## The Axiom Of Identity *Challenged*

### The Axiom Of Identity *Challenged*

In set theory there is a construction called a cartesian product.This takes two sets and puts the elements in relation to each other. So if we have A={a;b} and B={c;d} then the cartesian product of these sets A X B would be all the orderd pairs that link all the elements in both sets together. A X B would be {(a,c);(a,d);(b,c);(b,d)}

as mentioned this is a way of relating two sets. So this cartesian product would also be a relation. Any subset of this cartesian product would also be a full blooded relation, so R= {(a,d);(b,d)} would also be a relation (made from a subset of A X B.

An equivalence relation is a particular subset of the cartesian product of a set with itself.

Now equivalence relations come in many flavours like this subset of the natural numbers with themselves. When we consider the ordered pair (2,4) we can see that it is equivalent to (1,2) in the sense that they give the same remainder when the first divides the second.More importantly the remainders both have the same causal profile which should be obvious .It is "0". The same "0" you shouldnt divide by ...etc.Another name for an equivalence relation is a congruence relation so all equilateral triangles are equivalent in an obvious way but only if you ignore other respects from shape (e.g. size)

now we come to the axiom of identity which expresses equivalence between things. This uses the equivalence relation known as equality, "=".

Let us scruitanize this relation more closely. We would like perhaps to say it represents equality in all respects or a total equality.But do total equivalence relations exist?

If two things were totaly equivalent then odering them would violate this totality because one would come before the other. How can (a,a) be total if the first "a" comes before the other one. The very definition of a relation is "orderd pair" .How can the two things we want to totaly equate , totaly equate when they differ in any respect like order?And how can they be related if they are not part of an oderd pair.

But their both from the same set you say.Its the exact same "a". Look at the word underlined there same. You have just invoked *another* equivalence relation that orders that set with that set.In order for that set to be that set it has to be part of an orderd pair that relates it with it-self.So no matter what we do ,we still keep running into order. At first we might be tempted because of this to say that the relation (A,A) through reverse induction *causes* the set "A" to be, but i think you could get away with removing time from this type of causality.

So we are forced to conclude that order (differentiantion) is a fundamental property in judging equivalence which makes total equivalence impossible. How can the two be totaly equivalent when they differ in respect of order.

Then as a last great attempt we could try this;

Since if two sets are a subset of each other then they are the same set ,then perhaps we could imagine "a preceding a" then "swap" them around and have "a preceding a".

My response would be if those two things are equivalent then your attempt doesnt work because you in effect havent swapped anything around. If you say that they not equivalent and that you have swaped the "a's" then i raise the point that what does;

"a preceding a" not equal to "a preceding a" ,

mean if not

A is not equal to A.

Whichever way you want to make sense of equivalence , total equivalence is a "married batchelor" .

as mentioned this is a way of relating two sets. So this cartesian product would also be a relation. Any subset of this cartesian product would also be a full blooded relation, so R= {(a,d);(b,d)} would also be a relation (made from a subset of A X B.

An equivalence relation is a particular subset of the cartesian product of a set with itself.

Now equivalence relations come in many flavours like this subset of the natural numbers with themselves. When we consider the ordered pair (2,4) we can see that it is equivalent to (1,2) in the sense that they give the same remainder when the first divides the second.More importantly the remainders both have the same causal profile which should be obvious .It is "0". The same "0" you shouldnt divide by ...etc.Another name for an equivalence relation is a congruence relation so all equilateral triangles are equivalent in an obvious way but only if you ignore other respects from shape (e.g. size)

now we come to the axiom of identity which expresses equivalence between things. This uses the equivalence relation known as equality, "=".

Let us scruitanize this relation more closely. We would like perhaps to say it represents equality in all respects or a total equality.But do total equivalence relations exist?

If two things were totaly equivalent then odering them would violate this totality because one would come before the other. How can (a,a) be total if the first "a" comes before the other one. The very definition of a relation is "orderd pair" .How can the two things we want to totaly equate , totaly equate when they differ in any respect like order?And how can they be related if they are not part of an oderd pair.

But their both from the same set you say.Its the exact same "a". Look at the word underlined there same. You have just invoked *another* equivalence relation that orders that set with that set.In order for that set to be that set it has to be part of an orderd pair that relates it with it-self.So no matter what we do ,we still keep running into order. At first we might be tempted because of this to say that the relation (A,A) through reverse induction *causes* the set "A" to be, but i think you could get away with removing time from this type of causality.

So we are forced to conclude that order (differentiantion) is a fundamental property in judging equivalence which makes total equivalence impossible. How can the two be totaly equivalent when they differ in respect of order.

