A hidden assumption

[ArmillarySphere]

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.

The result that you say is independent is what you claim X is.

A set is not the union of its members.
A set may be considered as the union of a group of other sets, each containing a single member of the set. However, this definition seems hardly useful.

1) X = {A} u {B} u {C}

2) UX = A u B u C

These two statements are not equivalent.

 

[doronshadmi]

The result that you say is independent is what you claim X is.

A set is not the union of its members.
A set may be considered as the union of a group of other sets, each containing a single member of the set. However, this definition seems hardly useful.

1) X = {A} u {B} u {C}

2) UX = A u B u C

These two statements are not equivalent.

No,

{A},{B},{C},A,B,C are all independent of {A,B,C}, because X is not less than "{" and "A,B,C" and "}".

You get X by {A} u {B} u {C} but not because they are its members, and we are talking about the union of the members of X={A,B,C} U X={A,B,C} = X={A,B,C}.

 

[ArmillarySphere]

No,

{A},{B},{C},A,B,C are all independent of {A,B,C}, because X is not less than "{" and "A,B,C" and "}".

You get X by {A} u {B} u {C} but not because they are its members,

This is true. They're proper subsets of X, not members.

and we are talking about
the union of the members of X = {A,B,C} U X

My bolding. This equality does not hold. "The union of the members of X" does not mean X u X. It means U X, which is an entirely different thing.

 

[PixyMisa]

{A},{B},{C},A,B,C are all independent of {A,B,C}, because X is not less than "{" and "A,B,C" and "}".

I had always considered "meaningless" to be an absolute. A statement either has some meaning, or it is meaningless. One statement can't be more meaningless than another.

I was wrong.

 

[Apathia]

I had always considered "meaningless" to be an absolute. A statement either has some meaning, or it is meaningless. One statement can't be more meaningless than another.

I was wrong.

Ah Pixy, you are breaking free of the bonds of "serial thinking!" lol

This is Lewis Carroll material here. I'll going to follow this thread. There seems to be much of an ur-muchness going on here!

 

[jsfisher]

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

No, Q is exactly { {R}, {S} }.

You missed the point. The sequence of characters, "{ {R}, {S} }", is just a notational convenience to express what has been identified as the set Q. The braces and commas and such are actually part of Q and...oh, what's the point.

...meaningless aside...

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} }.

Since...then? There is no logical connective between the two. The union of {R} and {S} is {R, S}. That is not the union of the members of {R} and {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.

There. Fixed it for you.

You really need to work on that reading comprehension thing. You are missing so much.

 

[drkitten]

No, Q is exactly { {R}, {S} }.

Unless, of course, you decide to bind it to something else as well?

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

True. so far.

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

Wrong but irrelevant.

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

True so far.

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} }

Wrong. If Q is { {R}, {S} }, then the members of Q are {R} and {S}.

{R} U {S} is {R,S} which is not the original definition of Q.

Now, if you create a singleton set from every element of Q : for example, create the further sets {{R}} and {{S}}, then their union is { {R}, {S} }, which is Q. But this requires a further step, since neither {{R}} nor {{S}} is a member of Q.

The notation UQ refers to the union of all elements of Q -- but in order use this notation, the elements of Q must all themeslves be sets. This should be a strong clue that, in general, UQ != Q, since if the elements of Q are not sets, UQ isn't even defined.

(Furthermore, if UQ did =Q, there woudn't really need to be a need for the UQ notation, would there?)

{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.

No, {R,S} is the union of the members of Q. Which is distinct from, but related to, Q. Alterantively, it is the union of {R} and {S}, which is distinct from the union of the members of {R} and {S}. You're introducting an additional level of set-membership (or dropping it) and your results are therefore wrong.


And, yes, this is true corn-dog territory. She said so herself:

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).

"Parallel thinking" is therefore a method of effecting an arbitrary transformation on an expression via the fallacy of equivocation. Let E be an expression and x a variable free in E. Bind x to E and to any other expression of your choice ["simultaneously (by using a parallel thinking, for example)"]. Assert that that E and the new expression are both equivalent to x and therefore to each other. Q.e.d.

 

[Apathia]

The Empty Set:

{ }

But is it really empty?
Actually it's chock full of virtual elements:

{v, v1, v2, vn}

Formost among the virtual elements is the "gme"
(Grocho Marx Element)
that refuses to be a member of any set that would have it as a member.

Something gives me the impression that Doron would have something to tell us about virtual members of sets.
If not,
Happy April Fools to all!

 

[doronshadmi]

You missed the point. The sequence of characters, "{ {R}, {S} }", is just a notational convenience to express what has been identified as the set Q. The braces and commas and such are actually part of Q and...oh, what's the point.

Q is not less than {{R},{S}} where the outer braces is its whole side and {R},{S} is its parts side (the empty set is that has no parts).

This is the organic paradigm of the concept of set, based on the solution of Hilbert's 24th problem, that you and your friend here are unaware of it, because you are using a reductionist step-by-step approach, which is nothing but a particular case of the organic paradigm, which is based on parallel\serial complementation.

You have no chance to understand my theory by using only a reductionist step-by-step approach.

