|R| = |NxN|

[wtf]

On 2nd thought this is impossible to be true in linear logic.

If set A is countable (e.g in 1:1 bijection with N) then mapping A to N fully exhausts N.

There is no "room" left. You can't exhaust N again. |N| + |N| > |N|

Alternatively, if A ∪ B is countable (assuming A and B aren't the empty set) then neither A nor B are countable.

There simply isn't such a thing as "resource exhaustion" in a classical setting.
If I have 1 apple and I give you 2 - how many apples do I have left?

There exists no such operation.

The even naturals are countable, as are the odd naturals, and their union the natural numbers is countable.

In the general case, say that

a1, a2, a3, ... and b1, b2, b3, ... are countable sets. Then we enumerate their union as

a1, b1, a2, b2, a3, b3, ...

Countable plus countable is countable.

[wtf]

I object to completed infinities. Potential infinities - have at them. They are just Turing Machines which will never terminate.

On the whole here, you utterly failed to engage (or even read or understand) a word I said. You tossed out a pile of irrelevant jargon that does not apply to the situation at hand.

And with regard to your statement that you object to compleed infinities, in your OP you had no problem iterating an infinitary process to exhaust all the real numbers. Those two things are inconsistent with each other. I can't imagine what you're thinking.

They are always approaching - never arriving. Limits...

Yet you managed to exhaust all the real numbers via an infinitary process that you didn't bother to cleardly define.

I mean the exact same thing you mean when you apply the SUCC operator inductively/recursively to define the Naturals e.g

1. 0 is a Natural number
2. If n is a Natural number, then n + 1 is a Natural number

The number of times you need to call/apply the SUCC operator in order to define all elements of N is identical to the cardinality of N.

The logical implication of this is that the process of defining N will not halt.

You don't even know what halt means. If you give me a natural number, like 47 or googolplex, I can give you a TM that halts on that number after finitely many steps. You clearly don't even understand what it means for a TM to halt.

1/3 = .3333... is a computable number because there's a TM that, given n, halts and outputs the n-th digit. "Print 3" will do as such a TM. A TM is not required to print all the digits at once. You clearly don't know what you are talking about. I'm doing my best to explain it to you, but I doubt you'll take this to heart.

To be contrasted with a co-inductive/co-recursive (top-down?) definition:

Code: Select all

data CoNat : Set where
  zero : CoNat
  succ : ∞ CoNat → CoNat

I looked up coinduction and it does not have the remotest relation to the process you talked about in the OP, magically enumerating the reals by endlessly appending the antidiagonal to the list.

That's a different type-signature. I'm sure you don't need me to explain the difference between universal algebras; and universal coalgebras to you?

Sure you do. Go ahead and explain it to me. And then explain the relevance to your process of generating all the reals.

The curx is that if I define the naturals that way I get to declare some undefined element n

n : CoNat
n = succ (♯ n)

And then I get to play with symbolic AND numeric representations in a single notation e.g "{0, 1, 2, ..., n-2, n-1, n, n+1, n+2, ...}"

Which has f*ckall to do with anything.

Surely you differentiate between numbers with a definite value e.g 1,2,3; and objects with indefinite value e.g n-2, n-1, n, n+1, n+2.

You're just flailing, man. You didn't understand a word I wrote. You don't understand Cantor diagonalization. You said you had some infinitary process to generate all the reals, but when challenged, you retreat to irrelevant jargon.

As a way to communicate informally - sure. Can we agree this is an inductive/recursive but NOT a co-inductive/co-recursive definition?

Why don't you tell me in your own words what a co-inductive definition is, and what relevance it has to anything I wrote or to your own OP.

Can we agree that a co-inductive definition of the above would be something like:

e.g every Real number the anti-diagonal to an incomplete enumeration of R.

I can't agree till you explain what you mean.

Yeah, but that's just a vacuous tautology. Countable (by definition) means "1:1 bijection with the set of naturals".
You are already pre-supposing a completed infinity - the set of naturals.

You deny the set of naturals yet you claim to generate the reals. How's that work?

All I am presupposing is a recursive/co-recursive process which defines Nat such as 0, 1,2...; or a CoNat such as inf, inf+1, inf+2...
Sure - such objects can be arbitrarily large. The can be potentially infinite, but they are not completed infinities.

Gobbledygook you don't even understand. Feel free to explain to me what you are talking about.

It's not impossible for this to be true, but I see no reasont why YOU believe it's true. In your world-view "countable" means "There exists 1:1 bijection with N"

You appear to be ignorant of the most basic facts of higher math.

If you claim (by definition!) that L is countable then there is a 1:1 bijection between L and N BY DEFINITION.
ALL elements in L are mapped to ALL elements in N.
Theorem 1: There are NO elements in N which are NOT mapped to elements in L.

