Deeper than primes - Continuation

[jsfisher]

According to his one level reasoning, so?

No, according to ZFC.

 

[jsfisher]

"There exists AND does not exist set X" is a contradiction at the objective (platonic) level of discovery (or in other words, the objective (platonic) level is not discovered, which has no impact on the objective (platonic) existence of set X).

"There exists a set X" is not a complete statement. Neither is its complement; neither is the conjunction of the two. You cannot correctly say the conjunction is a contradiction simply because you never got far enough constructing the statement.

On the other hand, the statement, "for some X, X is a set" is a complete statement (a true one, in fact), and the statement, "for some X, X is a set and X is not a set" is a false (complete) statement. There are no levels to it being false, though.

"There exists set X" such that "it is AND it is not its own member" is a contradiction at the subjective (non-platonic) level of invention, which has an impact on X's subjective invented identity, but it does not have any impact on X objective level of discovery (X exists as a platonic object even if its non-platonic subjective level is not well-defined).

Curious use of quotations marks, Doron. Do you not understand the meaning they convey? At any rate, the statement, "there exists set X such that X is a member of X and X is not a member of X", is a false statement. There are no levels to it being false.

If there were levels, then you would be able to identify two such levels wherein the truth value for the statement differ.

 

[doronshadmi]

"There exists a set X" is not a complete statement.

"There exists a set X" is a complete objective platonic level of set X, which is discovered.

"There exists AND does not exist a set X" is contradiction that prevents the discovery of the complete objective platonic level of set X.

Yet, platonic objects exist independently of their discovery.

"there exists set X (the objective platonic level of discovery) such that (the non-platonic level of invention) X is a member of X and X is not a member of X", is a false statement at the subjective level of non-platonic invention.

If there were levels, then you would be able to identify two such levels wherein the truth value for the statement differ.

The truth value at the objective platonic level of set X (which is discovered) is only tautology, and it is indeed different from the truth value at subjective non-platonic level members (which is invented), which is not only a tautology.

The discovered is independent on the invented, but the invented depends on the discovered.

Your one level reasoning is not useful in order to deduce the difference between the discovered and the invented.

As a result you can't be aware of the inaccessibility of the invented to the discovered, as deduced by the following example:

{||}=0 , {|{}|}=1 , {|{},{{}}|}=2 etc. , or {|{}, {{}}, {{},{{}}}, ... |} < aleph0 (where aleph0 is at the objective level (notated here by the outer "{" and "}"), where no number of members (where members are at the subjective level) is accessible to it).

The outer "{" and "}" notates the notion of actual infinity, which is at the complete objective platonic level of set X, whether X has or does not have members, and if X is non-empty, then no number of its members (which are at the subjective non-platonic level of invention) is accessible to the level of actual infinity.

The

 

[doronshadmi]

No, according to ZFC.

jsfisher, this is no more than your interpretation of ZFC, which unable to deduce the difference between the platonic and the non-platonic, and how they are usefully used by ZFC.

 

[jsfisher]

...platonic objects...

So, despite all the pretense of discussing set theory and other actual mathematics, you really are just wallowing in Shadmized philosophy.

I will sleep better tonight now knowing all the defects in Mathematics you have alleged are actually just your philosophic musing. Mathematics remains unblemished by your assault.

 

[jsfisher]

jsfisher, this is no more than your interpretation of ZFC.

Nope. Nine little axioms don't leave much room for interpretation, and it wouldn't be Mathematics if they did.

 

[realpaladin]

According to his one level reasoning, so?

No, according to the strict and defined definition of ZFC.

It is time to fire up a new 'Doron Shadmi's errors' resumé I guess.

Doron, if you augment your reasoning statements in a way that makes it clear that *you* need to modify them so *your* theory works, then we all *have* to wait and see where you take us.

But as long as you try to make definitions of others seem 'broken', we can do naught but correct you...

And all of this is just rehashing the rehash that was rehashed 7 years ago.

We, the few that stick to this thread, have seen just about every argument you have made and we are eager to see where it will take us.

 

[doronshadmi]

Mathematics remains unblemished by your assault.

Wrong, Mathematics is based on Philosophy, whether you ignore it or not.

"There exists set X" is only a tautology, where what comes after "such that" is not only a tautology.

 

[realpaladin]

So, despite all the pretense of discussing set theory and other actual mathematics, you really are just wallowing in Shadmized philosophy.

I will sleep better tonight now knowing all the defects in Mathematics you have alleged are actually just your philosophic musing. Mathematics remains unblemished by your assault.

Wrong, Mathematics is based on Philosophy, whether you ignore it or not.

"There exists SET X" is only a tautology, where what comes after "such that" is not only a tautology.

Amazing... Doron manages to write a response that almost, but not entirely, completely shoots past the target...

