What is the Collatz Problem and how can it be solved?

[matt grime]

Ok, I'm beginning to see the picture.


1. The binary tree is what i said it is - the infinite thing above. At the k'th level you number the nodes from 1 to 2^k using binary expansions (this means by the way that the only consistent choice for the initial node is 1). You then superimpose those green lines and blue dotsby saying join dot r with dot t iff at least one of the following hold:

r=2t
t=2r
r=3t+1
t=3r+1

ok.

What youve observed is that at any level, it must be that there is a number that is sent outside that level to the next one, for instance you need to construct level k+1 and higher to indicate all th points where 2^k - 1 gets sent/can come from



let's address what you are claiming this implies.

In order to provide a proof of the Collatz conjecture in ZF, one must have all the Natural numbers. However it is in the assumption of the Collatz conjecture that the naturals exist and are a set in ZF. If we are not assuming the axiom of infinity, then the natural numbers do not form a set in the theory we are using, and hence Collatz is undefined in that theory. However, all the axiom of infinity states is that the natural numbers form a set, it does not imply the Collatz conjecture s true. The truth of the Collatz conjecture is not therefore equivalent to the axiom of infinity.

We must presume the axiom of infinity is in our set theory to define the Collatz conjecture, that is all.

the naturals exist, whether or not you chose to call them a set and do so independently of the set theory you use, there is nothing in what you've written to suggest that the assumption the naturals form a set is equivalent to Collatz being true or not.


I think the key here is your views on axiomatic set theory and the natural numbers, nothing to do with Collatz. The axiom of infinity just states that the naturals are an (inductive) set in ZF(C).

 

[Organic]

Matt,

You are close so maybe in this post you will understand my proof.

Please pay attention to the fact that only Collatz-like sequences take you always from some n to another n in the micro tree.

If you combine all these "movements" to one direction (let us choose the positive direction) then by collatz-like sequences you get the "If n exists then (at least) n+1 exists" which is equivalent to ZF axiom of infinity.

 

[matt grime]

But, Organic, the Collatz Iteration is defined on N already, that is all the n's in N 'exist' already. You're finding problems where there aren't any, and in particular you still seem to think that the axiom of infinity is what makes N exist, when that isn't what it states. The axioms of a group say do not force idnetity elements to exist, do they?

 

[Organic]

Matt,

Can N exists without the axiom of infinity?

Please show me how you can say that there are inifinitly many n's in N without the axiom of infinity?

 

[matt grime]

yes, they do exist without the axiom of infinity. This is what I've been attempting to explain since the year dot (English phrase meaning from the very beginning) all the axioms of *a* set theory do is tell you what collections of things constitute sets. It as an attempt to put the theory of sets onto a firm footing so that we can avoid things like the Russell paradox - the collection of all sets that do not contain themselves 'exists' but it is not a set (in ZFC). If you take all the other axioms of ZF(C) apart from the axiom of infinity, then there is no way to force the Natural numbers to be a set - that is it is not possible to deduce the existence of an infinite set from the other axioms alone.

There is some philosophical debate as to what one means by 'exist', but the fact that the greeks were able to talk about the natural numbers (and the rationals etc) without knowing the axiom of infinity should tell you something! The existence philosophy boils down to 'is there some higher plane where '1' genuinely exists in the same way that the computer in front of you exists, or is it that we force it to exist in this universe and it is purely an invention of this universe' this is the so-called idea of the platonic realm. Actually that wasn't a very good example, but if you want to learn about it try looking up the word 'qualia' which is the idea that oneness is some higher concept independent of 1, and that 1 is just some realization of it. But please don't think you can use that to do anytihng with 'infinity' with because, as we established earlier, the idea that 'infinity' exists is a myth, infinite is just 'not finite' which has associated implications, and useful properties.

 

[matt grime]

Just seen the edit.


Suppose that there are only finitely many natural numbers ( a natural number is one obtained by adding 1 repeatedly), then let M be the largest. M+1 is a natural number by definition, and strictly larger than M. This isn't the axiom of infinity. Don't like that one?

Clearly 2 is a natural number, so M>1 if it exists, yet then M*M > M # so M can't exist.

The axiom of infinity means that there is an infinite set, that is all.

 

[Organic]

Matt,

By saying: "yes, they do exist without the axiom of infinity."