Without a paradigm-shift in your mind, what I say is indeed meaningless, wrong, ...etc. from your point of view.

I am looking at my theory from your point of view and I get the "content" of { }.

 

[drkitten]

You have no chance to understand my theory by using only a reductionist step-by-step approach.

This is true. In fact, it's much weaker than necessary. No one -- including yourself -- has any chance of to understand your hteory by using any approach whatsoever, because there is no content there to understand.

Without a paradigm-shift in your mind, what I say is indeed meaningless, wrong, ...etc. from your point of view.

Again, this is unnecessarily weak. With or without a paradigm shift in any mind, what you say is indeed meaningless, wrong, ...etc from any point of view whatsoever.

 

[doronshadmi]

No one -- including yourself -- has any chance of to understand your hteory by using any approach whatsoever, because there is no content there to understand.

As I said, your reductionist step-by-step approach enables you to see exactly nothing, and so is your conclusion.

Thank you for demonstrating my argument about you.

Here is the relevant part from an old post of mine:

My mathematical framework is a complementation between VSL and ASL, where Mathematics of the past 2500 years is mostly based on ASL.

Here is a comparison between ASL and VSL, based on Dr. Linda Kreger Silverman's research (please look at the difference between what is called Visual Special Learning and Auditory Sequential Learning ( http://www.visualspatial.org/Articles/intro.pdf , http://www.visualspatial.org/Articles/idvsls.pdf ). ):

[qimg]http://www.geocities.com/complementarytheory/VSL-ASL.jpg[/qimg]

You are an ASL-only that has problems with parallel thinking but he is good at serial (step-by-step) thinking.

I am more VSL than ASL.

In my opinion Mathematics will be developed to ASL\VSL framework.

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

That has no parts can be understood only if we are aware of the complement that has parts.

That has parts can be understood only if we are aware of the complement that has no parts.

In both cases we are using That\parts complementation.

That is invariant whether it has or does not have parts.

On the contrary a part is variant because it is chanced by exist\not-exist.

Changeability is a fundamental property of a part, because it cannot be simultaneously in more than one state.

I called changeability Locality.

Stability is a fundamental property of That, because it is the same whether a part exists or not.

I called stability Non-locality.

By examine 2-valued Logic I have noticed that Non-locality\Locality complementation stands in the basis of it.

Let us use N for Nothing and S for Something.

Let us exchange False by N and True by S.

NN
NS
SN
SS

What is called That is invariant (symmetric) whether it has parts (SS) or does not have parts (NN).

What is called a part is variant (asymmetric) weather it is (NS) or (SN).

All what is written above cannot be known unless Non-locality\Locality complement each other.

The concept of Membership is not exceptional.

Some details (that are not understood by serial-only thinkers) can be seen here:

http://www.internationalskeptics.com/forums/showpost.php?p=3567213&postcount=193

http://www.internationalskeptics.com/forums/showpost.php?p=3568369&postcount=194

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

No, {R,S} is the union of the members of Q. Which is distinct from, but related to, Q. Alterantively, it is the union of {R} and {S}, which is distinct from the union of the members of {R} and {S}. You're introducting an additional level of set-membership (or dropping it) and your results are therefore wrong.

Q = {{R},{S}} which is a whole\parts complementation where "{" and "}" are the notation of the whole side of Q and "{R},{S}" is the notation of the parts side of Q.

This notion is not for serial-only thinkers.

 

[jsfisher]

As I said, your reductionist step-by-step approach enables you to see exactly nothing, and so is your conclusion.

That is ever so convincing coming from someone that cannot see the difference in meaning between "union of the sets A and B" and "union of the members of the sets A and B".

Yet another example of that reading comprehension thing? Get help, doronshadmi.

...gibberish...

Q = {{R},{S}} which is a whole\parts complementation where "{" and "}" are the notation of the whole side of Q and "{R},{S}" is the notation of the parts side of Q.

What's your point? No one here has disputed that "{ ... }" is used as a mathematical notation for "a set with ... for members".

 

[PixyMisa]

That has no parts can be understood only if we are aware of the complement that has parts.

That has parts can be understood only if we are aware of the complement that has no parts.

These two sentences are ungrammatical and devoid of meaning.

In both cases we are using That\parts complementation.

Meaningless.

That is invariant whether it has or does not have parts.

Meaningless.

On the contrary a part is variant because it is chanced by exist\not-exist.

Ungrammatical and meaningless.

Changeability is a fundamental property of a part, because it cannot be simultaneously in more than one state.

A shotgun blast of undefined terms.

I called changeability Locality.

So?

Stability is a fundamental property of That, because it is the same whether a part exists or not.

That what? What that? Stability of what? What part?

I called stability Non-locality.

Again, so?

By examine 2-valued Logic I have noticed that Non-locality\Locality complementation stands in the basis of it.

No. This is complete nonsense.

Two-valued logic is two-valued logic. You just took two meaningless labels and attached them to the two values. This proves nothing.

Let us use N for Nothing and S for Something.

Let us exchange False by N and True by S.

See?

NN
NS
SN
SS

Yes. You have just discovered that two times two is four. Congratulations!