What is L here? The only L I know is Godel's constructible universe. Is that what you mean? Or something else?

If you add another countable set S to L - what are the elements in S being mapped to in N ?!? There are no unmapped elements remaining in N!
And if |L| is order-isomorphic to |N| then it's impossible for |L ∪ S| to be order-isomorphic to |N|, unless S is an empty list.

Stop. Just stop. If I have all the even numbers, that's a countable set. If I toss in 47, it's still countable, just via a different bijection. You need to go learn basic set theory.

Sure, there MAY exist some other bijection, between L ∪ S and N. But it's absolutely not the same bijection as the one defined into existence.

It doesn't need to be the same. What gives you the idea that it does?

And you are merely asserting the existence of such a bijection without constructing it.

Pretty easy to do.

Non-sequitur. Using the exact same logic there cannot be a complete list of all the naturals either.

The identity map on the naturals is a bijection showing the naturals are countable.

I am not trying to defeat it. I am exploiting Cantor's argument and applying it to the Naturals.
I am constructively proving the statement NO enumeration of the Naturals is countably infinite.

Well that's just absurd. Every enumeration of the naturals is countably infinite. In fact the existence of ANY enumeration of the naturals shows by definition that the naturals are countably infinite.

Are you just a troll after all this? Do you take yourself seriously, denying the most obvious mathematical truths and waving your hands at half-understood jargon you picked up on Wikipedia?

Any recursive/corecursive enumeration of N has necessarily fewer elements than N thus impossible to be a bijection.
Because there are no such things as completed infinities.

Ok whatever. I don't think further conversation is helpful here.

I can also define CoInfinity

Code: Select all

data CoInfinity : Set where
  infinity : CoInfinity
  succ : ∞ CoInfinity → CoInfinity

-- Base infinity (analogous to ω)
ω : CoInfinity
ω = infinity

-- Successor to infinity (analogous to ω + 1)
ω+1 : CoInfinity
ω+1 = succ (♯ ω)

And I can give you the union of Nat, CoNat, CoInfinity

Code: Select all

data ExtendedNumber : Set where
  finite : ℕ → ExtendedNumber
  conat : ∞ ExtendedNumber → ExtendedNumber
  infinity : ExtendedNumber
  succ-infinity : ∞ ExtendedNumber → ExtendedNumber

Knock yourself out.

Why do you belive that an object which is countably infinite (1:1 bijection with N) ∪ another object which is countably infinite produces another countably infinite object?

Because I can prove it.

If a1, a2, a3, ... is an enumeration of one set; and

b1, b2, b3, ... is an enumeration of another set; then by interleaving the terms,

a1, b1, a2, b2, a3, b3, ...

is an enumeration of the union of the two sets.

As a concrete example, 0, 2, 4, ... enumerates the evens; 1, 2, 3, ... enumerates the odds; and 0, 1, 2, 3, ... enumerates the naturals.

You really need to understand this.

Sure. And if we don't assume the Axiom of Choice? n

Then the cardinality of the reals might not be an Aleph number, and it is not cardinally equivalent to any ordinal.

There appears to be a bijection between the Ordinals, Naturals and Transfinite Cardinals because all three are defined in identical manner:
initial element + successor function.

0:Nat <-> Inf:CoNat <-> ω:CoInfinity
succ 0 <-> succ (inf) <-> succ(ω)

The ordinals are a proper class. Skepdick my friend, please go learn some basic set theory. You claim a bijection between the naturals and the class of ordinals? How can I take you seriously? Are you a troll or a fool? Or just so egotistical that you refuse to learn anything?

By the way, what happened to the limit ordinals? Did you know there are uncountable ordinals? Did you know there's a Wiki article on ordinals? Go read it.

https://en.wikipedia.org/wiki/Ordinal_number

There is no bijection on a co-recursive definition because Nat is embedded in CoNat, and CoNat is embedded in CoInfinity.

Why don't you explain these concepts so I have some idea what you are talking about? Since you just claimed the natural numbers and the class of ordinals are in bijective correspondence, I already know that you have NO IDEA what you're talking about. Do you have ANY idea how ignorant that statement of yours is?

Yeah but the ordinal of the naturals is also uncountable. Again - by Cantor's argument.

When you don't understand something, you retreat to utter bullshit instead of trying to learn. What you just wrote is bullshit. The ordinal of the naturals in their usual order is ω, the first transfinite ordinal. It's countable. It's in bijective correspondence to the naturals. In fact, it IS the set of naturals in their usual order.

This applies equally to the Natural numbers.

Yeah yeah. In other words, no.

Even though we can order-enumerate all the naturals we can not do this using a LIST. Because a transfinite ordinal (such as infinity:CoNat) is NOT a list.

