The Axiom Of Identity *Challenged*
Posted: Sun Jun 28, 2015 11:44 am
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" .