Then as a last great attempt we could try this;

Since if two sets are a subset of each other then they are the same set ,then perhaps we could imagine "a preceding a" then "swap" them around and have "a preceding a".

My response would be if those two things are equivalent then your attempt doesnt work because you in effect havent swapped anything around. If you say that they not equivalent and that you have swaped the "a's" then i raise the point that what does;

"a preceding a" not equal to "a preceding a" ,

mean if not

A is not equal to A.

Whichever way you want to make sense of equivalence , total equivalence is a "married batchelor" .

### Re: The Axiom Of Identity *Challenged*

YOUR minds ARE SIMPLE, YOU JUST DONT GET IT. IS THIS THE BEST you CAN DO silence?

THIS IS A PROOF THAT THERE IS A GOD

I'd like to see someone take me on and win....pathetic you all are...

THIS IS A PROOF THAT THERE IS A GOD

I'd like to see someone take me on and win....pathetic you all are...

- Arising_uk
**Posts:**11964**Joined:**Wed Oct 17, 2007 2:31 am

### Re: The Axiom Of Identity *Challenged*

How is the above an ontological proof of 'God'? Are you making a version of Leibniz's argument?

p.s.

I was puzzled how you can have a Cartesian Product(CP) of a set of one element {a} as {a,a}? As I'd have thought you needed two sets {a} {a} to get {a,a}. If it was possible I'd have thought CP of one element set {a} would be {a} or maybe {a} and empty set {Ø} which would be {a,Ø} so still basically {a}.

With respect to the idea of 'absolute identity'' between two objects in philosophy, I tend to agree with Leibniz and think it an impossibility by the fact of them being two objects making the difference.

p.s.

I was puzzled how you can have a Cartesian Product(CP) of a set of one element {a} as {a,a}? As I'd have thought you needed two sets {a} {a} to get {a,a}. If it was possible I'd have thought CP of one element set {a} would be {a} or maybe {a} and empty set {Ø} which would be {a,Ø} so still basically {a}.

With respect to the idea of 'absolute identity'' between two objects in philosophy, I tend to agree with Leibniz and think it an impossibility by the fact of them being two objects making the difference.

### Re: The Axiom Of Identity *Challenged*

There could be a lot of reasons people don't respond. Isn't it simple-minded for YOU to think that the ONLY reason is that people have simple minds?Moyo wrote:YOUR minds ARE SIMPLE, YOU JUST DONT GET IT. IS THIS THE BEST you CAN DO silence?

No it's not. That's ridiculous. And the fact that you're responding this way, demonstrates faults in your logic -- therefore, you are in no position to realistically claim proof of anything.Moyo wrote: THIS IS A PROOF THAT THERE IS A GOD

Calling everyone pathetic simply reveals/reflects the world YOU live in. And THAT is what is pathetic -- the fact that you cannot imagine anything more. There's much, much more... and it does not have to speak your language.Moyo wrote:I'd like to see someone take me on and win....pathetic you all are...

Seems like a lot of people who cannot get others to agree with them on their terms and in their specific language, think everyone else are fools. And it is THAT POSITION that looks the most foolish of all, does it not? Because it narrows down everything into such a tiny pinpoint of possibility. How can there be complete truth in that?

I think any idea of winning or having THE definitive answer is just part of a game. Some people play some games... other people don't. There are plenty of games to choose from. In this immense universe that we cannot even fathom the limits and levels of, how could it be reasonable for anyone to think that THEIR game reigns supreme over and above others?

- Arising_uk
**Posts:**11964**Joined:**Wed Oct 17, 2007 2:31 am

### Re: The Axiom Of Identity *Challenged*

Not quite true really as we are fathoming the limits reasonably well, so up to the speed of light and down to the planck length. Makes it about 13 billion(never sure is this is a UK or US billion?)light years big.Lacewing wrote:... In this immense universe that we cannot even fathom the limits and levels of, ...

- Hobbes' Choice
**Posts:**8363**Joined:**Fri Oct 25, 2013 11:45 am

### Re: The Axiom Of Identity *Challenged*

No this is proof that there is Cartesian sets.Moyo wrote:YOUR minds ARE SIMPLE, YOU JUST DONT GET IT. IS THIS THE BEST you CAN DO silence?

THIS IS A PROOF THAT THERE IS A GOD

I'd like to see someone take me on and win....pathetic you all are...

We already won.

### Re: The Axiom Of Identity *Challenged*

Awww..cuteArising_uk wrote:I was puzzled how you can have a Cartesian Product(CP) of a set of one element {a} as {a,a}? As I'd have thought you needed two sets {a} {a} to get {a,a}. If it was possible I'd have thought CP of one element set {a} would be {a} or maybe {a} and empty set {Ø} which would be {a,Ø} so still basically {a}.