Troll or fool. Or relatively smart guy with an ego that doesn't allow you to try to engage with thngs you don't understand.

This holds for Natural numbers also - on a co-inductive/co-recursive definition. You can generate more and more naturals using anti-diagonals too. GIven the right encoding/representation.

There are no infinitely long naturals, which is why you can't diagonalize a list of naturals.

This holds for N also on a co-inductive definition.

Feel free to explain this in your own words so I don't think you're just typing in buzzwords you don't understand.

You are continuously equivocating "countably many" and "uncountably many"

I'm doing no such thing. You haven't understood a word I've written, and your ego won't allow you to engage in learning mode.

I am applying succ to 0 an arbitrarily large number of times. I call this "infinity" in CoNat.
I am applying succ to Infinity an arbitrarily large number of times. I call this ω in CoInfinity

You have to explain what "arbitarily large" means. I gave you the vocabulary of ordinals. That brings precision. Your idea is vague and ill-defined, that's why you're confusing yourself.

Like I said - I have no problem with "infinitary processes". I have a problem with COMPLETED infinitary processes.

Whatever. You wore me out with this one. You started a new thread to deny Cantor or to enumerate the reals or to talk about coinfinity, as if you have the vaguest notion of what that might mean. You've had your fun and you've trolled me. You win. If you say something reasonable I might engage. If I don't engage, you can be happy that you trolled me. Again.

An infinitary process which "eventually enumerates all Naturals" is nonsense.
For such a process would need to write down 1, and 2; and 3.... in sequence. On infinite tape wihch has a cardinality equivalent to N.

You are so wrong it's embarrassing. Not only for your finitism, which would be fine if you actually understood it. But because you don't understand what it means for a TM to halt.

When will such a process complete?
If it hasn't yet completed - then it's not yet a 1:1 bijection with N.

My process of letting you troll me has completed.

[wtf]

ps -- I wanted to call this out separately.

I have never heard of "coinfinite." I looked it up. The ONLY reference I found was on Mathoverflow, a site for professional mathematicians, who had never heard of it either.

Their best guess was that the word is a misspelling of cofinite, which is a set whose complement is finite.

So I call BS.

https://mathoverflow.net/questions/3873 ... inite-mean

[Skepdick]

The even naturals are countable, as are the odd naturals, and their union the natural numbers is countable.

In the general case, say that

a1, a2, a3, ... and b1, b2, b3, ... are countable sets. Then we enumerate their union as

a1, b1, a2, b2, a3, b3, ...

Countable plus countable is countable.

Once again. That's not impossible to be true, but you have no reason to believe it's true.

The proof of countability for the Evens produces the enumeration (0,0), (1, 2), (2, 4).... e.g it's the map f(x) = 2x
and exhausts all of N.

The natural number 0 maps to the even number 0, the natural number 1 maps to even number 2, the natural number 2 maps to the even number 4 ....

Given that the even numbers are countable what does the odd number 1 map to in the Naturals?

Where is your proof? Where is your bijection from (2x) ∪ (2x + 1) to (x) ?

a1, a2, a3, ... and b1, b2, b3, ... are countable sets. Then we enumerate their union as

a1, b1, a2, b2, a3, b3, ...

That's a really strange enumeration. I would've thought the union produces a1, a2, a3, ... ∪ b1, b2, b3, ...

[wtf]

That's a really strange enumeration. I would've thought the union produces a1, a2, a3, ... ∪ b1, b2, b3, ...

You agree it's an enumeration. One is all I need. I can of course reorder the set {a1, a2, a3, ..., b1, b2, b3, ...} as shown. Sets have no inherent order, and I can reorder them any way I find convenient.

And it's only strange the first time you see it. Interleaving is a standard technique. You can use the same idea to show that the real line and the x-y plane are in bijection, by interleaving the bits of an ordered pair of reals.

Once again. That's not impossible to be true, but you have no reason to believe it's true.

It's not a belief. I've proven it for you twice.Three times now including this post.

The proof of countability for the Evens produces the enumeration (0,0), (1, 2), (2, 4).... e.g it's the map f(x) = 2x
and exhausts all of N.

The natural number 0 maps to the even number 0, the natural number 1 maps to even number 2, the natural number 2 maps to the even number 4 ....

Given that the even numbers are countable what does the odd number 1 map to in the Naturals?

Where is your proof? Where is your bijection from (2x) ∪ (2x + 1) to (x) ?

You use a different bijection. Lets say the naturals are countable:

1 <-> 1

2 <-> 2

3 <-> 3

etc

Now I add an extra element, say pi, to the naturals. Do I still have a bijection? Of course:

1 <-> pi

2 <-> 1

3 <-> 2

etc.

