Cont: Deeper than primes - Continuation 2

[doronshadmi]

Opss...

 

[doronshadmi]

Do Gödel's First Incompleteness Theorem imply the inconsistency of Actual Infinity?

Ok, it is about time to gather the last discussion into one post.

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According to Modern Mathematics (where the majority of mathematicians agree about the notion of actual infinite sets, as established mostly by George Cantor) an inductive set (as given by ZF(C) Axiom Of Infinity) has an accurate cardinality, which implies that it is complete (no one of its members is missing).

In other words, by ZF(C) Axiom Of Infinity there exists at least one infinite AND complete set (if we agree with the notion of actual infinity, as mostly established by Cantor).

Now, assume a complete set of infinite axioms (according to the reasoning of actual infinity, as established mostly by Cantor and agreed by the majority of modern mathematicians).

But by Gödel's First Incompleteness Theorem such set of axioms must be inconsistent as follows:

Set A is a set of infinitely many axioms (where each axiom is written by finitely many symbols) which is established by using ZF(C) Axiom Of Infinity on ZF(C) itself, such that Infinity is taken in terms of Platonic Infinity (By Platonic Infinity there exists a set of infinitely many things as a complete whole (without using any process)).

Some example: The infinite set of all natural numbers is taken in terms of Platonic infinity.

Now all we care is about the set of all infinitely many wffs (in terms of Platonic Infinity) that are established in A.

Each wff has some Gödel number, where at least one of these wffs, called G, states "There is no number m such that m is the Gödel number of a proof in A, of G" (since G needs a proof, it is not an axiom but a theorem).

Since all wffs are already in A and all Godel numbers are already in A (because Infinity is taken in terms of Platonic Infinity) there is a Gödel number of a proof of G in A, which contradicts G in A, exactly because A is complete (as shown) and therefore inconsistent.

So the problem is actually the notion of a complete set of infinity many things in terms of Platonic Infinity, and in order to save the consistency of A, ZF(C) Axiom Of Infinity is taken in terms of Potential Infinity (process is used, exactly as done in case of Gödel's First Incompleteness Theorem).

But then ZF(C) Axiom Of Infinity can't be used in order to establish sets in terms of Platonic Infinity (for example: the notion of The infinite set of all natural number is logically inconsistent).

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Gödel was a Platonist (he agreed with Actual infinity in terms of Cantor (which is actually Platonic Infinity)) and his main motivation behind his Incompleteness Theorems was to logically demonstrate that formal systems that are strong enough in order to deal with Arithmetic, can't be complete AND consistent and also can't prove their own consistency (which means that many "interesting" formal systems can't deal with Platonic realms).

But Gödel's First Incompleteness Theorem also proves that the very notion of Actual infinity in terms of Platonism (which is also Actual infinity in terms of Cantor) does not hold logically.

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There is a non-interesting solution about the discussed subject, as follows:

G states: "There is no number m such that m is the Gödell number of a proof in A, of G"

If G is already an axiom in A (where A is an infinite set of axioms, such that Infinity is taken in terms of Platonic Infinity) it is actually a wff that is true in A, which does not have any Godel number that is used in order to encode G's proof (since axioms are true wff that do not need any proof in A).

But then no proof is needed and mathematicians are out of job (therefore it is an unwanted solution).

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Also please be aware of the following:

1) If ZF(C) Axiom Of Infinity is not necessarily taken in terms of Platonic Infinity, then ZF(C) Axiom Of Infinity is taken in terms of Platonic Infinity OR Not (useless tautology).

2) If ZF(C) Axiom Of Infinity is not necessarily taken in terms of Platonic Infinity, then it can't be used in order to establish even the set of all natural numbers (which means that N (and |N|) is not necessarily established by ZF(C)).

 

[jsfisher]

I mean that A is a set of infinitely many axioms which is "strong enough" in order to deal with Arithmetic, where Infinity is taken in terms of Platonic Infinity.

How do you get from

There exists a set which contains the empty set and for every member of the set it also contains the member's successor^*​

to

A is a set of infinitely many axioms which is "strong enough" in order to deal with Arithmetic, where Infinity is taken in terms of Platonic Infinity.​

That's a breath-taking leap, even for you. Also, nowhere did you "use" the axiom "on ZF(C) itself."


[SIZE=-2]^* 'Successor' has been defined various ways for different uses of the axiom in different set theories. ZF and ZFC share a specific successor function, S, where S(x) = x U {x}.[/SIZE]

 

[doronshadmi]

There exists a set which contains the empty set and for every member of the set it also contains the member's successor

All we care is that such set is taken in terms of Platonic Infinity.

That's a breath-taking leap

I agree with you, modern mathematicians indeed take such axiom in terms of Platonic Infinity.

