Deeper than primes

[dlorde]

...
From this part we understand that if A is equal to B then we have two names for the same object. But we can't conclude that because a given set has more than one name, then an element of that set is identical to that set, because the two names are the set, where no one of the names is an element of that set.

Yawn.

I thought I'd leave it a few months and see what's changed, but it's much the same. Like coming back from holiday and feeling like you'd never been away..

 

[jsfisher]

"Oh, my." is the name of your box, and so many years you are closed in your box, until you do not have the reasoning to get things beyond it.

Evasion noted. When you finally have some result, do let us know. Your inconsistent assertions and assumptions don't count as a result.

 

[doronshadmi]

No member of a given set is identical to that given set, and any attempt to force identity between a member of a given set to that set fails, because it is resulted by infinite regression that prevents the exact identification of the given set, for example:

A = {1,2,A}

If set {1,2,A} is identical to member A, then set A = {1,2,{1,2{1,2,{1,2,{1,2,{…}}}}}}

S = {

}( http://www.umpi.maine.edu/info/nmms/mirrors.htm )

 

[doronshadmi]

Evasion noted. When you finally have some result, do let us know. Your inconsistent assertions and assumptions don't count as a result.

Closed box reasoning has been noted.

Your agreed reasoning, which asserts that a member is identical to its set, does not hold water.

 

[doronshadmi]

Yawn.

I thought I'd leave it a few months and see what's changed, but it's much the same. Like coming back from holiday and feeling like you'd never been away..

Did you follow each post since your previous visit, in order to support your conclusion?

 

[The Man]

If A has elements that are identical to A, then there is no difference between A and the element of A.

First off set “A” has elements that are identical to set “A”, that’s what makes it set “A”. The issue is again when set “A” has itself as member and in that case there is no difference between set “A” and that member.

In that case A = element of A, which leads to infinite regression, as follows:

So you don’t understand the word “regression”, how surprising.

A = {A}, where an element of A is identical to A, so since the element of A is identical to A, then the element of A = {A}, and in this case A = {{A}}, but since the element of A is identical to A, then the element of A = {{A}}, and in this case A = {{{A}}}, etc… etc… ad infinitum.

The bracketed notation is not the only way to define a set or an element of a set, though it seems to be one that you particularly like to focus on.

In other words, there is no identity between A and an element of A, which is a simple fact that the agreed reasoning (that you quote form wiki, which is also accepted by you) can't get.

Again a set that has itself as a member can still have other members than just itself. You seem to be the only one having some kind of an “identity” crisis here.

EDIT:


So what?, you are the one that asserts that there is no difference between A's self identity and A's element.

Once again stop simply trying to posit aspects of your own failed reasoning onto others.

Once again your lack of direct perception, fails you.

Once again your “direct perception” fails you.

We have been over this before, however if you simply want to continue asserting the problems and paradoxes of naïve set theory, that actually do not exist, exactly as proper classes do not exist.

Ah so now the “problems and paradoxes of naïve set theory” “actually do not exist”. Why how absolutely, well, naïve of you, not to mention ignorant, since you’re the one that keeps harping on those very “problems and paradoxes of naïve set theory”.

Again The Man, since no element of a given set is identical to that set, then there is no problem, because the assertion that en element of A is identical to A, can't be satisfied, exactly because it is based on infinite regression, as clearly shown above.

Doron the assertion is “satisfied” when the set is defined to include itself as a member or is defined as an element of itself. Again this creates some of the problems and paradoxes of naïve set theory. That you would simply like to be satisfied with naïve set theory, is just your problem as it does not make those problems and paradoxes go away at your mere request.

Let us take this part from wiki ( http://en.wikipedia.org/wiki/Proper_subset#proper_subset ):

No, let’s look at the actual quote instead of your truncated nonsense

If A is a subset of B, but A is not equal to B (i.e. there exists at least one element of B not contained in A), then
• A is also a proper (or strict) subset of B; this is written as
or equivalently
• B is a proper superset of A; this is written as

Sorry the symbols did not copy over, but what is evident is that it is specifically referring to a "proper (or strict) subset" and a "proper superset" which exclude a set from being a member of itself.