I don't need the same bijection every time

In your example I can enumerate the even numbers unioned with {1} as

1, 0, 2, 4, 6, 8, 10, ...

In other words

0 <-> 1

1 <-> 0

2 <-> 2

3 <-> 4

etc.

If I have a bijection from the naturals to a1, a2, a3, ... then I also have a bijection to

x, a1, a2, a3, ...

where x is any arbitrary new element.

Haven't you seen the Hilbert hotel story? You can always "slide each guest one room up" to make an extra empty room.

You can always stick any finite number of new elements in front of a list and it's still a countable list.

And you can insert countably many new elements by interleaving them:

x1, a1, x2, a2, x3, a3, ...

Still a list. Still obviously bijected with the natural numbers.

Countable plus countable is countable. Evens unioned with odds are the naturals.

[Skepdick]

That's a really strange enumeration. I would've thought the union produces a1, a2, a3, ... ∪ b1, b2, b3, ...

You agree it's an enumeration. One is all I need.

Enumeration. Infinite datastream. Potato/potatoh.

I can of course reorder the set {a1, a2, a3, ..., b1, b2, b3, ...} as shown.

Are you doing this reordering before; or after you've finished enumerating not one but TWO consecutive infinite datastreams?

How many times are you going to use "..." in a set ?!?!? Why don't you just use "..." ... many times?!?!
1,...,2,...,3,...,4,...,

And also, I am super curious how you are going to be doing this "reordering" of supposedly "infinite" sets. Isn't that just a sorting algorithm?
Can you describe how you would seek to b1; and then where you plan on moving it to and how?

Sets have no inherent order, and I can reorder them any way I find convenient.

Eeeehhh. That's a cute sleight of hand. Are you now saying N has no ordering?!? Because throughout this chat you keep aluding to N with its usual order.. No matter - I am happy to work with that. Give me all the elements of N WITHOUT ORDERING.

Let me know when you are going to stop equivocating...

If you are going to assume an ordering - I expect an order-preserving map.
And if you are going to be ordering arbitrary sets without inherent order - I expexpect you to explain how you plan on doing that without the axiom of choice.

And it's only strange the first time you see it. Interleaving is a standard technique.

Yeah. It's called the ZIP function. It combines two (or more iterables) into an iterable.
It has implicit ordering. Interleaving A with B is not the same as interleaving B with A.

Left is not Right.

You can use the same idea to show that the real line and the x-y plane are in bijection, by interleaving the bits of an ordered pair of reals.

So order matters now? Until it doesn't matter. You seem to be hedging your bets.

It's not a belief. I've proven it for you twice.Three times now including this post.

None of your "proofs" were property or structure preserving.

You use a different bijection.

Ahhhh DIFFERENT bijection. Which one?

Now I add an extra element, say pi, to the naturals. Do I still have a bijection? Of course:

1 <-> pi

2 <-> 1

3 <-> 2

Contradiction. You are using sets. If pi is a natural number then pi is already in the original enumeration, so this is no longer a bijection.
It's now an injection from N into the multiset {pi, 1,2,3,...}.

I don't need the same bijection every time

I know that. Where's your new bijection?

In your example I can enumerate the even numbers unioned with {1} as

False equivalence. That's the union of a potential infinity (even numbers) with a finite subset of the odd numbers {1}.

When you union SOME odds (finite set) with ALL evens (potentially infinite set) you will get 1,3,5,7, 0,2,4,6,...
When you union ALL odds (potentially infinite set) with ALL evens (potentially infinite set) you will get 1,3,5,7,..., 0,2,4,6,...
In fact - the set you are going to get is {..., 7, 5, 3, 1, 0, 2, 4, 6, ... } because all you are doing is prepending new elements.

Also.. what a strange union you have there... In this union ANY even number > ANY odd number.

Haven't you seen the Hilbert hotel story? You can always "slide each guest one room up" to make an extra empty room.

Yeah, I have. It's an amazing feat of human self-deception. Given the set of even numbers {0,2,4,6,...} how many rooms must 0 move up, so that ALL Odd numbers can move in before it?

Countable plus countable is countable. Evens unioned with odds are the naturals.

0,2,4,8,..., 1,2,3,5,... = 1,2,3,... ?!?!?

ROFL.

And then there's the set difference... If Odd ∪ Even = N then it trivially follows that N \ Odd = Even.

But if |N \ Odd | = |Even| then Odd must be the empty set.

[wtf]

Enumeration. Infinite datastream. Potato/potatoh.

This is exactly the kind of thing you post that makes me think you're a troll, a fool, or can't even remember what we're talking about.

The subject at hand was your asking me why I "believe" that the union of countable sets is countable, or in terms of cardinality, that countable + countable = countable.

