation, right? In other words Hamkins is not just casting vague innuendo. He's providing a technical argument that falsifies a commonly held belief. Just to be clear about this.

But one would not claim that 'definabie' itself is a predicate in the theory. — GrandMinnow

Correct, that one of Hamkins's points.

I only mentioned a certain syntactical fact - regarding formulas of a certain form. I wouldn't say in set theory itself, about set theory, that there exists a definable something or other. To even speak of that "something' is to speak of an object that exists per a set theoretic model, — GrandMinnow

I'm afraid I couldn't quite glean the meaning of this.

but indeed, as we well know, set theory (if it is consistent, which I take as a "background" assumption) does not prove the existence of a model of set theory. (Though, I'm not expert enough to defend against possible other complications in the matter.) — GrandMinnow

No, that's not right. Set theory is consistent if and only if there's a model. That's Gödel's completeness theorem. If it's consistent there's a model, and if there's a model it's consistent.

I changed this post greatly:
So to make the argument work that there are only countably many definable real numbers, maybe something like this in a set theoretic meta-theory for set theory: — GrandMinnow

Remember, there are only countably many real numbers as seen from outside a Hamkins model; but within the model there are uncountably many reals (by Cantor's theorem) and they're all definable. Tricky stuff.

Let 'Rx' be the set theory formula 'x is a real number'. Let M be any model of set theory such that any subset S of the universe of M satisfies 's is countable' if and only if S is countable. Let D (the set of definable reals) be the subset of the universe of M by D = {d | there exists a formula F of set theory such that (E!x(Fx & Rx) is a theorem of set theory & d satisfies Fx and d satisfies Rx). Then D is countable. — GrandMinnow

There is a model of set theory in which everything is definable. In particular, each real number is definable. In such a model the real numbers are countable as seen from outside the model; but they are uncountable as seen from inside the model. You are using the word countable without regard for the fact that countability is a relative property. I think your argument is ambiguous because of that. In any Hamkins model, the reals are uncountable (inside the model) yet they're all definable. And that's because they're "really" countable as seen from outside the model.

tl;dr: There is a model of set theory in which all real numbers are definable. Such a model is actually countable, as seen from the outside; nevertheless within the model, Cantor's theorem is true and the reals are uncountable.

set theory (if it is consistent [...]) does not prove the existence of a model of set theory.
— GrandMinnow

No, that's not right. Set theory is consistent if and only if there's a model. That's Gödel's completeness theorem. — fishfry

No, it is right. Yes, set theory is consistent if and only if there is a model of set theory. But if set theory is consistent then set theory itself doesn't prove that it has a model;. That's Godel's 2nd Incompleteness Theorem.

No, it is right. Yes, set theory is consistent if and only if there is a model of set theory. But if set theory is consistent then set theory itself doesn't prove that it has a model;. That's Godel's 2nd Incompleteness Theorem. — GrandMinnow

Oh I see, you're right. My parser got confused.