From this part we understand that if A is equal to B then we have two names for the same object. But we can't conclude that because a given set has more than one name, then an element of that set is identical to that set, because the two names are the set, where no one of the names is an element of that set.

Again Doron that a set is an element of itself is not ‘concluded’ it is asserted in the definition of that set or as an element of that set.

 

[epix]

If Condition A leads to a contradiction via a paradox

AND

Condition A is conceptually identical to Condition B

THEN

a construction of a 4-D object suffers from a contradiction.


Q: What is the particular paradox/contradiction?

A:


knock...knock...knock...

Ask Doron. I'm busy.

How did You know that I wanted to ask You a question?

I'm like omniscient, aren't I?

So Doron couldn't solve it?

No. His face became pale upon considering the question.

In that case your visit is fully justified, coz if Doron can't eat what he cooked up, then I and My Omniscience is the only paved path leading to the solution. Let me see the rendition again . . . Oh, yes. A set that contains itself leads to a paradox and consequent contradiction. That means that a hyperspace "cube" is a contradictory CONSTRUCT.

I'm not deaf.

Sorry. (Clue no worky. Sigh.) An empty set has dimension -1. That's where we start.

Doron never mentioned that.

That's because Doron has never constructed anything -- he is a pure theorist. Here, lets go step by step:
http://www.maa.org/editorial/knot/tesseract.html

Okay. All done. But where is the contradiction?

No, we are not done. If we were, there would be no contradiction, coz the contradiction is this: You can construct a hypercube and at the same time you can't. Just take the piece of paper with the 4-D object rendered and fold it now into a real 4-D object.

That's not possible.

Well, that's the essence of the contradiction. You can construct a hypercube, which is a cube within a cube, or C = {C}, and at the same time you can't, unless . . .

Unless what?

Unleees . . .

 

[epix]

No member of a given set is identical to that given set, and any attempt to force identity between a member of a given set to that set fails, because it is resulted by infinite regression that prevents the exact identification of the given set, for example:

A = {1,2,A}

If set {1,2,A} is identical to member A, then set A = {1,2,{1,2{1,2,{1,2,{1,2,{…}}}}}}

S = {

}( http://www.umpi.maine.edu/info/nmms/mirrors.htm )

S( )?

Six = {Nine, Six, Nine, Six, Nine, Six, ... }

 

[doronshadmi]

First off set “A” has elements that are identical to set “A”

No, this is a false statement, because it leads to infinite regression, which prevents the strict identity of the considered set.

that’s what makes it set “A”

No, that what makes it a mathematical object that does not have strict identity, because of the infinite regression.

The issue is again when set “A” has itself as member and in that case there is no difference between set “A” and that member.

No, a given set is identical to itself, but no one of its members is identical to it, otherwise we get infinite regression.

Since we are talking about sets that have strict identity, then it is obvious that no one of their members is identical to them.

By understanding this simple and straightforward notion, both Russell's paradox and the need for proper classes are avoided.

Furthermore, no collection of members is identical to the existence of a given set, because the existence of a given set is beyond the existence of the members that define it, whether the considered set is empty or non-empty.

Doron the assertion is “satisfied” when the set is defined to include itself as a member or is defined as an element of itself.

The Man, your agreed assertion can't distinguish between "identical to" and "defined by", and can't get the independent existence of a given set from a given included collection w.r.t it, exactly as 1D space exists independently of any collection of 0D spaces along it.

The bracketed notation is not the only way to define a set or an element of a set, though it seems to be one that you particularly like to focus on.

As you see, I use both letters and bracketed notation, where the bracketed notation provides the notion of different levels of a given set that the agreed reasoning about sets simply ignores (only the first level of a given set is considered and it is clearly seen by the agreed use of Cardinality).

No, let’s look at the actual quote instead of your truncated nonsense

No The Man, you have missed my argument about the assertion that subset A is identical to Set B, and I show that this is nothing but giving different names to the same thing, which gives the illusion that a member of a given set is identical to that set.