Infinity is taken in A (which is strong enough to deal with Arithmetic) in terms of Platonic Infinity, such that all the infinite wffs are already in A, exactly as ZF(C) Axiom Of Infinity defines an infinite set as a complete whole (Platonic Infinity).

jsfisher, please move on to http://www.internationalskeptics.com/forums/showpost.php?p=12786763&postcount=3382.

Thank you.

 

[jsfisher]

All we care is that such set is taken in terms of Platonic Infinity.

Who is this "we" of which you speak?

All "we" (= the set of not you) care about is what the axiom actually asserts, not this extra baggage and bogus inferences you insist upon.

So, again, the Axiom of Infinity is very specific in what it asserts. How do you get from there, in small, logical steps, please, to the rather distant conclusion you allege?

 

[doronshadmi]

So, again, the Axiom of Infinity is very specific in what it asserts.

By Platonic Infinity there exists a set of infinitely many things as a complete whole.

The Axiom of Infinity establishes a set in terms of Platonic Infinity (if Platonic Infinity is rejected, then The Axiom of Infinity can't establish even the infinite set of all natural numbers (which means that N (and therefore |N|) is not established even in the abstract sense)).

Let A (which is a formal system) be the set of all infinite (in terms if Platonic Infinity) wffs (axioms (that do not need to be proven) OR theorems (that need to be proven)) that are encoded by Gödel numbers in A, such that A is strong enough in order to deal with Arithmetic.

In that case what is the cardinality of all the infinite Gödel numbers in A (where infinity is taken in terms of Platonic Infinity)?

Before you answer to this question, please be aware of the following:

Each wff is encoded by a Gödel number, where at least one of these wffs, called G, states "There is no number m such that m is the Gödel number of a proof in A, of G" (since G needs a proof, it is not an axiom but a theorem).

Since all wffs are already in A and all Gödel numbers are already in A (because Infinity is taken in terms of Platonic Infinity) there is a Gödel number of a proof of G in A, which contradicts G in A, exactly because A is complete (in terms of Platonic Infinity) and therefore inconsistent.

 

[jsfisher]

The Axiom of Infinity establishes a set in terms of Platonic Infinity....

The Axiom of Infinity establishes a set in terms of Mathematics. The properties the set has are based in Mathematics. How you may relate to that philosophically is of no interest to Mathematics.

You have told us that your set, A, is based on "using the ZF(C) Axiom of Infinity on ZF(C) itself."

So, a very reasonable question than is what do you mean by that. So far, you have failed to give it any meaning, yet it is fundamental you this set, A, you banter about.

 

[doronshadmi]

You have told us that your set, A, is based on "using the ZF(C) Axiom of Infinity on ZF(C) itself."

By "using the ZF(C) Axiom of Infinity on ZF(C) itself." I mean that A is a formal system with infinitely many wffs in terms of Platonic Infinity, which is strong enough in order to deal with Arithmetic.

Now please answer to http://www.internationalskeptics.com/forums/showpost.php?p=12788750&postcount=3386 but before that please explain what do you mean by "The Axiom of Infinity establishes a set in terms of Mathematics." (especially the highlighted part)?

Thank you.

 

[jsfisher]

By "using the ZF(C) Axiom of Infinity on ZF(C) itself." I mean that A is a formal system with infinitely many wffs in terms of Platonic Infinity, which is strong enough in order to deal with Arithmetic.

You are telling us something about what you think you get by "using the ZF(C) Axiom of Infinity on ZF(C) itself." You first need to tell us what "using the ZF(C) Axiom of Infinity on ZF(C) itself" all by itself means.

Many sets have the characteristics you ascribe to your set, A. Would any such set do (in which case, all this blather about the Axiom of Infinity is extraneous)?

 

[doronshadmi]

You are telling us something about what you think you get by "using the ZF(C) Axiom of Infinity on ZF(C) itself." You first need to tell us what "using the ZF(C) Axiom of Infinity on ZF(C) itself" all by itself means.

Many sets have the characteristics you ascribe to your set, A. Would any such set do (in which case, all this blather about the Axiom of Infinity is extraneous)?

A, by itself, is exactly a formal system with infinitely many wffs in terms of Platonic Infinity (exactly as a set with infinitely things is established by ZF(C) Axiom Of Infinity), which is strong enough in order to deal with Arithmetic.

There can be many sets like A, but A is enough in order to address the discussed subject.

Please explain what do you mean by "The Axiom of Infinity establishes a set in terms of Mathematics." (especially the highlighted part)?

 

[jsfisher]

A, by itself, is exactly a formal system with infinitely many wffs in terms of Platonic Infinity (exactly as a set with infinitely things is established by ZF(C) Axiom Of Infinity), which is strong enough in order to deal with Arithmetic.

There can be many sets like A, but A is enough in order to address the discussed subject.

Please explain what do you mean by "The Axiom of Infinity establishes a set in terms of Mathematics." (especially the highlighted part)?