You maybe use a vary deep platonic realm.

Well, I do not accept your platonic point of view, because without the axiom of infinity all you have is the empty set by the empty set axiom.

 

[matt grime]

No, they exist by induction, they form a set in ZF by the axiom of infinity. If you want a set theoretic proof for these things try the peano postulates.

I suppose saying they exist inductively sounds wrong doesn't it?


We can count, we know what 1 dog is, 2 dogs, etc, the counting things are natural numbers, we get them by adding 1 repeatedly. Now the question is can we form the SET of natural numbers? Well, the collection of natural numbers exists, and is clearly infinite, that they are a set in ZF is useful, and exactly the reason why the axiom of infinity is there, in ZF they for an inductive set. That's all, I'm not appealing to any platonic realm, only that we can count and have labels for this counting. Are you getting this yet? That it is whether or not we can call it a set, that's all? You can define classes of arbitrary things, they just don't turn out to behave as sets morally ought to if our logic is to be consistent. Set theory is just a formalization of what we know ought to be true, in a way that makes it consistent - the class of sets that do not contain themselves is not a set, yet it still exists.

I knew I'd regret telling you abuot the platonic realm. You already seem to have misunderstood what was admittedly a bad explanation of it; the explainer must take the blame.

 

[Organic]

Peano postulates and ZF axoim of infinity are the same induction.

 

[matt grime]

Peano just puts the naturals on a well founded set theory footing. Induction is there, it is independent of the Axiom of infinity, which after all requires induction, to demonstrate that there is actually an infinite set.


The natural numbers exist, that they form a set is all we are saying with the axiom of infinity.

 

[Organic]

Without this induction you can have one and only one element.

This induction does not define the internal structure made by Von Neumann.

My proof combines between Von Neumann recursive method and the ZF induction.

By this recursive(micro)-inductive(macro) point of view we can check the invariant symmetry that exists in the base of the examined elements.

 

[matt grime]

Originally posted by Organic
Matt,

By saying: "yes, they do exist without the axiom of infinity."

You maybe use a vary deep platonic realm.

Well, I do not accept your platonic point of view, because without the axiom of infinity all you have is the empty set by the empty set axiom.

But how can the empty set exist? By your logic it can't. Because the existence of the empty set in your logic requires the axiom of the empty set, which requires the empty set to exist... oh no, your circular reasoning goes horribly wrong! Have you read anything that talks about set theory at all, ever? Get a book from the library, you might learn something. And when you understand it, try again.

 

[matt grime]

OF course if you don't like that, then how about this (which i noted earlier but you ignored)

if one wants to prove that Collatz iterations of all natural numbers reach one, then you must be presuming that the natural numbers all already exist to even define the Conjecture, so there can be no issue in presuming they exist to prove it. Anyway, you haven't proved anything, well apart from that you don't know what you're talking about, yet still feel confident enough to tell me I'm wrong about everything.

 

[Organic]

Nothing circular here.

The axiom of the empty set and the empty set are the same.

The axiom of infinity use this axiomatic existence as its input and using induction to define its pruducts.

Von Neumann Hierarchy uses the same axiomatic existence as its input and using recursion to define its products.

My proof combines between Von Neumann recursive method and the ZF induction.

By this recursive(micro)-inductive(macro) point of view we can check the invariant symmetry that exists in the base of the examined elements.

 

[matt grime]

You think the axiom of the empty set is the empty set? Wow, that's a new one. You know you really ought to learn some maths, Organic if you're going to persist in this.

Erm, there is no such thing as ZF induction as far as I'm aware. Please post some reference for this.

Inducion is a mathematical technique, using the existence of the 'inductive' set that the axiom of infinity requires, one can construct an infinite set. And? How does this bear any relation to anything here?

At least accept that the existence of the Natural numbers is implicit in the statement of the Collatz conjecture. Therefore assuming their existence in the proof is acceptable!


That you do not seem to understand the ideas of set theory can be left alone - go and get a book and learn about them. The axiom of infinity does not state the naturals exist, it states that there must be an infinite set.

 

[Organic]

Let me put it this way, the ZF axiom of the empty set uses x
where x is something and then it says that for any x(=something),
x(=something) not in some set X.

This is nothing but a negative and complicated way to say positively and simply: "there exist set X with no elements" notated as {}.