You doubt the proposition.

I proved it for you several times.

Now instead of saying, "Oh I see. You're right. And not only that, I can make the exact same proof work in the language of streams," you make this mindless non-sequitur that doesn't mean anything.

Why? Are you so oppositional that you can't even realize you've agreed with me? You're so obsessed with your monomania that you can't even converse rationally? What is your deal, brother Skepdick?

Are you doing this reordering before; or after you've finished enumerating not one but TWO consecutive infinite datastreams?

If you'd like to do this proof in the language of datastreams, you just interleave them. Are you seriously telling me you don't understand how one could interleave two datastreams? It's child's play, right? One from stream A, one from stream B. Another from stream A, another from stream B. etc.

And what this shows, interestingly, is that if you say you reject completed infinities, I can agree with you for sake of discussion and the proof goes through. That is, even if we reject the axiom of infinity, countable + countable is still countable. Given two datastreams we just interleave them.

How many times are you going to use "..." in a set ?!?!? Why don't you just use "..." ... many times?!?!
1,...,2,...,3,...,4,...,

Huh? What's between 1 and 2? Your notation doesn't designate anything sensible unless you explain it. I'll wait.

And also, I am super curious how you are going to be doing this "reordering" of supposedly "infinite" sets. Isn't that just a sorting algorithm?

I can reorder an infinite set any way I like. Here's a famous reordering of the natural numbers:

1, 2, 3, 4, ..., 0

We just define 0 to be greater than every other natural. This is an order type called ω + 1.

I'm sorry you missed out on elementary set theory, but being snarky about your own ignorance is a bad look.

Can you describe how you would seek to b1; and then where you plan on moving it to and how?

Are you seriously telling me you don't know how to interleave two streams? I don't believe you. I'm meeting you more than halfway by talking to you in the language of streams, but you seem ignorant even of your own framework.

Exercise: Write a program to interleave two streams.

Eeeehhh. That's a cute sleight of hand. Are you now saying N has no ordering?!? Because throughout this chat you keep aluding to N with its usual order.. No matter - I am happy to work with that. Give me all the elements of N WITHOUT ORDERING.

N has a natural ordering: 0, 1, 2, 3, ...

I showed you one alternate ordering: 1, 2, 3, 4, ..., 0

Here's another, called the "even-odd ordering":

0, 2, 4, 6, 8, ..., 1, 3, 5, 7, 9, ...

That's an order type called, for fairly obvious reaons, ω + ω.

There are many possible orderings on the natural numbers. We call the usual one, the usual one. An order is just a particular relation; in other words in the usual ordering, (3,5) is true and (5,3) is false. That's just another way of saying that 3 < 5 is true and 5 < 3 is false.

You can put all kinds of relations on a set. Some relations represent orders. You can read up on the subject here. Bottom line, given the set (that is, unordered collection) of natural numbers, we can put various order relations on them; one of which is called the "usual" order and is the one we learned in grade school.

https://en.wikipedia.org/wiki/Order_theory

Let me know when you are going to stop equivocating...

Let me know when you're going to stop snarking to cover up your ignorance.

If you are going to assume an ordering - I expect an order-preserving map.
And if you are going to be ordering arbitrary sets without inherent order - I expexpect you to explain how you plan on doing that without the axiom of choice.

You don't have the foggiest notion what the axiom of choice is if you think I need it to order a set. But the question you asked here doesn't actually make any sense. Perhaps if you have something in mind you can pose your question more clearly. As stated it's not a well-formed question.

Yeah. It's called the ZIP function. It combines two (or more iterables) into an iterable.

So why are you pretending not to follow?

It has implicit ordering. Interleaving A with B is not the same as interleaving B with A.

Makes no difference to the countability argument.

So order matters now? Until it doesn't matter. You seem to be hedging your bets.

You're a very disingenuous fellow. I do not get the sense that I'm conversating with someone who's interested in learning or in achieving mutual understanding; only someone who wants to be right about ... something or other.

None of your "proofs" were property or structure preserving.

I have no idea what that means in this context. I know what structure preserving means, this isn't it.

Ahhhh DIFFERENT bijection. Which one?

The one I showed you several times already.

Contradiction. You are using sets. If pi is a natural number then pi is already in the original enumeration, so this is no longer a bijection.
It's now an injection from N into the multiset {pi, 1,2,3,...}.

Pi is the real number pi 3.14159... I'm using it as example of a non-natural adjoined to the naturals.

I know that. Where's your new bijection?

I showed it to you already. Read my earlier post.

False equivalence. That's the union of a potential infinity (even numbers) with a finite subset of the odd numbers {1}.

What of it? I assume you can interleave an infinite stream with a finite one.

