A hidden assumption

nathan said:
ahem, I wouldn't want to derail this truly fascinating discussion in any way -- no really I wouldn't. But I feel compelled to point out that 2's primality is important when considering binary approximations of decimal approximations of real numbers (and decimal approximations of binary approximations of real numbers).

Carry on.

:popcorn1
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That would be a nathan's famous corndog of course.

I think the transform function here would go like this:

Reality >{X} > Fantasy where {X} = mathematical jargon

Tis but a rough scaffolding feel free to tear it down. If you dare without threatening the foundations of your sanity. lol

I shall now have to soak my hands in vinegar!
 
We seem to have achieved understanding, for the moment.

Now if only you could express what all this symmetry/assymetry, researchable, complementation and whatnot is about so that people can understand, we may have a discussion. And please don't handle all of them at once, as you seem to have done so far. One is plenty.
 
doronshadmi said:
My argument is wrongly understood by you if you think that I claim that 2 exclusively belongs to any one of the titles that you used above.
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That's not the problem. At least, not the main problem.

In that case 2 is independent of any special membership.
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This is the problem. It's complete nonsense.

The number 2 is a member of infinitely many sets. It's never independent of those sets, indeed, the very notion is contrary to the concept of a set.

[So is the case of E={A,B} or F={A,C,D}

X={E,F}

E or F are independent of any special membership like X, exactly as 2 is independent of any special membership.
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Nope. You can consider E and F without bothering to take into account all the sets that they are members of, so long as the sets are not relevant to the theorem under consideration. That doesn't make them independent of those sets. You're just ignoring that fact for the moment.

So the result of E U F is not exclusively related to X, unless we explicitly determine that E and F are X members.
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E U F isn't X, exclusively or otherwise, and we have already explicitly determined that E and F are members of X, because you just said so.

[quote[In this case we are not dealing with E U F = {A,B,C,D} but we deal with {E,F}U{E,F}={E,F}=X[/quote]
So now you accept that sets aren't the union of their members?
 
There is a problem of Time response ...

Can anyone knows how this post can be erased?
 
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There is a problem of Time response ...

Can anyone knows how this post can be erased?
 
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jsfisher said:
No, its set memberships are simply properties of 2. The number has many, many properties, and it is independent of none of them.
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Exactly as a placeholder does not depend on any term that replaces it.

jsfisher said:
By the way, your comments of "exclusive" vs "explicitly" shows a shift in your position. You did in fact say and mean "exclusive" before, culminating in the severed body part analogy. You statements and examples were clear. In effect, you were claiming a sort of Heisenberg Uncertainty Principle for set theory. You are now back-pedaling.
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What I said is this:

If (by mistake, because of my English problems) I wrote "explicitly" and meant "exclusive" then I apologize for the confusion that I made.
 
Alas, I canot. I lack the following definitions:

* researchable
* complementation
* symmetry

And, for that matter, what the supposed hidden assumption is...
 
doronshadmi said:
When a name is used in some general definition, it must be understood as no more than a placeholder ( http://en.wikipedia.org/wiki/Placeholder ).
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Your own citation calls you a liar.

Wikipedia said:
In disciplines involving formal languages, particularly mathematics and logic, the placeholder concept is replaced by the more carefully defined concepts of free variables and bound variables. In computer science placeholders are sometimes referred to as metasyntactic variables.
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In particular, once you say "E={A,B}" and "F={A,C,D}", E and F are no longer free variables and may not be manipulated arbitrarily. Similarly, once you say that "AA=E",
AA becomes a bound variable and may no longer be used to represent X (since X is distinct from E).

Similarly, once GNARF is defined as an unmarried male, the free variable GNARF becomes bound to a specific use for that specific discourse....

This is a fundamental notion of general mathematical definitons.
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... and your misunderstanding of such a fundamental notion is one reason why your pseudotheory is gibberish.
 
doronshadmi said:
Bravo!

Now explain it to jsfisher (he has some problem to understand the concept of placeholder)
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Not necessary. It is a concept I understand. It was your notation and use of language that was misleading.


Be that as it may, why do you continue to focus on X U X? It has nothing to do with the ZFC Axiom of Union nor with your incorrect assertion that a set is equal to the union of its members.

(By the way, for any set X, X = X U X. That is something of an identity.


ETA: The = was meant to be an equivalence symbol. I didn't have one handy, so I just underscored the equal sign.)
 
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doronshadmi said:
What I said is this:

If (by mistake, because of my English problems) I wrote "explicitly" and meant "exclusive" then I apologize for the confusion that I made.
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Oh, gee! You really meant that! I was giving you the benefit of the doubt, that you flopped the word order, but, no!, you really did mean "exclusive".

So you actually think that taking an interest in an object's membership in one set has to be at the exclusion of any other membership property the object may have.
 
drkitten said:
In particular, once you say "E={A,B}" and "F={A,C,D}", E and F are no longer free variables and may not be manipulated arbitrarily. Similarly, once you say that "AA=E",
AA becomes a bound variable and may no longer be used to represent X (since X is distinct from E).

