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**Set Theory**

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

And there in lies the problem. You are operating under what is called (I swear, this is the actual term, I'm not trying to be insulting), "naive set theory". It is the concept that any collection can be a set. Naive set theory is called so because it is, well, naive. It leads to contradictions and paradoxes.

On the other hand, I am going off of ZF set theory, which is the just about the standard when it comes to set theory in mathematics (although there are others). In ZF set theory, you can't just claim something is a set. The set must be constructed through the use of existing sets. The only set that is guaranteed to exist is the null set (which I will be denoting θ simply because of laziness). But through precise axioms, we can take the null set and form sets such as {θ} and {{θ}} or {θ, {{θ}}}. In ZF set theory, the Axiom of Union is used to define what we typically think of when we see union (though said in a completely different form).

So you need union to state that your set {1, 2, 3, 4, 5, 6} exists before you can ever even talk about it being a set. And this is the reason why we can never have union be an operator. Because the sets need to exist before we can call it an operator, but we need union to exist before we can say the sets exist.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**Sekky****Member**- Registered: 2007-01-12
- Posts: 181

I started before you posted, and that isn't without loss of generality, because we assume U to be the set of all sets, so it makes no difference whether the set exists, because it won't be included in here.

As for ZFC, yadda yadda blah blah Trussian. It still holds because it's not predicate dependent, union can still be expressed as 2-ary. The axiom of union is present for essentially that purpose, to axiomate that such a set is possible to exist. The union operator however is only named because it's serves the purpose that the union axiom expresses is possible.

*Last edited by Sekky (2007-02-23 13:47:53)*

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,559

Guys, this is such a great discussion, but it is in the Formulas section, so due for deletion at some point!

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Sekky, we're starting to beat around the bush here. Please come up with a binary operator definition for the union of two (possibly different) sets.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**Sekky****Member**- Registered: 2007-01-12
- Posts: 181

Ricky wrote:

Sekky, we're starting to beat around the bush here. Please come up with a binary operator definition for the union of two (possibly different) sets.

Hello? The definition is right there in my last post!

Defined as binary

*Last edited by Sekky (2007-02-26 02:12:22)*

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**Sekky****Member**- Registered: 2007-01-12
- Posts: 181

Ricky wrote:

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.

*Last edited by Sekky (2007-02-26 07:05:06)*

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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.

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**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

defining union from AxA to A?

***edit*: this is wrong!**

*Last edited by kylekatarn (2007-02-26 11:29:29)*

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

In your definition, how is A defined?

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?

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**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

I see your point, the definition doesn't seem correct after all.

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: (

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Oddly enough, I didn't think I had a point. I was really just trying to understand your definition. But perhaps it can be saved.

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.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,585

Remove this after cleanup:

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.

-------------------------

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

*Last edited by John E. Franklin (2007-02-26 13:22:03)*

**igloo** **myrtilles** **fourmis**

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**Sekky****Member**- Registered: 2007-01-12
- Posts: 181

Ricky wrote:

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.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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.

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.

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**Sekky****Member**- Registered: 2007-01-12
- Posts: 181

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.

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**Sekky****Member**- Registered: 2007-01-12
- Posts: 181

Ricky wrote:

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.

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**medivijaysagar****Member**- Registered: 2008-01-09
- Posts: 1

hi ganesh, first of all i would like to thank you for this great stuff. Any guy will find it very useful in any mode of his life. i fell lucky to find this site. Thanks once again and keep going........

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,071

Thanks, medivijaysagar!

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

Welcome to the forum!!!

Character is who you are when no one is looking.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,071

**Cartesian Products**

1)

2)

Character is who you are when no one is looking.

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**DeanPemberton****Member**- From: USA
- Registered: 2014-10-30
- Posts: 21
- Website

Set theory is a mathematical theory of well determined collections, sets of objects that are called members or elements. Set theory is very simple theory but the concept of it is little bit difficult.

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**loveMath****Member**- Registered: 2015-02-18
- Posts: 3

Sorry I am new here. If possible, can anybody help me? My question is: given the two sets A={x} and B={{x}} are they equal?

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No, they are not.

Is having a carrot in a box and a carrot in a box in a box the same?

Please make a new thread in the help me section

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**loveMath****Member**- Registered: 2015-02-18
- Posts: 3

thanks I will make a new post. I just did know how.

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