Cont: Deeper than primes - Continuation 2

[doronshadmi]

Doron,

Does there exist one other person, just one, that agrees with your, um, novel views on Mathematics? There was your associate Moshe Klein, but even he abandoned you after the treatment you gave him here.

Just curious.

jsfisher,

You are far from really being curious. All you wish is to pack math in your standard box.

Oh, by the way, Mathematics continues to work despite all your criticism of it.

It can be improved by using {X} as an optional successor of X as very simply given in http://www.internationalskeptics.com/forums/showpost.php?p=11337788&postcount=1723.

 

[doronshadmi]

You'd be better served focusing on the claim rather than the claimant.

Already done all along this thread, but by using jsfisheretics it surely can't be seen.

Doron interprets "X OR ~X" as a literal statement in English, "X is true or not X is true." The OR is taken as a conjunction, not a dyadic boolean operator.

Doronetics isn't about getting anything right. It is all about claiming everything else is wrong. More economical that way.

Does there exist one other person, just one, that agrees with your, um, novel views on Mathematics?

Says the hypocrite.

 

[jsfisher]

jsfisher,

You are far from really being curious. All you wish is to pack math in your standard box.

No, my desire is that Mathematics continue to work...and it does.

It can be improved by using {X} as an optional successor of X as very simply given in http://www.internationalskeptics.com/forums/showpost.php?p=11337788&postcount=1723.

You claim many things, but you demonstrate no improvement.

By the way, do you realize there are more set theories than just ZFC? Do you realize there are other versions of the axiom of infinity besides the one used in ZFC? Do you realize one of the variations uses the condition if X is a member of the set, then {X} is a member of the set? Do you know why ZFC doesn't use that variation?

 

[doronshadmi]

No, my desire is that Mathematics continue to work...and it does.

Well, your desire is that Mathematics continue to work under your artificial restrictions.

You claim many things, but you demonstrate no improvement.

By using jsfisheretics in order to comprehend doronetics, no improvement can be seen.

EDIT:

Do you realize one of the variations uses the condition if X is a member of the set, then {X} is a member of the set?

Please provide this variation, which also shows that this set is complete.

 

[jsfisher]

Well, your desire is that Mathematics continue to work under your artificial restrictions.

The restrictions, as you call them, are neither artificial nor are they mine.

By using jsfisheretics in order to comprehend doronetics, no improvement can be seen.

Doronetics is your own personal fantasy shared by no one. It is fair to name it as such. I have no personal version of Mathematics, so I have no idea why you would attempt to present Mathematics as if it were.

http://www.internationalskeptics.com/forums/showpost.php?p=11315376&postcount=1664 clearly demonstrates that jsfisheretics have no case.

How does (mis-)presenting the ZF(C) version of the Axiom of Infinity "clearly demonstrate" anything about other versions?

 

[doronshadmi]

The restrictions, as you call them, are neither artificial nor are they mine.



Doronetics is your own personal fantasy shared by no one. It is fair to name it as such. I have no personal version of Mathematics, so I have no idea why you would attempt to present Mathematics as if it were.

You are the one who used the words "my desire is that Mathematics continue to work...and it does" so please do not play your "non personal" game about Mathematics.

How does (mis-)presenting the ZF(C) version of the Axiom of Infinity "clearly demonstrate" anything about other versions?

1) You are the one that (mis-)presenting the ZF(C) version of the Axiom of Infinity, as clearly seen in http://www.internationalskeptics.com/forums/showpost.php?p=11315376&postcount=1664.

2) I have corrected the reply on this part in my previous post, so please reply to the correction.

 

[jsfisher]

You are the one who used the words "my desire is that Mathematics continue to work...and it does" so please do not play your "non personal" game about Mathematics.

I have a desire for world peace, too, but that doesn't make it my personal edition.

Mathematics is not mine. Doronetics is exclusively yours.