ZF induction is my mistake, it has to be the ZF axiom of infinity induction.

You don't see the invariant symmetry that exists in both Von Neumann
recursion(micro level) and Collatz sequences (macro level).

The ZF axiom of infinity products (if n exists then n+1 exists implies N members) are in the level of the macro tree, where Collatz sequences exists, therefore we cannot distinguish between them and the Collatz sequences because in the macro level they are the same elements.

Eech n in the induction macro level has its own internal unique structure produced by Von Neumann recursion.


When we examine the invariant Binary tree that stands in the basis of both Collatz sequences and Von Neumann recursion, then and only then we can see that Collatz sequences and ZF axiom of infinity are the same iteration.

 

[matt grime]

Oh. My. God. For the god knows how many'th time, what is the ZF axiom of infinity induction?. You are the only person I have ever known use it, Google for the phrase. You and responses to you were the only hits the last time I did it.

There is the axiom of infinity which states there must be a set W satisfying a certain property: if y is in W then y u {y} is in W. There is absolutely no mention of induction there. Now, using induction we can conclude that there is a set which does not have finite cardinality. That's it. It is not the ZF axiom of infinity induction.

For some reason you seem to think that it is the axioms of ZF that force things to exist. Exercise: just using the axioms, how do you see that the class of all continuous Real valued functions on a compact hausdorff topological space exists?

The class exists, independently of ZFC. Which is just a formal attempt to put the theory of sets onto a firm mathematical footing. These things all exist albeit in some mathematical sense anyway.

Read a book on set theory, please, I'm begging you.

 

[Organic]

Matt,

Don't you see that "if y is in W then y u {y} is in W" is exactly the same as "if n is in N then n+1 is in N"?

"if n is in N then n+1 is in N" is an induction.

Please look at the other things that I wrote in the previous post.

Thank you.

 

[matt grime]

You are missing the subtle point about what the word induction means here, in particular it is properly the principle of mathematical induction which states that, suppose P(r) is a set of statements indexed by N. Then if P(n) => P(n+1) and P(1) is true, it follows P(r) is true for all r in N. What ever short hand you may have for it, and whatever the superficial similarity, the axiom of infinity that if y is in W, then yu{y} is in W is not the principle of mathematical induction. Using the principle of mathematical induction one concludes that the axiom of infinity states there is a set whose cardinality is not finite. Roughly speaking the set theory one uses must have an inductive set. These are subtle issues, but you're lack of precision with mathematical language causes you to say things that are wrong, often very wrong - you have said repeatedly that a set is equal to an inequality for instance. This is a common error you commit - saying that two different things are equal or the same.

In particular if n is in N then n+1 is in N is an 'inductive step' in a proof by induction.

Yes there are similarities, and that is how the axiom of infinity comes to be in the form it is - there are equivalent formulations (inductive sets).

Anyway, enough of this. Back to the main point.

Your proof. what stage is at at now. It is on at least its third rewrite isn't it. Each time you insist it is a proof, and each time you have to rewrite it cos you realisze you've made another mistake, and each time you're as certain that it is a proof. How about relabelling it as an idea cos you don't seem to understand what a proof is.

 

[Organic]

1) I rewrote it because i saw that you don't understand it.

2)"if n is in N then n+1 is in N" is not an 'inductive step' but an induction forced by an axiom (the ZF axiom of infinity).

3) Please this time read what is below and write your remarks:

Let me put it this way, the ZF axiom of the empty set uses x
where x is something and then it says that for any x(=something),
x(=something) not in some set X.

This is nothing but a negative and complicated way to say positively and simply: "there exist set X with no elements" notated as {}.

If you don't see the invariant symmetry that exists in both Von Neumann recursion(micro level) and Collatz sequences (macro level) then you can't understand my proof.

The ZF axiom of infinity products (if n exists then n+1 exists implies N members) are in the level of the macro tree, where Collatz sequences exists, therefore we cannot distinguish between them and the Collatz sequences because in the macro level they are the same elements.

Eech n in the induction macro level has its own internal unique structure produced by Von Neumann recursion.


When we examine the invariant Binary tree that stands in the basis of both Collatz sequences and Von Neumann recursion, then and only then we can see that Collatz sequences and ZF axiom of infinity are the same iteration.