When you union SOME odds (finite set) with ALL evens (potentially infinite set) you will get 1,3,5,7, 0,2,4,6,...
When you union ALL odds (potentially infinite set) with ALL evens (potentially infinite set) you will get 1,3,5,7,..., 0,2,4,6,...
In fact - the set you are going to get is {..., 7, 5, 3, 1, 0, 2, 4, 6, ... } because all you are doing is prepending new elements.

You lost me. I'm out of patience here.

Also.. what a strange union you have there... In this union ANY even number > ANY odd number.

Clearly you're not understanding. But even if it were true, so what? Countability doesn't depend on order.

Look. If you claim it's false that the union of countable sets is countable, at some point I have to wish you well in life and bow out of this fruitless conversation, same as I get limited but finite amusement from talking to flat earthers. I like flat earthers, as it happens. But only to a point. Likewise with someone denying the most very basic facts about countable sets. Facts that are just as true even in the language of streams, using essentially the exact same proof strategies.

Yeah, I have. It's an amazing feat of human self-deception. Given the set of even numbers {0,2,4,6,...} how many rooms must 0 move up, so that ALL Odd numbers can move in before it?

Ok. Elementary set theory's not for everyone, I suppose. Pearls before swine, or leading a horse to water. Nothing personal. You want to play games, I find it tiresome. Especially since you already conceded my point in the language of streams long ago. If I can interleave two infinite streams to make one infinite stream, what is the problem?

0,2,4,8,..., 1,2,3,5,... = 1,2,3,... ?!?!?

Well yes. Why do you think otherwise? The left and right sides are different as ORDERED sets; but they are exactly the same as sets.

ROFL.

Like I say. Snark at things you don't understand. Which is odd in this case, since I'm not doing advanced set theory. When they teach set theory in high school or in "Discrete math" class, the first thing they tell you is that sets are independent of order. {a, b, c} = {b, a, c}.

So what problem do you have with

{0, 2, 4, 6, ..., 1, 3, 5, 7, ...} = {0, 1, 2, 3, 4, ...}?

They're different ORDERED sets; but per high school set theory, they're the exact same set. You can see that by picking any element on the left, and seeing it's also on the right; and vice versa. By the axiom of extensionality, two sets with the same elements are the same set.

https://en.wikipedia.org/wiki/Axiom_of_extensionality

You could say something like, "Ok I get it," or, "Thanks for clarifying that."

Can you see how just snarking off every other paragraph makes my job tedious and disincents me from engaging with you?

And then there's the set difference... If Odd ∪ Even = N then it trivially follows that N \ Odd = Even.

But if |N \ Odd | = |Even| then Odd must be the empty set.

How the hell do you figure that? If I have all the natural numbers and cross out every odd number, what's left are the even numbers.

Ah ... are you perhaps making a cardinality argument? You have infinitely many things and you subtract infinitely many so you must have subtracted all of them?

But this argument is falsified by your own even/odd argument. Euclid proved there are infinitely many primes. If we remove all the primes from the natural numbers, what's left? The composites, right?

So you can't subtract cardinals. You just proved that yourself.

countable minus countable is not defined unless you tell me exactly what the sets are. It might be countable (as in naturals minus odds) or it might be empty (as in naturals minus naturals) or it might be 47 (as in naturals minus everything except 0 through 46).

You just proved that yourself. Please think about Euclid's proof of the infinitude of primes. If we remove the infinitely many primes from the natural numbers, we're left with the composites. Right?

[godelian]

There exists NO complete enumeration of ANY infinite set

What about the axiom of infinity?

https://en.wikipedia.org/wiki/Axiom_of_infinity

In axiomatic set theory and the branches of mathematics and philosophy that use it, the axiom of infinity is one of the axioms of Zermelo–Fraenkel set theory. It guarantees the existence of at least one infinite set, namely a set containing the natural numbers. It was first published by Ernst Zermelo as part of his set theory in 1908.[1]

It is indeed an axiomatized unjustified belief, but it somehow seems to make sense to believe this. Of course, I will not attempt to justify that belief because -- if a justification existed -- it would not be an axiom.

Repeatedly invoking the SUCC function won't produce all the natural numbers either.

In ZF set theory, it is not produced by successive invocation of the SUCC function but by successive invocation of the union operator on Von Neumann set numerals representing the natural numbers:

In the formal language of the Zermelo–Fraenkel axioms, the axiom is expressed as follows:[2]

In technical language, this formal expression is interpreted as "there exists a set 𝐼 (the set that is postulated to be infinite) such that the empty set is an element of it and, for every element x of 𝐼, there exists an element y of 𝐼 consisting of just the elements of x and x itself."