What is called That is invariant (symmetric) whether it has parts (SS) or does not have parts (NN).

What is called a part is variant (asymmetric) weather it is (NS) or (SN).

That's a XOR function. So?

All what is written above cannot be known unless Non-locality\Locality complement each other.

It's a XOR function. With meaningless labels.

The concept of Membership is not exceptional.

Sure. We told you that.

Q = {{R},{S}} which is a whole\parts complementation where "{" and "}" are the notation of the whole side of Q and "{R},{S}" is the notation of the parts side of Q.

Is Q a set, or isn't it? If it's a set - and you are using set notation - then everything else you said is irrelevant. If it's not a set, then everything else you said is... irrelevant.

So, what's it to be?

 

[doronshadmi]

That's a XOR function. So?

Not at all.

By XOR connective X,Y prevent each other.

I am talking about X,Y complementation that one of is particular cases is prevention.

X=NXOR Y=XOR ( and I am using XOR(X,Y) )

My logic is NXOR\XOR complementation.

Your logic is closed under the XOR dichotomy, where any two things prevent each other.

You may say: so your point of view is closed under complementation, where any two things complement each other.

My answer is:

No, my point of view is opened under complementation because NXOR\XOR complementation enables XOR as some particular case.

On the contrary, XOR dichotomy does not enable NXOR\XOR complementation.

 

[PixyMisa]

Not at all.

By XOR connective X,Y prevent each other.

No. You have attached labels to a XOR truth table. That's all you have done. Nothing "prevents" anything else.

I am talking about X,Y complementation that one of is particular cases is prevention.

That doesn't mean anything.

X=NXOR Y=XOR ( and I am using XOR(X,Y) )

Nor does that.

My logic is NXOR\XOR complementation.

Nor that.

Your logic is closed under the XOR dichotomy, where any two things prevent each other.

Again, this is nonsense. XOR can be viewed as an inequality function (and XNOR as equality). That says nothing at all about anything preventing anything else.

You may say: so your point of view is closed under complementation, where any two things complement each other.

No thanks.

My answer is:

No, my point of view is opened under complementation because NXOR\XOR complementation enables XOR as some particular case.

Meaningless.

On the contrary, XOR dichotomy does not enable NXOR\XOR complementation.

On the counter-contrary, if you wire the output of a XOR gate to both inputs of a second XOR, you get an XNOR (what you call NXOR). So as far as this has any meaning - which is not very far at all - it's wrong.

 

[drkitten]

Not at all.

By XOR connective X,Y prevent each other.

I am talking about X,Y complementation that one of is particular cases is prevention.

X=NXOR Y=XOR ( and I am using XOR(X,Y) )

My logic is NXOR\XOR complementation.

Your logic is closed under the XOR dichotomy, where any two things prevent each other.

You may say: so your point of view is closed under complementation, where any two things complement each other.

My answer is:

No, my point of view is opened under complementation because NXOR\XOR complementation enables XOR as some particular case.

On the contrary, XOR dichotomy does not enable NXOR\XOR complementation.

If that had been meaningful, it would have been wrong.

 

[doronshadmi]

What's your point? No one here has disputed that "{ ... }" is used as a mathematical notation for "a set with ... for members".

My point is simple.

There is no dichotomy between Set and Member concepts.

The empty set has no members (we are using both concepts without any dichotomy).

A non-empty set has members (we are using both concepts without any dichotomy).

If X is a set then it cannot be defined without its members (or their absence, which is not the absence of the concept of Member).

A set is AND(Function, the concept of Member), otherwise it does not hold.

If S is a set then the union function is :

Union(AND(Function, the concept of Member))

If you omit the concept of Member from the union of S with itself then S is undefined.

"S is the union of its members" (or their absence, which is not the absence of the concept of Member).

"S is its own union" is meaningless because S without the concept of Member is undefined.

Every fundamental concept must explicitly be expressed, without any hidden assumptions.

 

[doronshadmi]

If that had been meaningful, it would have been wrong.

The punch line of a fundamentamathematicalist:

"X is wrong even if I do not get it!"

 

[jsfisher]

My point is simple.

...meaningless stuff...

None of this had anything to do with my comment about your prior post.

It's that reading comprehension thing, still, isn't it?

 

[PixyMisa]

There is no dichotomy between Set and Member concepts.

What do you mean "dichotomy"? Set and member are two different concepts.

The empty set has no members

True.

A non-empty set has members

True.

If X is a set then it cannot be defined without its members (or their absence, which is not the absence of the concept of Member).

True.

A set is AND(Function, the concept of Member), otherwise it does not hold.

Gibberish.

If S is a set then the union function is :

Union(AND(Function, the concept of Member))

More gibberish.

If you omit the concept of Member from the union of S with itself then S is undefined.

If S is undefined, then S is undefined. Yes.

"S is the union of its members" (or their absence, which is not the absence of the concept of Member).

NO!

"S is its own union" is meaningless because S without the concept of Member is undefined.

Unless you are restating the law of identity, this is meaningless.

Every fundamental concept must explicitly be expressed, without any hidden assumptions.

Explicitness is valuable, but coherence is essential.