Sorry the symbols did not copy over, but what is evident is that it is specifically referring to a "proper (or strict) subset" and a "proper superset" which exclude a set from being a member of itself.

The Man, no strict subset means that there is no difference at all between "subset" and "set". You can use as many names as you which, but is does not change the fact that we are dealing with the identity of a set to itself, where names like "sub…" or "element of" or "member of" have no significance at all.

Again Doron that a set is an element of itself is not ‘concluded’ it is asserted in the definition of that set or as an element of that set.

Your assertion is based on the inability to distinguish between "defined by" or "definition" and "identical to" or "identity".

Identity and Definition are not the same concept but your used agreed reasoning (that can't grasp http://www.internationalskeptics.com/forums/showpost.php?p=6321305&postcount=11503) can't get it.

 

[doronshadmi]

S( )?

Six = {Nine, Six, Nine, Six, Nine, Six, ... }

Six = {Nine, {Nine, {Nine, Six, Nine, Six, Nine, Six, ... }, Nine, {Nine, Six, Nine, Six, Nine, Six, ... }, Nine, {Nine, Six, Nine, Six, Nine, Six, ... }, ... }, Nine, {Nine, {Nine, Six, Nine, Six, Nine, Six, ... }, Nine, {Nine, Six, Nine, Six, Nine, Six, ... }, Nine, {Nine, Six, Nine, Six, Nine, Six, ... }, ... }, Nine, {Nine, {Nine, Six, Nine, Six, Nine, Six, ... }, Nine, {Nine, Six, Nine, Six, Nine, Six, ... }, Nine, {Nine, Six, Nine, Six, Nine, Six, ... }, ... }, ... } and you are invited to replace Six by {Nine, Six, Nine, Six, Nine, Six, ... } ad infinitum ... and get infinite regression that prevents the strict id of set Six ...

 

[epix]

Six = {Nine, {Nine, {Nine, Six, Nine, Six, Nine, Six, ... }, Nine, {Nine, Six, Nine, Six, Nine, Six, ... }, Nine, {Nine, Six, Nine, Six, Nine, Six, ... }, ... }, Nine, {Nine, {Nine, Six, Nine, Six, Nine, Six, ... }, Nine, {Nine, Six, Nine, Six, Nine, Six, ... }, Nine, {Nine, Six, Nine, Six, Nine, Six, ... }, ... }, Nine, {Nine, {Nine, Six, Nine, Six, Nine, Six, ... }, Nine, {Nine, Six, Nine, Six, Nine, Six, ... }, Nine, {Nine, Six, Nine, Six, Nine, Six, ... }, ... }, ... } and you are invited to replace Six by {Nine, Six, Nine, Six, Nine, Six, ... } ad infinitum ... and get infinite regression that prevents the strict id of set Six ...

There is a simple remedy:

1. A denotes a set.
2. {A} denotes a member of the set.
Therefore
3. A cannot expand within {}, because A doesn't equal {A}

It's all matter of an arbitrary decision how to define the elements of an abstract construct. I wonder if there has been a natural event similar to the unwanted expansion of sets.

Btw, if the cardinality of the set is finite and defined, 'A' cannot expand.

 

[doronshadmi]

There is a simple remedy:

1. A denotes a set.
2. {A} denotes a member of the set.
Therefore
3. A cannot expand within {}, because A doesn't equal {A}

It's all matter of an arbitrary decision how to define the elements of an abstract construct. I wonder if there has been a natural event similar to the unwanted expansion of sets.

Btw, if the cardinality of the set is finite and defined, 'A' cannot expand.

Some corrections:

1. Set A denotes {A}.
2. A that included in {A} denotes a member of {A}.
Therefore
3. A that included in {A} doesn't equal {A}.

It's all matter of Complexity, which prevents a member of A to be identical to A, and there is no arbitrary decision here.

 

[The Man]

No, this is a false statement, because it leads to infinite regression, which prevents the strict identity of the considered set.

Really, so what element does set “A” have that is not identical to an element of set”A”?

No, that what makes it a mathematical object that does not have strict identity, because of the infinite regression.