This formula can be abbreviated as:

∃ I ( ∅ ∈ I ∧ ∀ x ( x ∈ I ⇒ ( x ∪ { x } ) ∈ I ) ) .

Some mathematicians may call a set built this way an inductive set.

The construction of inductive sets is tolerated in ZF set theory. Hence, the notion of complete induction of the set of natural numbers is tolerated by the language of ZF.

However, I also think that we should not try to construct the infinite set of natural numbers in PA by using the SUCC function.

For a starters, sets are not a native concept in PA. It is the natural numbers that are the native objects in PA. In the following mathematics.stackexchange.com post, the respondents explain why the axiom of infinity cannot be added to PA:

https://math.stackexchange.com/question ... y-in-peano

Consequently, adding your proposed axiom to PA does nothing at all, and all unprovable sentences over PA remain unprovable over the new system.

Peano arithmetic already proves "There is no largest natural number," so adding this statement to PA just yields PA again.

[Skepdick]

There exists NO complete enumeration of ANY infinite set

What about the axiom of infinity?

https://en.wikipedia.org/wiki/Axiom_of_infinity

In axiomatic set theory and the branches of mathematics and philosophy that use it, the axiom of infinity is one of the axioms of Zermelo–Fraenkel set theory. It guarantees the existence of at least one infinite set, namely a set containing the natural numbers. It was first published by Ernst Zermelo as part of his set theory in 1908.[1]

It is indeed an axiomatized unjustified belief, but it somehow seems to make sense to believe this. Of course, I will not attempt to justify that belief because -- if a justification existed -- it would not be an axiom.

Repeatedly invoking the SUCC function won't produce all the natural numbers either.

In ZF set theory, it is not produced by successive invocation of the SUCC function but by successive invocation of the union operator on Von Neumann set numerals representing the natural numbers:

In the formal language of the Zermelo–Fraenkel axioms, the axiom is expressed as follows:[2]

In technical language, this formal expression is interpreted as "there exists a set 𝐼 (the set that is postulated to be infinite) such that the empty set is an element of it and, for every element x of 𝐼, there exists an element y of 𝐼 consisting of just the elements of x and x itself."

This formula can be abbreviated as:

∃ I ( ∅ ∈ I ∧ ∀ x ( x ∈ I ⇒ ( x ∪ { x } ) ∈ I ) ) .

Some mathematicians may call a set built this way an inductive set.

The construction of inductive sets is tolerated in ZF set theory. Hence, the notion of complete induction of the set of natural numbers is tolerated by the language of ZF.

However, I also think that we should not try to construct the infinite set of natural numbers in PA by using the SUCC function.

For a starters, sets are not a native concept in PA. It is the natural numbers that are the native objects in PA. In the following mathematics.stackexchange.com post, the respondents explain why the axiom of infinity cannot be added to PA:

https://math.stackexchange.com/question ... y-in-peano

Consequently, adding your proposed axiom to PA does nothing at all, and all unprovable sentences over PA remain unprovable over the new system.

Peano arithmetic already proves "There is no largest natural number," so adding this statement to PA just yields PA again.

All inductive definitions use some variant of SUCC. Even the inductive definition of Set.

Naturally - because that's what recursion is.
Base case: be it 0; or the empty set. Distinction without a difference.
SUCC(BASE)

Here's the definition of the naturals

Code: Select all

data ℕ : Set where
  zero : ℕ
  suc  : ℕ → ℕ

Here's the definition of an Inductive set.

Difference. What difference?

Code: Select all

[quote]
record IndSet (S : Set) : Set₁ where
  field
    empty : S
    succ : S → S

The axiom of infinity states that there exists a set that:
1. Contains the empty set (initial element)
2. Is closed under successor operation

The Peano axioms state that:
1. 0 is a number (initial element)
2. All numbers are closed under the successor operation.

It's all iterations of the exact same nonsense. induction/recursion. Base case -> house of cards.

The question remains: How many times do you need to call SUCC to complete the construction of an infinite object?

Yes - I can give you a function to give you the N-th successor to any initial object.
No I can't give you a function to give you ALL sucessors to any initial object.

Seen one inductive data type - seen them all: https://en.wikipedia.org/wiki/Intuition ... tive_types

Every subset has a minimal element. Blah blah... https://en.wikipedia.org/wiki/Well-founded_relation

And then you can do induction. over the induction. over the induction. Over the induction....Over the induction.
Apparently ad infinitum.

And you can induce over the infinite induction also ad infinitum. Infinitely. Because infinities are are infinitely infinite.

There's infinitely many infinities, don't you know?

Except there's no set of all infinities. Because that's too large. That's a proper class, don't you know.

[godelian]

The question remains: How many times do you need to call SUCC to complete the construction of an infinite object?
And you can induce over the infinite induction also ad infinitum. Infinitely. Because infinities are infinite.