Its a CP of a set onto itself.This CP is necessary for the set to have an identity

This identity would be expressed this way ; set A = Set A.

The Axiom of identoty ; A = A & A != !A

Is implicit in every system but it contains a viscious infinite regress.

### Re: The Axiom Of Identity *Challenged*

Since the axiom of identity is invalid no thing aor statement is meaningful .therfore the statement.Arising_uk wrote:How is the above an ontological proof of 'God'?

"There is no God"

is meaningless and so cannot be true (or have any property)

### Re: The Axiom Of Identity *Challenged*

Hobbes' Choice wrote:No this is proof that there is Cartesian sets.Moyo wrote:YOUR minds ARE SIMPLE, YOU JUST DONT GET IT. IS THIS THE BEST you CAN DO silence?

THIS IS A PROOF THAT THERE IS A GOD

I'd like to see someone take me on and win....pathetic you all are...

We already won.

Soo...now that you've put down your leggos.

### Re: The Axiom Of Identity *Challenged*

I think the word you're looking for is viscous, "having a thick, sticky consistency."Moyo wrote: Is implicit in every system but it contains a viscious infinite regress.

ps ... Ok that was just intended as a little pun ... but I see you are the OP of this thread and I want to see if I can shed any light on the ordered pair business. I didn't follow everything you wrote but here's how it's generally understood in math.

First, you are correct that a binary relation on a set A is actually a subset of the Cartesian product A X A. And, that the equality relation is the set of pairs (a,a) where a ranges over A. I think we agree on this much.

As I understand it, you are concerned that the notation (a,a) does not distinguish between the one copy of 'a' and the other. But it does. The first coordinate is one 'a', and the second coordinate is the other. The concept of an ordered pair can be formally defined as a particular set, so there is no problem.

So if I have some relation 'R' such that R consists of all and only the pairs (a,a) as a ranges over A; then it's always the case that xRx for any x in A, and if x and y are different, then it's NOT the case that xRy. So R is in fact the equality relation, defined exactly as the set of ordered pairs (a,a).

Last edited by wtf on Wed Oct 14, 2015 11:48 pm, edited 2 times in total.

### Re: The Axiom Of Identity *Challenged*

wtf wrote:I think the word you're looking for is viscous, "having a thick, sticky consistency."Moyo wrote: Is implicit in every system but it contains a viscious infinite regress.

wtf is meaningless too.

### Re: The Axiom Of Identity *Challenged*

The following statement;

"Moyo will lose this debate"

is meaningless so it cant happen.

Lets see you break out of that

"prenup".

"Moyo will lose this debate"

is meaningless so it cant happen.

Lets see you break out of that

"prenup".

### Re: The Axiom Of Identity *Challenged*

Check my edit.Moyo wrote: wtf is meaningless too.

### Re: The Axiom Of Identity *Challenged*

First i want to give you a heads up that i've gone through all the permutations of how this thread will go. I win every time. Now to sup with the masses...wtf wrote:So if I have some relation 'R' such that R consists of all and only the pairs (a,a) as a ranges over A; then it's always the case that xRx for any x in A, and if x and y are different, then it's NOT the case that xRy. So R is in fact the equality relation, defined exactly as the set of ordered pairs (a,a).

See the underlind above. How can those two a's be equal when they differ in "order" or any respect. Yaaaaawn.

### Re: The Axiom Of Identity *Challenged*

Oh that's a perfectly good question. The answer is that order is not an inherent part of the definition of a set. So the set {a,b,c} has no order on it at all. It's exactly the same set as {b,c,a} and all the other permutations.Moyo wrote: See the underlind above. How can those two a's be equal when they differ in "order" or any respect. Yaaaaawn.

Order is a secondary characteristic of the elements of a set. For example take a set of school kids. You can line them up by height, by alphabetical order of last name, alpha order of first name, reverse alpha order of last name, etc.

So if I ask some particular kid, "What is your order," in one order she might be number 5 and in other order she might be number 7. The order imposed on a set says nothing about the individual members. The order is only about the order.

You are correct that (a,a) declares that a is the first coordinate and a is the second coordinate; however:

* This is perfectly legal in set theory;

* The fact that the symbol 'a' may appear on either the left or the right of an '=' is not a problem, and says nothing about the "order" of anything.

Bottom line, if you are fifth in line at the movie theater or you're 1st in line to sup with the masses ... either way it's still you. You are always you, regardless of what your order is in some queue. And 'a' is always 'a', regardless of whether it's the first or second coordinate of an open pair.

Last edited by wtf on Thu Oct 15, 2015 12:31 am, edited 1 time in total.

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