Doron, JSFisher never said what your response alludes to.

I for one would welcome some rigour in your replies.

 

[doronshadmi]

Nope. Nine little axioms don't leave much room for interpretation, and it wouldn't be Mathematics if they did.

Yep. They are constructed by platonic and non-platonic levels of existence.

 

[realpaladin]

Yep. They are constructed by platonic and non-platonic levels of existence.

Wrong. They are not. There is no proof for that, anywhere.

 

[doronshadmi]

Curious use of quotations marks, Doron. Do you not understand the meaning they convey? At any rate, the statement, "there exists set X such that X is a member of X and X is not a member of X", is a false statement. There are no levels to it being false.

Let's use Cantor's Theorem in order to explicitly demonstrated the difference between the platonic level of existence and the non-platonic level of existence.

A = {a, b, c, ...}

P(A) = {{}, {a, b, c, ...}, ...}

By Cantor's theorems its is proved that |A|<|P(A)|, by using two steps:

1. |P(A)| can't be < |A| (for example, there are at least such pairs: (a,{a}), (b,{b}), (c,{c}), ...).

2. There is at least one member of P(A) (let's call it S), which is defined as "the set of all A members, which are not members of the P(A) members that are paired with them", so given some A member (let's call it J) to be paired with S, if J is not a member of S, then according to S definition, J is a member of S, but then according to S definition, J is not a member of S.

Since S is well-defined and since J membership with S leads to contradiction (J is a member AND not a member of S), it is concluded that S is not paired with any A member, which enables to prove that |P(A)|>|A|.

In other words, J can't be defined as some A member that is paired with S (which is a well-defined P(A) member).

Yet, the platonic objective level of existence (notated by the outer "{" and "}" in both sets) is independent on the non-platonic level of existence of the members these sets, and this notion is expressed in terms of cardinality as follows:

{|{}, {a, b, c, ...}, ...|} > {|a, b, c, ...|} where no number of members in both sets is accessible to the platonic objective level of existence of these sets.

Actually the platonic level is a common objective existence for both non-platonic subjective level of existence of members, where only at this level a contradiction like J membership, enables to conclude that J existence is false (which has an influence on P(A) and A identity, but not on P(A) and A platonic level of common objective existence).

Again, the platonic objective level of set is a discovered tautology, where the non-platonic subjective level of members is an invetion that is not necessarily tautology (as can be seen in the case of J).

 

[jsfisher]

Let's use Cantor's Theorem in order to explicitly demonstrated the difference between the platonic level of existence and the non-platonic level of existence.

Don't bother. Cantor's Theorem is forged in Mathematics--rigorous, consistent, and well-defined. Shadmized philosophy, not so much. Whether you accept or agree with Cantor on a philosophic basis has no impact on the math.

...
2. There is at least one member of P(A) (let's call it S), which is defined as "the set of all A members, which are not members of the P(A) members that are paired with them"
...

And this shows another reason you shouldn't be commenting on Cantor's Theorem. You cannot even express the proof correctly. Much like your mindless repetition of the Axiom of Infinity, you omit rather important detail and you insert things that simply aren't there. And you get things ass-backwards.

 

[realpaladin]

Let's use Cantor's Theorem in order to explicitly demonstrated the difference between the platonic level of existence and the non-platonic level of existence.

<ERRONEOUS TRAIN OF THOUGHT CUT OUT>

Again, the platonic objective level of set is a discovered tautology, where the non-platonic subjective level of members is an invetion that is not necessarily tautology (as can be seen in the case of J).

And again, the above only is of the category 'observation' (for want of a better word).

Doron never reaches anything near the category 'conclusion'.

If one were to take the whole of the 7+ year thread of Doron's meandering philosophizing, one could only conclude that Organic Mathematics never has lifted off (one may even be so bold as to conclude it never reaches the launchpad...).

Does anyone see anything of worth in this current strain of Doronetics?

 

[doronshadmi]

you omit rather important detail and you insert things that simply aren't there.

Please demonstrate your claim in details, according to what I wrote in here

...
2. There is at least one member of P(A) (let's call it S), which is defined as "the set of all A members, which are not members of the P(A) members that are paired with them"
...

Here are two examples of such P(A) member, that is not paired with any A member:



example 1:

{} (which is some P(A) member) is defined as "the set of all A members, which are not members of the P(A) members that are paired with them", by the following example:

{ a , b , c , ... }
  ↕   ↕   ↕    
{{a},{b},{c}, ... }

Since all A members are members of the P(A) members that are paired with them, we get {} (the empty set) as the P(A) member that is not paired with any A member.



example 2:

{a,b,c,...} (which is some P(A) member) is defined as "the set of all A members, which are not members of the P(A) members that are paired with them", by the following example:

{ a , b , c , ... }
  ↕   ↕   ↕    
{{b},{c},{d}, ... }

Since all A members are not members of the P(A) members that are paired with them, we get {a, b, c, ...} (set A) as the P(A) member that is not paired with any A member.


