# Thread: x and y cannot occur simultaneously

1. ## x and y cannot occur simultaneously

Can you show in general xϵy and x=y cannot occur simultaneously, ‘ϵ’ arbitary except that it is not ‘=’?

You certainly can for sets (xϵx).
You certainly can for order (x<x).

EDIT: I'm tempted to say, given x*y and x=y for some x, x*x is only possible if '*' is the same as '='.

Any takers?

2. ## Re: xϵy

Does ϵ mean Relation? In the course I did involving relations we used a funny shaped R.
I don't understand your proof for sets and order. The sets one looks like an example where xϵy and x=y CAN occur simultaneously. And I'm not sure if the ϵ in the sets example is meant to still be the arbitrary symbol or if it is meant to mean subset, if it is the arbitrary symbol then you really didn't show anything.
Your example for order is only one example of many possibilities for what ϵ could be. What if ϵ is "Less than or equal to" then x<=y and x=y can occur simultaneously.

3. ## Re: xϵy

If xϵy means set membership, then you get xϵx (x is a member of the set x) which gives x={A,x} -> x={A,{A,x} -> x = {A,{A,{A,{A,x}}} ->.....ie, undefined. xϵx (Set of all Sets, SϵS) leads to Russels (and Cantors) paradox. Since there is no such thing as xϵx there is no such thing as the paradox (ok, my opinion).

I just chose < as another example to show you can't have < and = at the same time. In other words x<x is impossible.

Since you bring up relationship, then I have to ask, what if xRy. If xRx for some x, do I have to conclude 'R' is the same as '='? If R is <or= then I certainly can have xRx. But that's not quite the same (?) because it's a combination of relationships. Sureley if 'R' is '<' 1 can't have xRx.

Off-hand it looks like whether or not I can have xRx depends on R.

But I seem to recall seeing a discussion somewhere that ϵ, even undefined, was not the same as relationship. Wish I could remember.

So I guess at this point x*x given * and some x doesn't necessarily imply * is the same as =.

Thanks

4. ## Relationship versus membership

Originally Posted by Shakarri
Does ϵ mean Relation? In the course I did involving relations we used a funny shaped R.
I don't understand your proof for sets and order. The sets one looks like an example where xϵy and x=y CAN occur simultaneously. And I'm not sure if the ϵ in the sets example is meant to still be the arbitrary symbol or if it is meant to mean subset, if it is the arbitrary symbol then you really didn't show anything.
Your example for order is only one example of many possibilities for what ϵ could be. What if ϵ is "Less than or equal to" then x<=y and x=y can occur simultaneously.
R can’t be defined without ϵ. R=ϵ would be a circular definition, so ϵ has to be independent of R, ie, ϵ can’t be a relationship..

This is how I got there in case anyone is interested (it’s educational):

If A and B are sets, a Relation from A to B, is a subset of AXB.

Examples:
1) A={x,y,z}, B={1,2,3}, AXB={(x,1), (x,2), (x,3), (y,1)……..(z,3)}.
R ={(x,1), (x,2), (z,2), (z,3)}, and we say xR1,xR2,zR2,and zR3.

2) A={1,2}, B={1,2}, AXB={(1,1), (1,2), (2,1), (2,2)}
R ={(1,1), (2,2)}: Equal Relationship: 1=1 and 2=2.
R = {(1,2)}: Less Than relationship: 1<2.
R = {(2,1)}: Greater Than relationship: 2>1.

xϵA means x is a member of A.

Now suppose x is any member of A and y is any member of B, then a relation could be expressed as xRy which looks like xϵy in its most general form.

So could xϵy in its most general form be a relationship? No, because you need xϵy to define a relationship.

5. ## Re: Relationship versus membership

Originally Posted by Hartlw
R can’t be defined without ϵ. R=ϵ would be a circular definition, so ϵ has to be independent of R, ie, ϵ can’t be a relationship..

This is how I got there in case anyone is interested (it’s educational):

If A and B are sets, a Relation from A to B, is a subset of AXB.

Examples:
1) A={x,y,z}, B={1,2,3}, AXB={(x,1), (x,2), (x,3), (y,1)……..(z,3)}.
R ={(x,1), (x,2), (z,2), (z,3)}, and we say xR1,xR2,zR2,and zR3.

2) A={1,2}, B={1,2}, AXB={(1,1), (1,2), (2,1), (2,2)}
R ={(1,1), (2,2)}: Equal Relationship: 1=1 and 2=2.
R = {(1,2)}: Less Than relationship: 1<2.
R = {(2,1)}: Greater Than relationship: 2>1.

xϵA means x is a member of A.

Now suppose x is any member of A and y is any member of B, then a relation could be expressed as xRy which looks like xϵy in its most general form.

So could xϵy in its most general form be a relationship? No, because you need xϵy to define a relationship.
You could probably create something similar using Allegories from Category Theory, but having never studied allegories myself, I would be ill-equipped to explain how.

6. ## Re: xϵy

slipeternal. Thanks, Thanks for the tip.

Category theory - Wikipedia, the free encyclopedia

definition: morphism: a->b where a and b are objects and -> is a map from a to b.
definition: identity morphism: x->x.

“From the axioms, it can be proved that there is exactly one identity morphism for every object.”

ie, you can’t simultaneously have xϵx and x=x.

ie, the fundamental primitive concepts of axiomatic set theory xϵy and x=y are incorrect because you can’t have xϵx.

So the Fundamental Atomic Concepts of Axiomatic Set Theory should be modified to:

x=y and xϵy unless x=y.

(just playing it safe with notation because the only thing I have at my disposal when I use “ϵ” is “=.” That’s why I wrote them in that order.)

7. ## Re: xϵy

It is not possible to prove that $$x\not \in x$$ It is true when x is an ordinal but is a bit restriction on the set.

This is a statement which is independent from standard set theory. However, set theorists have felt it is appropriate to therefore include it into the list of axioms of set theory. Today it is known as the "axiom of well-foundation".