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**Set Theory**

**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 26,648

**Set:-** A well defined collection of objects is called a set.

There is no repetition of elements in a set, each element appears only once.

**Elements of a set:-** The objects in a set are called its members or elements.

If an element a is prsent in setA, it is denoted by

If the element a is not present in set A, it is denoted by

**Description of a set**

**I. Roster mehod:-**

In this method, a list of all the objects of the set is made and they are put within braces.

Example 1:- If A is the set of the first eight prime numbers,

A={2, 3, 5, 7, 11, 13, 17, 19}

Example 2:- If B is the set of all the counting numbers between 20 and 30, B={21, 22, 23, 24, 25, 26, 27, 28, 29}

**II. Set Builder form:-**

In this method, the properties common to all the elemnts of the set are listed.

It is written in the form{x:x has properties P}

Example 1:- If B={3,5,7,9,11}, it is written as

B={x:x=2n+1 where n∈N, n<6} or

B={x|x=2n+1 where n∈N, n<6}

**Singleton set:-** A set consisting of a single elements is called a singleton set.

Example :- A={2} or A={x:x is an even prime number}

**Empty set:-** A set which has no elements is called an empty set. It is represented by {} or

**Finite and Infinite sets:-** A finite set is one in which the number of elements is finite.

Example:- A={x:x is a multiple of 5, x<1,000,000,000}

An infinite set is one in which the number of elements is not finite.

Examples:- Set of all points on the arc of a circle, Set of all concentric circles with a given centre,

{x|x∈Q, 0<x<1} (Q is the set of Rational numbers).**Equal sets:-** Two sets A and B are equal if every element of set A is in set B and everyelement of set B is in set A. It is denoted by

A=B.

**Cardinal Number of a set:-**The number of (distinct)elements in a set is called the Cardinal number of the set.

If A is the set, n(A) is the caridanl number of set A.

Example:- If A={0,1,2,3,4,5,6,7,8,9}, n(A)=10

**Equivalent sets:-**Two sets are said to be equivalent if n(A)=n(B), that is, if their cardinal numbers are equal.

Remark:- Equal sets are always equivalent but equivalent sets are not always equal.

Example:-

if A={1,2,3,4,5}, B={a,b,c,d,e}

A and B are equivalent sets because n(A)=n(B)=5, but A and B are not equal sets.

**Subset:-** Let A and B be two sets given in such a way that every element of A is in B, then it is said that A is a subset of B, written as

**Superset:-**If A is a subset of B, then B is a superset of A, denoted by

**Proper subset:-** If A is a subset of B and set A is not equal to set B, then A is called a proper subset of B, denoted by

Example, if A={1,2,3}, B={1,2,3,4,5,6,7}, then

**Comparable sets:-** Two sets A and B are comparable either if

A is a subset of B or B is a subset of A.

**Properties of Subsets:-**

1. {}, the empty set, is a subset of all sets.

2. Every set is a subset of itself.

3. The number of all subsets of a set containing n elements is

4. The number of all proper subsets of a set containing n elements is

5. The set of all subsets of a given set A is called the **Power set** of set A, denoted by P(A).

If A has n elements, P(A) has

**Operation on sets**

**Union of sets:-** The union of two sets A and B is the set of all eleements in A or in B or in both A and B. It is denoted by

Example:- if A={0,2,4,6,8,10} and B={1,2,3,4,5},

**Intersection of sets:-** The intersection of two sets A and B is the set of elements common in A and B.

It is denoted by

If A={0,2,4,6,8,10} and B={1,2,3,4,5},

**Difference of sets:-**

The difference of two sets A and B is defined as the elements present in set A not present in set B.

It is denoted by A-B.

If A={0,2,4,6,8,10} and B={1,2,3,4,5},

A-B={0,6,8,10} and B-A={1,3,5}

It should be remembered that

**Symmetric Difference of sets:-** The symmetric difference of two sets A and B is defined as

Symmetric difference is commutative.

If A={0,2,4,6,8,10) and B={1,2,3,4,5},

A-B={0,6,8,10}, B-A={1,3,5}

**Universal Set:-** A set which is a superset of all the given sets, denoted by U, is known as the Superset.

**Complement of a Set:-**

If set A is the subset of Universal set U, then the complement of A, denoted by

is the set of all the elements in the Universal set not in A.

Example:-

if U={a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z} and

A={a,e,i,o,u}, then

**Important results on Complements**

**Laws of Operations**

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 26,648

**Commutative Laws:-**

**Associative Laws:-**

**Distributive Laws:-**

**De Morgan's Laws:-**

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 26,648

**Number of elements in a set**

**Example**

Let

As per the formula,

Therefore, LHS=RHS.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 26,648

**Some Important Results**

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 26,648

**Cartesian Product **

The Cartesian Product of two sets A and B is defined to be the set of all ordered pairs with the first element in A and the second element in B. The Cartesian Product is denoted by

Example:- A={1,2,3}, B={a,b}

**Imortant Results of Cartesian products**

If A or B or both are infinite sets, A X B is also an infinite set.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**knvsp123****Member**- Registered: 2006-11-24
- Posts: 1

how can we say minimum no of elements of AUB IF N(A)=8 AND N(B)=5

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

you said: "

denotes an element is not in the set"this is wrong, the correct notation is

*Last edited by Sekky (2007-02-21 12:28:13)*

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 26,648

Hi Sekky,

Welcome to the forum!