I see you’ve missed the point. Set “A” has elements identical to set “A” even when set “A” is not itself an element of set “A”. Again that is what makes it set “A” in the first place that it has all elements in common (no more and no less) with set “A”.

No, a given set is identical to itself,

Ah, so you did get the point of my first statement, but simply choose to ignore it in your above statements?

but no one of its members is identical to it, otherwise we get infinite regression.

Again if it is one of its members then that member is identical to the set, and your “infinite regression” is just in your imagination.

Since we are talking about sets that have strict identity, then it is obvious that no one of their members is identical to them.

In this case we are talking about a set that has the “identity” of being an element of itself. That you would simply like to deny that “identity” does not make it go away.

By understanding this simple and straightforward notion, both Russell's paradox and the need for proper classes are avoided.

Which notion? The “strict” one or your naïve notions?

Furthermore, no collection of members is identical to the existence of a given set,

Doron a set is a “collection of members”.

because the existence of a given set is beyond the existence of the members that define it, whether the considered set is empty or non-empty.

Evidently the notion of a set is simply “beyond” you.

The Man, your agreed assertion can't distinguish between "identical to" and "defined by", and can't get the independent existence of a given set from a given included collection w.r.t it, exactly as 1D space exists independently of any collection of 0D spaces along it.

What “agreed assertion”? You have made plenty of assertions and most of them I have specifically not agreed with. Like the one you keep repeating about “"identical to" and "defined by"”.

As you see, I use both letters and bracketed notation, where the bracketed notation provides the notion of different levels of a given set that the agreed reasoning about sets simply ignores (only the first level of a given set is considered and it is clearly seen by the agreed use of Cardinality).

Doron a letter does not define a set it is simply used to represent a set and you equate your letter (set “A”) to all your various instances of different “bracketed notation”, thus asserting them as, well, equal. All you have is set “A” so your “different levels of a given set” are all just set “A”. Your “infinite regression” is just in your imagination as there is nowhere for you to regress to, you only have set “A”. Sure, in this case, it is both the set and the element of that set, you can go around that circle as many times as you like but you just end up back were you started with only set “A”.

No The Man, you have missed my argument about the assertion that subset A is identical to Set B, and I show that this is nothing but giving different names to the same thing, which gives the illusion that a member of a given set is identical to that set.

Doron you have missed the point of the section you truncated a quote from to claim some other “argument”. Evidently you are arguing just with yourself as it is your own different representations, in “bracketed notation”, of your set “A” for your imaginary “infinite regression” that is giving only you some kind of “identity” crisis.

The Man, no strict subset means that there is no difference at all between "subset" and "set". You can use as many names as you which, but is does not change the fact that we are dealing with the identity of a set to itself, where names like "sub…" or "element of" or "member of" have no significance at all.

Something you might actually want to consider with your imaginary “infinite regression” and “identity” crisis where you not only just represent your set “A” with different “bracketed notation” but still equate every iteration to your set “A”.

Your assertion is based on the inability to distinguish between "defined by" or "definition" and "identical to" or "identity".

Identity and Definition are not the same concept but your used agreed reasoning (that can't grasp http://www.internationalskeptics.com/forums/showpost.php?p=6321305&postcount=11503) can't get it.

Once again stop simply trying to posit aspects of your own failed reasoning onto others.

 

[epix]

Some corrections:

1. Set A denotes {A}.
2. A that included in {A} denotes a member of {A}.
Therefore
3. A that included in {A} doesn't equal {A}.

It's all matter of Complexity, which prevents a member of A to be identical to A, and there is no arbitrary decision here.

Obviously there is. You just did it. But your initial premise doesn't make sense. "Set A" cannot denote, or better define, {A}. It's just A that defines set {}, which is the definiendum, whereas A is the definiens. If you deal with two sets {} { }, you may consider to define them (categorical definition) as A and B, otherwise goulash. After it's done, you refer to {} as Set A and to { } as Set B, not as "that set and the other set," unless THAT SET = {} and THE OTHER SET = { }.

If the definiens, or the defining term C, includes the finite cardinality of a set {A B C D}, then the set cannot expand. It all depends on what the defining term C has under its hood. But the paradox had to be fixed other way, coz there were implications already known to the ancient Greeks.