 

[matt grime]

OK, let's ignore the infinity stuff, cos it is implicit in the definition of the Conjecture that we are proving it for all n in N.

Right, now you put the collatz tree in some sort of bijection with the binary tree. However, as each integer n occurs an infinite number times in the labelling on the binary tee, how are you deciding which one you want?

Seeing as you tend not to see things like this, here's a proof of that assertion. let r be fixed, the number 2^r-1 occurs on the k'th level for every k > r. So which of those are you picking?

I say some sort of bijection cos I don't see what is in bijection with what clearly. what are the two sets that are in bijection?

 

[Organic]

The Binary Tree is in bijection with itself, or if you like macro tree (the induction level) is in bijection with the micro tree (the internal recursion level).

 

[matt grime]

So, something is in bijection with itself? Stop presses. You claimed something about the n being in bijection with {} or something, please repost that.

 

[Organic]

Please overlap the example of number 4 in page 1(the internal recursion leval)
with the example of page 2 (the induction level).

http://www.geocities.com/complementarytheory/3n1proof.pdf

Dear Matt, I'm going to sleep. have a very good night, and thank you very very much.

 

[Hurkyl]

Some comments for all:

You belong to a community that express its ideas by words, I don't.

In my opinion, there is nothing in the tools themselves that determines if some idea is mathematical or not.

It's not the tools, it's how you use them. Mathematics can be done with pictures, but you have to use them the same way you use words and formulas; by following rules of deduction that let you deduce new pictures from old pictures.

The tricky thing is that pictoral arguments say absolutely nothing about things that get described by words and formulas unless you provide a mapping between picture world and word/formula world.

Sometimes when I've done "proofs" via deductions of pictures, I was only looking for results in picture-land and I was happy. Other times, I was working in word/formula-land and I carfeully constructed a picture-land that had a correspondence back to word/formula land so that my picture proofs really do translate into facts in word/formula-land.

In order to provide a proof of the Collatz conjecture in ZF, one must have all the Natural numbers. However it is in the assumption of the Collatz conjecture that the naturals exist and are a set in ZF.

It takes a bit of effort, but you can state the Collatz conjecture in ZF without ever mentioning N, thus the Collatz conjecture can be defined even if you deny the existence of N.

Of course, it does require you to define the term "natural number", but that can be done without the axiom of infinity. You just don't get the convenience of having a set of all natural numbers.

"If n exists then (at least) n+1 exists" which is equivalent to ZF axiom of infinity.

The axiom of infinity states (accepting your slight abuse of notation):

"There is a set, S, such that if n is in S, then n + 1 is in S".

Please try to understand the difference.


The statement "If n exists then (at least) n+1 exists" would be better said as "For any n, there exists a set we write as n+1". This fact is guaranteed by the axioms of the sum set and the axiom of the pair set:

Given n:
The axiom of the pair set proves the set {n} exists.
The axiom of the pair set proves the set {n, {n}} exists.
The axiom of the sum set proves that n U {n} exists
and "n U {n}" is what is really meant in set-theory land, not "n + 1"

Can N exists without the axiom of infinity?

The answer to this statement (interpreted mathematically) is yes.

ZF is an incomplete theory; to the best of my knowledge, the statement "N exists" is independant from the axioms of ZF without the axiom of infinity.

Thus N can exist without the axiom of infinity.

Please show me how you can say that there are inifinitly many n's in N without the axiom of infinity?

Without the axiom of infinity, I can prove that there does not exist a finite set that contains all of the natural numbers. Here's the sketch:

Suppose F is a set of all natural numbers. The function f(n) = n + 1 is a 1-1 mapping from F onto a proper subset of itself. Therefore F is infinite.

Therefore, no finite set can contain all natural numbers.

Peano postulates and ZF axoim of infinity are the same induction.

Nothing in the Peano postulates say anything about sets. ZF does nothing but postulate the existence of a set with a given property. Induction is a mathematical technique for proof and thus cannot be one or more axioms. None of the three things mentioned in your sentence can be the same.

Without this induction you can have one and only one element.

Tell me how many sets you want proved exist and I'll do it for you without any form of induction. I'll do two sets for you, as a freebie.

{} exists by the axiom of the empty set.
{{}} exists by the axiom of the pair set.

{} is unequal to {{}} because {} is in {{}} but {} is not in {}.

Thus, I've proven two sets, {} and {{}} exist.