I am waiting to your detailed demonstration of your quoted claim above.

 

[jsfisher]

you omit rather important detail and you insert things that simply aren't there.

Please demonstrate your claim in details, according to what I wrote in here

...
2. There is at least one member of P(A) (let's call it S), which is defined as "the set of all A members, which are not members of the P(A) members that are paired with them"
...

• Your set A is a specific set, yet Cantor's Theorem doesn't deal with any specific set.
• "members that are paired" -- what pairing would that be?
• "the set of all A members" -- an odd bit of gibberish. Did you mean "the set of all the members of A"? Wouldn't that be just the set A? Or maybe you haven't mastered the fine art of the comma.
• "(let's call it S)" -- Why? You don't mention S again.
• Where in the proof of Cantor's Theorem is "at least one member of P(A)" defined as you claim? (Order is important. A constructed set that is then shown to be a member of P(A) wouldn't be the same thing.)
• ...

 

[doronshadmi]

*Your set A is a specific set, yet Cantor's Theorem doesn't deal with any specific set.

{a,b,c, ...} is an example without loss of generality of set A, where {{}, {a, b, c, ...}, ...} is an example without loss of generality of set P(A) (the power set of A), so we no not deal with any specific set or its power set.

* "members that are paired" -- what pairing would that be?

See the examples (which are without loss of generality) in http://www.internationalskeptics.com/forums/showpost.php?p=10030370&postcount=3795.

* Or maybe you haven't mastered the fine art of the comma.

I agree with your remark, thank you. At has to be "the set of all A members that are not members of the P(A) members that are paired with them".

* "(let's call it S)" -- Why? You don't mention S again.

S is used here without a loss of generality as "the set of all A members that are not members of the P(A) members that are paired with them", and it is mentioned more than once in http://www.internationalskeptics.com/forums/showpost.php?p=10029127&postcount=3792.

* Where in the proof of Cantor's Theorem is "at least one member of P(A)" defined as you claim?

By using Cantor's theorem please conclude that |P(A)|>|A| without "at least one member of P(A) that is not paired with any member of A".

*...

*...

So jsfisher, thank you for your remark about my (,) wrong use, but since it was corrected it can be seen that you actually have no case (according to your attitude, papers like
http://www.math.umaine.edu/~farlow/sec26.pdf
http://www.math.ucla.edu/~hbe/resource/general/131a.3.06w/cantor.pdf
should not be written, so?).

 

[jsfisher]

{a,b,c, ...} is an example without loss of generality of set A, where {{}, {a, b, c, ...}, ...} is an example without loss of generality of set P(A) (the power set of A), so we no not deal with any specific set or its power set.

So, you agree this claim of "without loss of generality" was something you left out of the current presentation.

See the examples (which are without loss of generality) in http://www.internationalskeptics.com/forums/showpost.php?p=10030370&postcount=3795.

So, you agree these examples were something you left out of the current presentation.

I agree with your remark, thank you. At has to be "the set of all A members that are not members of the P(A) members that are paired with them".

Now that you have addressed the extraneous comma issue, perhaps you'd consider dealing with the gibberish aspect.

S is used here without a loss of generality as "the set of all A members that are not members of the P(A) members that are paired with them", and it is mentioned more than once in http://www.internationalskeptics.com/forums/showpost.php?p=10029127&postcount=3792.

So, you agree this other mention was something you left out of the current presentation.

By using Cantor's theorem please conclude that |P(A)|>|A| without "at least one member of P(A) that is not paired with any member of A".

Since Cantor's Theorem establishes that |P(A)| is strictly greater than |A|, there is nothing further needed to conclude |P(A)| > |A|.

However, this is an aside and unrelated to the point you were (mis-)responding to.

 

[doronshadmi]

So, you agree this claim of "without loss of generality" was something you left out of the current presentation.

I agree with you that I wrongly assumed that your professional level of mathematics enables you to understand that {a,b,c,...} or {{},{a,b,c,...},...} are expressions that are used "without loss of generality".

Now I realize that my assumption was wrong.

Since Cantor's Theorem establishes that |P(A)| is strictly greater than |A|, there is nothing further needed to conclude |P(A)| > |A|.

Cantor's Theorem establishes that |P(A)| is strictly greater than |A|, exactly because it establishes that there is P(A) member that is not paired with any A member.

 

[doronshadmi]

However, this is an aside and unrelated to the point you were (mis-)responding to.

Please direct me to the point I were (mis-)responding to.