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]]>It is a combined effort of all the members of this forum.

Welcome to the forum!!!]]>

Yet again, you post

an example. What I am asking for is a general definition. For example, a general definition for addition on the natural numbers is:Where S(n) is the successor function of n.

Ok, define addition for the set {{},{1},{2},{1,2}}

You can't, it makes no sense.

Define addition for the set {fish}

Define the union operation for the natural numbers.

The natural numbers are a set, no more or less specific than the set {{},{1},{2},{1,2}}

I've defined union on my set, you've defined addition on your set. There is nothing more or less specific or general about either of them. You have generalised nothing. You have defined one operation specific to the set. I have defined an operation specific to my set. My operation just happens to be applicable to all power sets.

]]>2. Take a set which contains the elements of sets A and B, and create a mapping. Problem is, you can't define a set, in general, which contains the elements of A and B without taking the union of them.

No, you don't take a set which contains the elements of sets A and sets B. You take the power set of any given set and sets A and B will be elements of that set as a result! If sets A and B don't exist, it's because the elements that constitute them didn't exist in the given set. Notice how the set is given? Because it's a given set. I can't stress this enough. You don't need to define for general A and B because A and B are derived from whatever set you decide to power in the first place. I love how you can easily assume a very composite predicate such as the successor function definition of the natural numbers, but not a simple axiom such as the power set. It exists for a reason, and this is it.

]]>Where S(n) is the successor function of n.

This defines addition for all natural numbers. I am looking for the same general definition for union. There are two ways I see that you can attempt to do so:

1. Take the universal set and create a mapping. Problem is, the universal set doesn't exist under ZFC.

2. Take a set which contains the elements of sets A and B, and create a mapping. Problem is, you can't define a set, in general, which contains the elements of A and B without taking the union of them.

]]>You have still yet to define union in terms of a binary operation. Please, answer the question. You state that union is a binary operation, all I ask is that you define it as such. I'm not looking for examples, I'm look for you to define union as a mapping from a set AxA to A.

Are you serious? You want me to list every possible tuple in existance? You're clinically insane, I'd like to see you do that for an infinite set.

Here it is for my example above

Here we go

There you go, have fun writing it out for every possible set imaginable. My previous posts are the method in which it is defined, not by the mapping. Are you telling me scalar addition is defined by writing out every single possible tuple and a permutation? Of course not, it's defined by a couple of predicate rules.]]>

Something from wikipedia....

Some philosophers take a more radical approach, holding that some contradictions are true, and thus a theory's being inconsistent is not always an indication that it is incorrect. This view, known as dialetheism, is motivated by several considerations, most notably an inclination to take certain paradoxes such as the Liar and Russell's paradox at face value.

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What if you guys just make two or more different threads on different kinds of set theory?]]>

If we have A be the universal set, then we define the binary operation of union:

This is defined over all of AxA in the same way we can take the union of any two sets. It is also uniquely defined, as all unions are.

Only one problem. The universal set contains all sets. So the universal set must contain itself, as it is a set. But this is not allowed by the Axiom of Regularity* (under ZFC set theory). So such a universal set can't exist.

*Axiom of Regularity states:

This is saying in a general way that A can not be in A.

Proof by contradiction. Assume A is in A. **Then A is in {A} intersect A.** There must exist a b in A such that b intersect {A} = null. Since the only element of {A} is A itself, it must be that b = A. So we replace "b intersect {A} = null" with **"A intersect {A} = null"**. So A is in {A} intersect A and A intersect {A} is null. Contradiction.

if X,Y are subsets, then 'U' cant be defined in AxA. , but in P[A]xP[A], cartesian product of the power sets of A, right ?

darn: (

]]>For it to be a binary operator, the mapping must be defined on all of AxA. You seem to be saying that X and Y are elements of A, no?

]]>***edit*: this is wrong!**

I'm sorry, but that is not a binary operation. A binary operation is a mapping from AxA to A.

Hello!? A binary operation is an operation whose arity is two

Which is exactly the same as saying a mapping from AxA to A

It maps a power set cross power set to the power set, it will always be closed.

You're arguing me back down the exact point I'm putting forward. Here is an extremely simple analogy.

This doesn't get much clearer, if this still doesn't make sense, I suggest you study group theory before even considering trying to reiterate the exact point I'm making as an attempt to argue back, because you're not demonstrating much ability by repeatedly saying "A CROSS A MAPS TO A LOLZ", because I've already proved that it can. Simply because the elements are sets doesn't make it any different to any other binary operation you would care to mention.

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