The axiom of the empty set and the empty set are the same.

The axioms of a mathematical theory are not the objects of that mathematical theory.

 

[matt grime]

And I thought I was the only one dumb/obstinate enough to try and explain stuff to Organic. I don't suppose by any chance you're a set theorist, are you Hurkyl? I'm one of those representation theory people and attempting to explain ZF and set theory is not something I'm particularly comfortable with - my statements here are reasonably informed opinions, but I'm absolutely convinced I've stated things that would have the genuine set theorist spitting blood. The use of 'model' throughout is not in the sense of model theory (or if it is it was a lucky guess).

 

[Hurkyl]

And I thought I was the only one dumb/obstinate enough to try and explain stuff to Organic.

I've spent probably a whole year trying to explain things to Organic. I'm on a break.

I don't suppose by any chance you're a set theorist, are you Hurkyl?

I don't really know enough about anything, yet, to have earned any sort of specialist title!

 

[Organic]

Hurkyl and Matt,

"There is a set, S, such that if n is in S, then n + 1 is in S".

Nothing in the Peano postulates say anything about sets. ZF does nothing but postulate the existence of a set with a given property. Induction is a mathematical technique for proof and thus cannot be one or more axioms. None of the three things mentioned in your sentence can be the same.

Before we can speak about a concept like set we have to define its existence.

For example:

ZF axiom of the empty set

There is a set, A, such that for any x, x not in A.



Also before we can use Peano postulates we have to define some "minimal input" to start with.

For example:

Peano's first postulate

1 is in N (also can be understood as: 1 is a natural number)



In both axiomatic systems, there are at least two basic concepts, which are:

1. A container (in the above examples they are A, N)
2. A content (in the above examples they are x, 1)



Now let us examine ZF axiom of infinity and Peano's second postulate:

ZF: There is a set, S, such that if n is in S, then n + 1 is in S.

Peano: If n is in N, then its "successor" n' is in N.

As we can see, both axioms are forcing induction on some content, which means that they are the same axiom.

The result of this forced induction is infinitely many elements called “the natural numbers”.

These infinitely many elements cannot be defined without the forced induction axiom.

Without the axiom of infinity, I can prove that there does not exist a finite set that contains all of the natural numbers. Here's the sketch:

Suppose F is a set of all natural numbers. The function f(n) = n + 1 is a 1-1 mapping from F onto a proper subset of itself. Therefore F is infinite.

Therefore, no finite set can contain all natural numbers.

Isn't f(n) = n + 1 equivalent to the axiom of infinity?

Tell me how many sets you want proved exist and I'll do it for you without any form of induction. I'll do two sets for you, as a freebie.

{} exists by the axiom of the empty set.
{{}} exists by the axiom of the pair set.

{} is unequal to {{}} because {} is in {{}} but {} is not in {}.

Thus, I've proven two sets, {} and {{}} exist.

Thank you Hurkyl for correcting my mistake here.

The axiom of the empty set and the empty set are the same.

In this address (http://www.cs.bilkent.edu.tr/~akman/jour-papers/air/node5.html ) I have found this:

“The Null Set Axiom guarantees that there is a set with no elements, i.e., the empty set {}. This is the only set whose existence is explicitly stated".

Again, let me put it this way, the ZF axiom of the empty set uses x
where x is something and then it says that for any x(=something),
x(=something) not in some set X.

This is nothing but an indirect and complicated way of what we can say directly and simply: "there exist set X with no elements" notated as {}.

(In my opinion any element whose existence is explicitly stated, is equivalent to the axiom that defines it).

"The Pair Set Axiom states the existence of a set which has a member when the only existing set is {}. So the set {{}} can now be formed now and we have two objects {} and {{}}. The application of the axiom repetitively yields any finite number of sets, each with only one or two elements.

It is the Sum Set Axiom which states the existence of sets containing any finite number of elements by defining the union of already existing sets. Thus
{{},{{}}}U{{{},{{}}}} = {{},{{}},{{},{{}}}}

However it should be noted that all these sets will be finite because only finitely many sets can be formed by applying Pair Set and Sum Set finitely many times.

It is the Axiom of Infinity which states the existence of at least one infinite set, from which other infinite sets can be formed. The set which the axiom asserts to exist is {{},{{}},{{},{{}}},{{},{{}},{{},{{}}}}}.