2) I have corrected the reply on this part in my previous post, so please reply to the correction.

I have no interest in monitoring your posts for their frequent content-altering edits. You did not correct your post; you altered its content in a significant way.

 

[doronshadmi]

I have no interest in monitoring your posts for their frequent content-altering edits. You did not correct your post; you altered its content in a significant way.

Do you realize one of the variations uses the condition if X is a member of the set, then {X} is a member of the set?

Please provide this variation, which also shows that this set is complete.

 

[jsfisher]

Please provide this variation, which also shows that this set is complete.

Define "complete" so we will all know that you are talking about.

 

[doronshadmi]

I have a desire for world peace, too, but that doesn't make it my personal edition.

Once again your restricted reasoning rises its head.

World peace may be achieved in case that it is really a daily desire of most persons in our planet (where you and me are probably such persons).

 

[zooterkin]

By doing that you actually determine that it is complete, exactly becuae by zooterkinetics Successor(x) = x u {x}.

Are you now saying that the set of natural numbers is complete?

 

[doronshadmi]

Are you now saying that the set of natural numbers is complete?

No, it is actually incomplete but you wrongly determine it as a complete set exactly because you are unaware of the option that {n} is the successor of n.

In other words, the set of natural numbers does not exist.

You can talk about the set of natural numbers as much as you like, but it does not change the mathematical fact that is does not exist.

 

[doronshadmi]

Define "complete" so we will all know that you are talking about.

{X}~$X and as a result set X is complete.

You claim the following:

If set X is a member of set S, then set {X} is a member of set S.

Since set {X} must be a singleton set, then set S must include all the singleton sets.

But given set S, set {S} is not its member and {S}$S is a possible option, or on other words, set S is incomplete.

So I'll ask you again:

Please provide this variation, which also shows that this set (called S or whatever) is complete (includes all the singleton sets).

 

[jsfisher]

{X}~$X and as a result set X is complete.

Ok. Something that isn't a part of Mathematics.

You claim the following:

If set X is a member of set S, then set {X} is a member of set S.

I made no such claim.

Since set {X} must be a singleton set, then set S must include all the singleton sets.

That doesn't follow from the premise. For example, if your set S is the smallest such set containing {}, then {{ {}, {{}} }}, a singleton set you may note, is not a member of S.

You really need to stop spouting claims that (a) aren't Mathematics, and (b) aren't true.

 

[Little 10 Toes]

No, it is actually incomplete but you wrongly determine it as a complete set exactly because you are unaware of the option that {n} is the successor of n.

In other words, the set of natural numbers does not exist.

You can talk about the set of natural numbers as much as you like, but it does not change the mathematical fact that is does not exist.

Yet, you can't define successor. And the set of natural numbers does exist. It existed before your idea, it currently exists, and it will continue to exist after you ideas are forgotten.

Ps: you still haven't defined successor.

 

[doronshadmi]

I made no such claim.

Really? Here it is:

Do you realize one of the variations uses the condition if X is a member of the set, then {X} is a member of the set?

That doesn't follow from the premise. For example, if your set S is the smallest such set containing {}, then {{ {}, {{}} }}, a singleton set you may note, is not a member of S.

Let's see:

X = {}

If {} is a member of the set, then {{}} is a member of the set, so our set is { {}, {{}} }.

Since you claim that form {X} is a member of the set, the member of { {{}, {{}}} } is such a form so the set is now { {}, {{}}, {{}, {{}}} } etc. ... af infinitum.

On the other hand, if by

if X is a member of the set, then {X} is a member of the set

we get a finite set (because {X} is not a general form of a member of the set ({ {}, {{}}, {{{}}}, {{{{}}}},... } is not the result of {X} as a general form of a singleton set) you can't claim that

if X is a member of the set, then {X} is a member of the set

is some other version of the axiom of infinity, as you wrongly do in

Do you realize there are other versions of the axiom of infinity besides the one used in ZFC? Do you realize one of the variations uses the condition if X is a member of the set, then {X} is a member of the set?