You first need to tell us what "using the ZF(C) Axiom of Infinity on ZF(C) itself" means.

 

[abaddon]

You offer your opinion, which does not actually logically deals with anything that is found in my posts.

dlorde is not alone in that opinion. You have done nothing but produce scads of navel gazing crapulence in this entire thread. We have reached, and passed the point where you have for protrtacted periods, happily posted to yourself, even at times arguing with yourself. People dip in now and then just to see how deep the rabbit hole has become. And that's it.

You have, for reasons unknowable, reduced yourself to an internet curiosity. Why? Who can tell. All we can do is observe it happening.

 

[doronshadmi]

You first need to tell us what "using the ZF(C) Axiom of Infinity on ZF(C) itself" means.

ZF(C) is an axiomatic formal system that is strong enough in order to deal with Arithmetic.

By using ZF(C) Axiom Of Infinity on ZF(C) itself, there is an axiomatic formal system A that has infinitely many wffs in terms of Platonic (or Actual) Infinity AND it is strong enough in order to deal with Arithmetic, exactly as ZF(C) enables to do it.

Now please reply to my question to you in http://www.internationalskeptics.com/forums/showpost.php?p=12789914&postcount=3390.

Thank you.

 

[doronshadmi]

dlorde is not alone in that opinion.

Opinions are not enough and also your reply is nothing but an opinion.

 

[jsfisher]

By using ZF(C) Axiom Of Infinity on ZF(C) itself,...

How?

 

[doronshadmi]

How?

"using ZF(C) Axiom Of Infinity on ZF(C) itself"

is the same as

"There is an extension of ZF(C), called A (which is strong enough in order to deal with Arithmetic (exactly as ZF(C) has this property)) where all of its infinitely many wffs are already in A, since Infinity (according to ZF(C) Axiom Of Infinity) must be taken in terms of Platonic (or Actual) Infinity" (otherwise even the infinite set of all natural numbers can't be established by this ZF(C) axiom).

Now please explain what do you mean by "The Axiom of Infinity establishes a set in terms of Mathematics." (especially the highlighted part)?

 

[jsfisher]

"using ZF(C) Axiom Of Infinity on ZF(C) itself"

is the same as

"There is an extension of ZF(C), called A (which is strong enough in order to deal with Arithmetic (exactly as ZF(C) has this property)) where all of its infinitely many wffs are already in A

There is nothing in the Axiom of Infinity that lets you conclude your set A exists. You also have not explained what "using ZF(C) Axiom Of Infinity on ZF(C) itself" means. Telling us the result you think you have isn't the same as telling us how you got it.

So, once again, what does "using ZF(C) Axiom Of Infinity on ZF(C) itself" mean?

 

[Little 10 Toes]

"using ZF(C) Axiom Of Infinity on ZF(C) itself"

is the same as

"There is an extension of ZF(C), called A (which is strong enough in order to deal with Arithmetic (exactly as ZF(C) has this property)) where all of its infinitely many wffs are already in A, since Infinity (according to ZF(C) Axiom Of Infinity) must be taken in terms of Platonic (or Actual) Infinity" (otherwise even the infinite set of all natural numbers can't be established by this ZF(C) axiom).

Now please explain what do you mean by "The Axiom of Infinity establishes a set in terms of Mathematics." (especially the highlighted part)?

Let's get rid of the extra wording...

"There is an extension of ZF(C), called A where all of its infinitely many wffs are already in A, since Infinity must be taken in terms of Actual Infinity."

This still doesn't make sense. First set A was a set, then a "formal system", now it's an extension. And it contains things that it already contained? And it must be taken?

 

[doronshadmi]

There is nothing in the Axiom of Infinity that lets you conclude your set A exists.

Wrong, for example: Without the Axiom of Infinity (where Infinity is taken in terms of Platonic (or Actual) Infinity) no axiom schema (for example: the ZF(C) Axiom Schema Of Replacement) exists.

"Given that the number of possible subformulas or terms that can be inserted in place of a schematic variable is countably infinite, an axiom schema stands for a countably infinite set of axioms." https://en.wikipedia.org/wiki/Axiom_schema#Finite_axiomatization

In other words, A (exits, exactly as given in http://www.internationalskeptics.com/forums/showpost.php?p=12790479&postcount=3396) or the ZF(C) axiom schema of replacement are both established by taking Infinity in terms of Platonic Infinity, since countably infinite means that there is bijection between, for example, the infinite set of all natural numbers (in terms of Platonic (or Actual) Infinity) and some given set.

 

[jsfisher]

Wrong....

That's nice, but you still haven't told us how you are actually applying the axiom on ZF(C). Keep in mind, the axiom simply states that there exists a certain set with certain properties. Nothing more.

So, once again, what does "using ZF(C) Axiom Of Infinity on ZF(C) itself" mean?