Thus, the ZF universe simply starts with the {} and extends to infinity. It can be noticed that cumulative hierarchy produces all finite sets and many infinite ones, but it does not produce all infinite sets.”


Well Pair Set Axiom and Sum Set Axiom are the basis of Von Neumann recursion.

If you don't see the invariant symmetry that exists in both Von Neumann recursion(micro level) and Collatz sequences(macro level) then you can't understand my proof.

The ZF axiom of infinity products (if n exists then n+1 exists implies infinitely many N members) are in the level of the macro tree, where Collatz sequences exists, therefore we cannot distinguish between them and the Collatz sequences because in the macro level they are the same elements.

Eech n in the induction macro level has its own internal unique structure produced by Von Neumann recursion.


When we examine the invariant Binary tree that stands in the basis of both Collatz sequences and Von Neumann recursion, then and only then we can see that Collatz sequences and ZF axiom of infinity are the same iteration.

General: any iteration that is based on root or exponent value 2, is equivalent to the ZF axiom of infinity iteration.

Please look again at: http://www.geocities.com/complementarytheory/3n1proof.pdf

Thank you,

Organic

 

[matt grime]

You seem to not be able to distinguish between different objects.

A function is not an axiom, a set and an axiom, etc.

You also appear to have invented another new term: ZF axiom of infinity products.

I don't really have the inclination to even think about what you mean in your new view of it.

 

[matt grime]

quote:
In my opinion any element whose existence is explicitly stated, is equivalent to the axiom that defines it

I missed that one. that's probably one of the more silly things you've said, and boy is that a hotly contested fight.

so many counter-examples and corrections spring to minf, but what's the point? You keep insisting you are correct and have this proof of something, and you keep on having to correct it and each time you come back just as vehement that we are wrong and you are right... hell some times you don't even correct it, yet still claim it is correct despite the evidence to the contrary (I wonder if you are still claiming a set is equal to an inequality...?)

What is it you are claiming anyway - that the Collatz conjecture is undecidable? Do you take this from the assertion that COnway proved some Collatz like conjectures are undecidable? did you even look at his proof? do you even know what undecidable actually means? Seeing as number theory uses ZF, and you claim that he Collatz conjecture is EQUIVALENT to the axiom of infinity, then you've probed Collatz is true (in ZF).

You haven't done that though.

Tell you what, you produce a nice picture that, say, proves the 4-subspace problem is wild, and maybe we'll take you a little more seriously. Until then I'll just keep pointing out the gross abuses of the English language and mathematics you are disguising your lack of ability in.

 

[Hurkyl]

Before we can speak about a concept like set we have to define its existence.

Why would you think that?

If I have a list of properties that something called a "set" obeys, I don't need to know anything about the existence of a "set" in order to reason about those properties.

(This is a familiar idea even in "everyday" logic; we often call it a "hypothetical scenario" in such a context)

In both axiomatic systems, there are at least two basic concepts, which are:

1. A container (in the above examples they are A, N)

There is no container concept in peano arithmetic. (There are also quite a few other theories that have no container concept)

The peano axioms give a list of properties for three undefined terms: the unary predicate "is a natural number", the constant symbol "1", and the unary function "successor".

"Natural number" is no more a container than "human" is; in laymen's terms, "Natural number" is just a name we give to a 'type' of thing; it is a 'category' (I mean this in the lay sense, not the technical sense).


An aside on notation: sometimes, we like to use notation that looks like set notation, even when there are no sets involved. To use a different example, we might choose to write the logical predicate "Z is an ordinal number" in the form $Z \in \mathrm{OR}$, even though there is not (and in ZF, cannot be) a set OR that contains all ordinal numbers.

ZF: There is a set, S, such that if n is in S, then n + 1 is in S.

Peano: If n is in N, then its "successor" n' is in N.

As we can see, both axioms are forcing induction on some content, which means that they are the same axiom.

First off, we have to assume that you intend the phrases "n + 1" and $n \cup \{ n \}$ to be synonymous.

Secondly, how can they be the same axiom?

The axiom of infinity guarantees the existence of a set, but I can't use the Peano axiom "If n is a natural number, then the successor of n is a natural number" to prove the existence of any set.


And, incidentally, the set guaranteed by the axiom of infinity cannot be proven to be an inductive set under the "n + 1" operation.