Similarly, once GNARF is defined as an unmarried male, the free variable GNARF becomes bound to a specific use for that specific discourse....

... and your misunderstanding of such a fundamental notion is one reason why your pseudotheory is gibberish.
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AA or BB are placehoders of some sets.

Let us say that these sets are non-empty.

Nothing in the definition of the union determines that AA or BB must be different or the same set.

Furthermore, nothing in the definition of the union prevents the ability to simultaneously define AA={A,B} U BB={A,C,D}={A,B,C,D} or
AA={{A,B},{A,C,D}} U BB = {{A,B},{A,C,D}} = {{A,B},{A,C,D}} since AA or BB are placeholders.

It is a problem if you are limited to a step-by-step serial thinking.

It is not a problem if you understand the definition of the Union in terms of a parallel thinking, as I do in this case.

Mathematics is parallel\serial thinking.
 
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jsfisher said:
It has nothing to do with the ZFC Axiom of Union nor with your incorrect assertion that a set is equal to the union of its members.
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What??

Do you still think that we are talking about ZFC Axiom of Union ?

Let me refresh your memory:

http://www.internationalskeptics.com/forums/showpost.php?p=3540669&postcount=88

Here is set X={{A,B},{A,C,D}}

The definition of the union: If AA and BB are sets, then the union of AA and BB is the set that contains all elements of AA and all elements of BB, but no other elements. The union of AA and BB is usually written "AA U BB".

AA or BB are placeholders of any given set.

I state that according to the definition of the union, the union of the members of set X is set X as follows:

AA={A,B} U BB={A,C,D} = {A,B,C,D} but {A,B} or {A,C,D} are not necessarily the members of {{A,B},{A,C,D}}=X

So {A,B,C,D} result is not necessarily related to X and it is the result of two different sets ({A,B} and {A,C,D}) .

But my argument cannot be but about a one and only one set called X.

In that case AA=BB=X={{A,B},{A,C,D}}U{{A,B},{A,C,D}} and the union of the members of set X is set X.
 
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doronshadmi said:
I state that according to the definition of the union, the union of the members of set X is set X as follows:

AA={A,B} U BB={A,C,D} = {A,B,C,D) but {A,B} or {A,C,D} are not necessarily the members of {{A,B},{A,C,D}}=X
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This must be some new and interesting meanings of "membership" and standard notation for sets with which I was previously unfamiliar. Under the meaning with which I am familiar, we write X = { {A,B}, {A,C,D} } to indicate that X is a set of the two members, {A,B} and {A,C,D}.

So {A,B,C,D} result is not necessarily related to X and it is the result of two different sets ({A,B} and {A,C,D}) .
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Unrelated to X (whatever that really means) except that those two sets are the members of set X.

But my argument cannot be but about a one and only one set called X.

In that case AA=BB=X={{A,B},{A,C,D}}U{{A,B},{A,C,D}} and the union of the members of set X is set X.
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Your argument continues to be junk. X is not a member of the set X, so why are you even considering X U X? That is simply not the union of the members of X.
 
jsfisher said:
This must be some new and interesting meanings of "membership" and standard notation for sets with which I was previously unfamiliar. Under the meaning with which I am familiar, we write X = { {A,B}, {A,C,D} } to indicate that X is a set of the two members, {A,B} and {A,C,D}.



Unrelated to X (whatever that really means) except that those two sets are the members of set X.



Your argument continues to be junk. X is not a member of the set X, so why are you even considering X U X? That is simply not the union of the members of X.
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Dear jsfisher,

This is beyond your current ability if you cannot grasp the fact that I am talking about a one and only one set, which is in union with itself, and this self union is based exactly on the definiton of the union between the members of AA=X and the members of BB=X.
as follows {{A,B},{A,C,D}}U{{A,B},{A,C,D}}={{A,B},{A,C,D}}.
 
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doronshadmi said:
...patronizing nonsense...I am talking about a one and only one set, which is in union with itself, and this self union is based exactly on the definiton of the union between the members of AA=X and the members of BB=X.
as follows {{A,B},{A,C,D}}U{{A,B},{A,C,D}}={{A,B},{A,C,D}}.
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Your premise was this: A set is equal to the union of its members.

Your premise is false. To prove it false, it is sufficient to show one example for which your premise does not hold.

Here is an example: Let Q be the set of members {R} and {S}. (We could express this as Q = { {R}, {S} }.)

The members of Q are {R} and {S}.
The union of the members of Q is {R} U {S}.
The union of {R} U {S} is {R, S}

Q != {R, S}
Q != the union of the members of Q
The premise does not hold for this example, Q.

Therefore, the premise is false.

Any questions?
 