It is an axiomatic belief that it is possible to obtain the set of the natural numbers constructed by completing the induction.
You seem to argue that this belief is unjustified, but that is exactly why it is an axiom.
You do not have to believe it. You can work with ZF-inf instead of ZF. That is perfectly legitimate.
I am not going to defend this belief, because it is not meant to be defended. It is meant to be accepted or rejected.

[Skepdick]

The question remains: How many times do you need to call SUCC to complete the construction of an infinite object?
And you can induce over the infinite induction also ad infinitum. Infinitely. Because infinities are infinite.

It is an axiomatic belief that it is possible to obtain the set of the natural numbers constructed by completing the induction.
You seem to argue that this belief is unjustified, but that is exactly why it is an axiom.
You do not have to believe it. You can work with ZF-inf instead of ZF. That is perfectly legitimate.
I am not going to defend this belief, because it is not meant to be defended. It is meant to be accepted or rejected.

You are basically preaching teleology.

What is the decision-procedure for accepting or rejecting any axiom-schema?

Axiomatics is Invention, not discovery.

Anything inductively constructed has inherent ordering. ZF-inf has too weak a notion of "equality". Element equivalence is sufficient.
Evern if ordering/structure is destroyed. This is information loss. And this is ultimately the purpose of any notation/formalism - it encodes information/structure. When you add an axiom you destroy all of that.

Go ahead and identify a single element in any unordered set.

[godelian]

The question remains: How many times do you need to call SUCC to complete the construction of an infinite object?
And you can induce over the infinite induction also ad infinitum. Infinitely. Because infinities are infinite.

It is an axiomatic belief that it is possible to obtain the set of the natural numbers constructed by completing the induction.
You seem to argue that this belief is unjustified, but that is exactly why it is an axiom.
You do not have to believe it. You can work with ZF-inf instead of ZF. That is perfectly legitimate.
I am not going to defend this belief, because it is not meant to be defended. It is meant to be accepted or rejected.

You are basically preaching teleology.

What is the decision-procedure for accepting or rejecting any axiom-schema?

Axiomatics is Invention, not discovery.

Anything inductively constructed has inherent ordering. ZF-inf has too weak a notion of "equality". Element equivalence is sufficient.
Evern if ordering/structure is destroyed.

This is information loss.

I accept your criticism on the fundamental nature of mathematics, at the core of which you will indeed find unjustified beliefs.

I do not insist that anybody should believe in the axiom of infinity. For me, it works. For other people, apparently, it doesn't.

Finitism is certainly a legitimate minority position in the philosophy of mathematics:

https://en.m.wikipedia.org/wiki/Finitism

Model theory would become impossible without infinitism. That would not work for me. I find model theory philosophically too attractive.

[Skepdick]

I accept your criticism on the fundamental nature of mathematics, at the core of which you will indeed find unjustified beliefs.

I do not insist that anybody should believe in the axiom of infinity. For me, it works. For other people, apparently, it doesn't.

Finitism is certainly a legitimate minority position in the philosophy of mathematics:

https://en.m.wikipedia.org/wiki/Finitism

Model theory would become impossible without infinitism. That would not work for me. I find model theory philosophically too attractive.

Finitism/infinitism is another hogwash dichotomy.

Is finite non-infinite; or is infinite non-finite? Yay! LEM again.

This is about the OTHER axiom. Identity.

Go ahead and tell me if two objects {x,y} and {a,b} are identical; or whether there's a bijection between them without knowing the relations between the elements. You can't even determine if they are sets or multisets.

[godelian]

I accept your criticism on the fundamental nature of mathematics, at the core of which you will indeed find unjustified beliefs.

I do not insist that anybody should believe in the axiom of infinity. For me, it works. For other people, apparently, it doesn't.

Finitism is certainly a legitimate minority position in the philosophy of mathematics:

https://en.m.wikipedia.org/wiki/Finitism

Model theory would become impossible without infinitism. That would not work for me. I find model theory philosophically too attractive.

Finitism/infinitism is another hogwash dichotomy.

Is finite non-infinite; or is infinite non-finite? Yay! LEM again.

This is about the OTHER axiom. Identity.

Go ahead and tell me if two objects {x,y} and {a,b} are identical; or whether there's a bijection between them without knowing the relations between the elements. You can't even determine if they are sets or multisets.

The foundations of mathematics are philosophically problematic. This problem became increasingly understood over a century ago, and it will never get solved.

[Skepdick]

The foundations of mathematics are philosophically problematic. This problem became increasingly understood over a century ago, and it will never get solved.

Something which doesn't exist cannot be problematic.

You can't solve a problem that you can't even define.