Isn't f(n) = n + 1 equivalent to the axiom of infinity?

As per a later statement you make, I would like to point out that the existence of the function I used here is given by the axiom of the power set and the axiom of subsets.

The axiom of power sets guarantees the existence of the set P(P(F)). (the power set of the power set of F)

P(P(F)) contains, among other things, all ordered pairs (a, b) where a and b are elements of F.

The axiom of subsets guarantees the existence of a set, call it f, which is the subset of P(P(F)) that contains only ordered pairs of the form (x, x + 1). This is the function we would more commonally write f(n) = n + 1.

More precisely:

$$f := \{ x | x \in \mathcal{P}(\mathcal{P}(F)) \wedge(\exists a \in F : x = (a, a+1) \}$$

(In my opinion any element whose existence is explicitly stated, is equivalent to the axiom that defines it).

My limited comprehension of your ideas may be at fault here, but this sounds awfully like confusing "x-content" with an "x-model".



Finally, I would like to make a comment on models.

I mentioned earlier that:

The peano axioms give a list of properties for three undefined terms: the unary predicate "is a natural number", the constant symbol "0", and the unary function "successor".

(I changed "1" to "0" because I prefer formulations of the natural numbers that include zero)

Allow me to emphasize that these are undefined terms of the theory of peano arithmetic. They are NOT "n is in N", "{}", and "n U {n}".


When we make these identifications, we are making what mathematicians call "a model of the natural numbers in ZF". A model, more or less, consists of taking the undefined terms of one theory and defining them in another theory. There are additional details, such as you have to verify that all of the axioms of the one theory are satisfied when you define the terms in the other theory.

 

[Organic]

If I have a list of properties that something called a "set" obeys, I don't need to know anything about the existence of a "set" in order to reason about those properties.

(This is a familiar idea even in "everyday" logic; we often call it a "hypothetical scenario" in such a context)

Any x in some theory cannot be but a model(X), therefore we always have to be aware to the combination of model() and X, where model() is the global state and X is some local state.

Model() is the container (global state).

X is the content. (some local state).

x is the product of model(X) (the combination or relations between global and local states).

Shortly speaking any theory is first of all model() or if you like an empty container (a global state) "waiting" to some X (some local state) to be its examined concept.

From this point of view, any theory must be aware to the relations between the global and the local, otherwise it cannot use its full potential.

If Peano Axioms is a theory then first of all is a model()(a container) that "needs" some X(a content) to deal with.

For example:

1 is in N (also can be understood as: 1 is a natural number)

x=model(X) where x cannot be but a combination of model()(= theory of numbers) and X(= the concept of a number).

It means that if we want to understand x(=1 in this case) we have to "put on the table" its combination (container-content) property.

Shortly speaking, 1 cannot be but 1 in N, whether you say it or not.

Therefore There is a container concept in Peano arithmetic.

Again, there can be a big problem for us to understand and develop deeper connections between so called different areas of research, if we don't take in account the global-local or container-content relations.

What I wrote here also can explain why ZF axiom of infinity and Peano first axiom are the same axiom.

And this axiom can be called "The forced-induction axiom".

My limited comprehension of your ideas may be at fault here, but this sounds awfully like confusing "x-content" with an "x-model".

All products of some theory are nothing but an x-model, an axiom is a product therefore an x-model.

I am talking about the hierarchy of dependency among these products.


The basic level is the axioms level, and on top of it there is the hierarchy of products, which can exist iff they do not contradict the axioms level.

But when we need an axiom that directly determines the existence of some element, it means that there is no hierarchy here but "the same lady with a different dress".

 

[matt grime]

Originally posted by Organic
x=model(X) where x=1 cannot be but a combination of model()(= theory of numbers) and X(= the concept of a number).

Therefore any product in any theory cannot be but a combination of container and content.

The first thing in there translates as

1= theoryofnumbers(conceptofanumber)

which is beyond words.

 

[Organic]

No it is simple.

x=thory_of_some_concept(THE_CONCEPT).

thory_of_some_concept() is the container.

THE_CONCEPT is the content.

x is the container-content relations.

 

[matt grime]

Er, I think you missed the joke again. you say concept_of_A_number, well which one is 1 the realization of the concept of. I was pointing out the imprecisoin of your terminology again.