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doronshadmi said:
AA or BB are placehoders of some sets.
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No, they're not. They're free variables. The term "placeholder" is inappropriate in this instance precisely because it leads you into the sort of confusion in which you are wallowing.

They will remain free variables UNTIL and UNLESS you bind them to some instance in the discourse.

Let us say that these sets are non-empty.

Nothing in the definition of the union determines that AA or BB must be different or the same set.
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Nope. As free variables, they have no properties (yet) whatsoever.


Furthermore, nothing in the definition of the union prevents the ability to simultaneously define AA={A,B} U BB={A,C,D}={A,B,C,D} or
AA={{A,B},{A,C,D}} U BB = {{A,B},{A,C,D}} = {{A,B},{A,C,D}} since AA or BB are placeholders.
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Nothing in the definition of the union perhaps, but the definition of free and bound variables prevents you from making simultaneous incompatible definitions.

As soon as you bind AA to {A,B}, it is no longer capable of simultaneously being {{A,B}, {A,C,D}}.

Let me guess: you weren't very good at secondary school geometry, were you?


It is a problem if you are limited to a step-by-step serial thinking.[/QUTOE]

It is not a problem if you understand the definition of the Union in terms of a parallel thinking, as I do in this case.

Mathematics is parallel\serial thinking.
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Yeah. Damn me for doing mathematics with only step-by-step serial thinking. I wonder what all the journal editors were thinking when they accepted my papers....
 
jsfisher said:
Here is an example: Let Q be the set of members {R} and {S}. (We could express this as Q = { {R}, {S} }.)
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No, Q is exactly { {R}, {S} }.

Any non-empty set is first of all a collection of distinct objects (called members).

Without its members no non-empty set can be defined.

{R} or {S} are non-empty sets exactly as { {R}, {S} } is a non-empty set.

Since the non-empty set Q cannot be defined without its members, then the union of the members of Q is set Q={ {R}, {S} } U Q={ {R}, {S} } = Q={ {R}, {S} }

{R} U {S} = {R,S} is the union between the members of the non-empty set {R} and the members of the non-empty set {S}, and this result has nothing to do with Q.
 
You are incorrect.

If X = { A, B, C}, then "the union of the members of X" is A u B u C, sometimes written as U X.

The set X may be harder to define without its members, but the members themselves exists happily without the set. A, B and C may participate in any expression that is valid for them without considering their membership in X.
 
ArmillarySphere said:
You are incorrect.

If X = { A, B, C}, then "the union of the members of X" is A u B u C, sometimes written as U X.

The set X may be harder to define without its members, but the members themselves exists happily without the set.
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Good, in that case the happy objects A,B or C and any result that is deduced from them,
is independent of X = {A, B, C}.

Again, X = {A, B, C} and both "{","}" and "A","B","C" must not be ignored.
 
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drkitten said:
No, they're not. They're free variables.
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The term free-variable is simply the result if a serial thinking, for example:

If X represents some term, then according to a serial thinking it cannot be used to represent another term simultaneously (by using a parallel thinking, for example).

There is nothing general here, but no more than some particular way of thinking.

Mathematics is based on parallel\serial complementation, where only-serial or only-parallel are nothing but some particular and extreme states of this complementation.

This parallel\serial complementation is exactly the answer to Hilbert's organic paradigm of the mathematical science.

After suggesting 23 problems, Hilbert spoke about the 24th problem, but his organic paradigm of the mathematical science went over the heads of the community of mathematicians.

Hilbert closed his 1900 famous lecture by these words:

“…The problems mentioned are merely samples of problems, yet they will suffice to show how rich, how manifold and how extensive the mathematical science of today is, and the question is urged upon us whether mathematics is doomed to the fate of those other sciences that have split up into separate branches, whose representatives scarcely understand one another and whose connection becomes ever more loose. I do not believe this nor wish it. Mathematical science is in my opinion an indivisible whole, an organism whose vitality is conditioned upon the connection of its parts.”

I think that Hilbert saved his most important notion and aired his view about it at the end of the lecture.

I also think that he did it for purpose, because generally a person put his most important notions at the end of his lecture, in order to not to be forgotten.

He explicitly says:

"The problems mentioned are merely samples of problems, yet they will suffice to show how rich, how manifold and how extensive the mathematical science of today is."

But then he immediately continues (I call it the 24th problem):

" … , and the question is urged upon us whether mathematics is doomed to the fate of those other sciences that have split up into separate branches, whose representatives scarcely understand one another and whose connection becomes ever more loose."

Then he expresses his most important notion (which is the solution to the 24th problem):

"I do not believe this nor wish it. Mathematical science is in my opinion an indivisible whole, an organism whose vitality is conditioned upon the connection of its parts.”
drkitten said:
Yeah. Damn me for doing mathematics with only step-by-step serial thinking. I wonder what all the journal editors were thinking when they accepted my papers....
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Let me guess: you and the journal editors are no more than the particular case of serial thinking, don't you?
 
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