 

[epix]

Again if it is one of its members then that member is identical to the set, and your “infinite regression” is just in your imagination.

Doron can actually do this due to the definition of an element A of a set S: A is an element of S if and only if A is not an element of A.

Here is a tautology that deals with a similar situation:

y = 2x - 5y

y + 5y = 2x
6y = 2x
y = (2x)/6

If the tautology doesn't tolerate 'y' on both sides, then there could be a problem with A = {A, B, C}, as it was, coz Godel got busy with it.

 

[jsfisher]

Closed box reasoning has been noted.

Your agreed reasoning, which asserts that a member is identical to its set, does not hold water.

Where did I agree to that? You are lying again, Doron. Is everything with you just made up stuff? So far, nothing you've presented is real.

 

[doronshadmi]

Really, so what element does set “A” have that is not identical to an element of set”A”?

Any given member that is included in set A, is not identical to set A.

The maneuvering with names like “subset” and “set” has no significance, because these names are actually the same object.

I see you’ve missed the point. Set “A” has elements identical to set “A” even when set “A” is not itself an element of set “A”.

Again if it is one of its members ...

I see you’ve missed the point. Set A and A as a member of set A are not identical, because set A has an additional level that A as a member of set A does not have, ad infinitum ...

In this case we are talking about a set that has the “identity” of being an element of itself.

In that case you get a set that is based on infinite regression, and infinite regression does not have strict identity. Again, any given set is identical to itself, also any given member is identical to itself, but no member of a given set is identical to that set.

You still miss the difference between “define by” or “defined as” and “identical to”.

Which notion? The “strict” one or your naïve notions?

The “strict” one (which asserts that a given member of a given set is identical to that set) is exactly the naive notion that leads to fantasies like Russell’s Paradox and proper classes.

It is all derived from the fantasy that the magnetite of existence of infinite collected members is identical to the magnitude of existence of a given collector, which leads to the inability to distinguish between “define by” or “defined as” and “identical to”.

Doron a set is a “collection of members”.

Set is a collector\collected framework that enables the existence of collections, whether they are empty or not.

Evidently the notion of a set is simply “beyond” you.

Evidently the notion of “Set is a collector\collected framework that enables the existence of collections, whether they are empty or not”, is simply beyond you.

What “agreed assertion”?

The traditional one.

All you have is set “A” so your “different levels of a given set” are all just set “A”

And this fact prevents a member of that set to be identical to that set.

A set is defined by its members, but it does not mean that any of these members is identical to that set.

...that is giving only you some kind of “identity” crisis.

The “crisis” is your fantasy that can’t distinguish between “Identical to” and “defined by” or defined as”.

Organic Mathematics has no “crisis” about the notion of the incompleteness of infinite collections.

On the contrary, the traditional notion has a “crisis” about this subject because it can’t distinguish between “Identical to” and “defined by” or defined as”.

The traditional notion uses frameworks like ZF(C) in order to avoid paradoxes that do not exist, and it also using a garbage can called “proper classes”.

... but still equate every iteration to your set “A”.

The fact that Set A is defined by these iterations, does not mean that any of these iterations is identical to set A.

Once again stop simply trying to posit aspects of your own failed reasoning onto others.

Once again stop simply trying to posit aspects of currently agreed failed reasoning onto others.

Where did I agree to that? You are lying again, Doron. Is everything with you just made up stuff? So far, nothing you've presented is real.

The Man, I see that also jsfisher does not agree with you about the identity of a member of set A to set A.

 

[doronshadmi]

Happy Birthday to you, The Man.



http://scitation.aip.org/getabs/ser...0001000299000001&idtype=cvips&gifs=yes&ref=no

 

[doronshadmi]

Slowly but surely OM notions are spread http://adsabs.harvard.edu/abs/2010AIPC.1232..299K .

 

[zooterkin]

Slowly but surely OM notions are spread http://adsabs.harvard.edu/abs/2010AIPC.1232..299K .

Really? This would be the same Moshe Klein who you chased away when he tried to understand OM, but in fact showed that even you did not understand it?