 

[jsfisher]

I made no such claim.

Really? Here it is:

Do you realize one of the variations uses the condition if X is a member of the set, then {X} is a member of the set?

That is where I commented about the existence of variations on the Axiom of Infinity. Notice, however, I didn't make the claim you said I did.

That doesn't follow from the premise. For example, if your set S is the smallest such set containing {}, then {{ {}, {{}} }}, a singleton set you may note, is not a member of S.

Let's see:

X = {}

If {} is a member of the set, then {{}} is a member of the set, so our set is { {}, {{}} }.

Since you claim that form {X} is a member of the set, the member of { {{}, {{}}} } is such a form so the set is now { {}, {{}}, {{}, {{}}} } etc. ... af infinitum.

Yeah, so? Nowhere does {{ {}, {{}} }} appear as a member of that set you are forming. Your claim the set must contain all singleton sets is flatly wrong.

ETA: Any by the way, there is no claim that "form {X} is a member of the set". If X is in the set, then {X} is in the set.

On the other hand, if by

if X is a member of the set, then {X} is a member of the set

we get a finite set

You'd get the set { {}, {{}}, {{{}}}, {{{{}}}}, ...}. How is that not an infinite set?

 

[doronshadmi]

You'd get the set { {}, {{}}, {{{}}}, {{{{}}}}, ...}. How is that not an infinite set?

I already wrote about this set in my previous post, but I was wrong and unclear about my claims, so here is my corrected and clearer reply about "if X is a member of the set, then {X} is a member of the set".

1) By using such set, the set of natural numbers can't be defined, since {0, 1, 1, 1, ...} is the best you can get.

2) Since {0,1} = {0,1,1,1,...} by the common notion of sets, what you call the set is actually {0,1} (since you are using the phrase "the set" this set can't be used in order to define the set of natural numbers).

3) If we take { {}, {{}}, {{{}}}, {{{{}}}}, ...} as some case of an inductive set, where {X} is a successor of X, then { {}, {{}}, {{{}}}, {{{{}}}}, ...} is infinite AND incomplete from within.

So no matter how we look at it, by "if X is a member of the set, then {X} is a member of the set" one can't define a set of natural numbers, where also { {}, {{}}, {{{}}}, {{{{}}}}, ...} is infinite AND incomplete set from within since {X} is a successor of X within it.

 

[doronshadmi]

4.1 Zermelo set theory

Infinity: There is an infinite set. Zermelo's original formulation asserted the existence of a set containing ∅ and closed under the singleton operation: {∅ ,{∅},{{∅}}, …}. It is now more usual to assert the existence of a set which contains ∅ and is closed under the von Neumann successor operation x --> x ∪ {x}. (Neither of these axioms implies the other in the presence of the other axioms, though they yield theories with the same mathematical strength).

http://plato.stanford.edu/entries/settheory-alternative/

So in both cases (Zermelo's original formulation OR Von Neumann successor operation) if {Z}$Z, then the involved set is infinite AND incomplete.


-------------

By more comprehensive view the the concept of collections, the void between the outer "{" and "}" is weaker than any amount of existing members, where the outer "{" and "}" is stronger than any amount of existing members.

By using this notion {||}=0 where |{}|= ∞

 

[doronshadmi]

By using {X} as a successor of X, any amount of members approaches but not reaches ∞, therefore such an amount can't be determined and therefore it is incomplete.

By not using {X} as a successor of X, no amount of members approaches ∞, therefore such an amount can be determined and therefore it is complete.

So no matter if a given non-empty set is "fed" externally or internally by {X}, in both cases the amount of its existing members can't be determined and therefore it is incomplete.

The following example in http://www.internationalskeptics.com/forums/showpost.php?p=11279727&postcount=1485 holds for both external or internal options.