Thanks for the post!

I do agree, the symbol you have given is used to denote 'does not belong to'. But the one I had given is also used.

Thanks again.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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

ganesh wrote:

Hi Sekky,

Welcome to the forum!

Thanks for the post!

I do agree, the symbol you have given is used to denote 'does not belong to'. But the one I had given is also used.

Thanks again.

No, it isn't, it's used to denote containment, not lack thereof. It's simply back to front, equally as every single other binary relation can be written back to front, as in subset containment or have you ever heard of < and >? The symbol means the same, but the relation is reflexed.

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

Actually this list should really be in the algebra formulas thread, since a set is just a degenerate algebra.

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

Sekky, this is a big world with many, many, mathematicians. Is it really that much of a surprise that two different mathematicians don't use exactly the same symbols?

Symbols are arbitrary. They can mean whatever you want them to. There are standards, but different people may conform to different standards.

Actually this list should really be in the algebra formulas thread, since a set is just a degenerate algebra.

Can you please define "degenerate algebra"? Wikipedia, MathWorld, and myself have no idea what you mean when you say that. Also, if you're going to go with that notion, then *every* thread in the formula section should be put into Sets, as sets are pretty much the basis of all math.

"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, this is a big world with many, many, mathematicians. Is it really that much of a surprise that two different mathematicians don't use exactly the same symbols?

Symbols are arbitrary. They can mean whatever you want them to. There are standards, but different people may conform to different standards.

Can you please define "degenerate algebra"? Wikipedia, MathWorld, and myself have no idea what you mean when you say that. Also, if you're going to go with that notion, then

everythread in the formula section should be put into Sets, as sets are pretty much the basis of all math.

He's using the same symbol to denote both the relation and the opposite relation, it makes no sense. The subset relation means the same back to front providing the relatives are switched, as does the less than symbol, as do any mathematical relations.

As for your query:

http://en.wikipedia.org/wiki/Algebraic_structure

and I quote: * Set: a degenerate algebraic structure having no operations.

Sets are not the "basis" of all mathematics as you put it, Sets are one of many algebraic structures, which is preciesley what algebra is: The theory of structures and relations. True that all algebraic structures have sets associated with them, but simply having a set does not base the entirety of mathematics. Morphisms, Operators, Relations ect, are not constructed from set algebras, they are associated alongside them to form more complex algebras. Hence the reason why when you strip everything away from an algebra, it becomes degenerate. Algebra is extremely fundamental yes, but it doesn't constitue everything.

Hope that clears things up for you

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 26,648

Sekky,

The formulas section is intended to help the members and all the students in general.

The beginners identify set theory as a separate branch, when these threads werte created, the idea was to put all formulas useful to students. Whether a thread was a subset of one or superset was not a point of concern.

Thanks for your ideas, and your contributions!

Ricky and me were posting the formulas, and we wanted to create a databse. I think we have done fairly well in our endeavor!

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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

Morphisms, Operators, Relations ect, are not constructed from set algebras

A morphism is a mapping. A mapping is simply a special subset of AxB, where the mapping would be from A to B.

An operator is a mapping from AxA->A, which is in turn a subset of (AxA)xA.

A relation is again a subset of AxB.

He's using the same symbol to denote both the relation and the opposite relation, it makes no sense.

That's like saying:

and

Are the same symbol. I mean, they are just flipped around, right?

"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

He's using the same symbol to denote both the relation and the opposite relation, it makes no sense.

That's like saying:

and

Are the same symbol. I mean, they are just flipped around, right?

Oh, right, we shifted into a parallel universe where everybody writes vertically.

You guys are just being awkward for awkward's sake. Either that or you're just trying to get the new guy off your forums.

Those are binary operators, which are commutative and therefore denoted such that it makes no different if you flip them horizontally because the operands won't need to change. we're talking about **non-symmetric relations** here.

*Last edited by Sekky (2007-02-22 19:15:29)*

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,686

In my experience there are alternative notations (which often lead to confusion, but that is how the world is).

Perhaps we could include a note to that effect?

And then, sorry to say, we will need to clean up the discussion according to http://www.mathsisfun.com/forum/viewtopic.php?id=3284

"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

Those are binary operators, which are commutative and therefore denoted such that it makes no different if you flip them horizontally because the operands won't need to change. we're talking about non-symmetric relations here.

Whats a binary operator? Intersection and union or inclusion? Cause neither are. A binary operator is a mapping from AxA to A.

And then, sorry to say, we will need to clean up the discussion according to http://www.mathsisfun.com/forum/viewtopic.php?id=3284

I will be doing so once this is resolved.

"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:

Whats a binary operator? Intersection and union or inclusion? Cause neither are. A binary operator is a mapping from AxA to A.

There's no way you can be THAT naive, you're obviously doing this on purpose, either for kicks or to chase me away from the forums, but I'll humour you in any case.

Given any set, both intersection and union map P|A| x P|A| -> P|A|, hence arity of two, hence a binary operation. Except if you'd actually thought about what you were saying, you would clearly see that they're clearly binary operators because **they take two parameters**. Stop quoting me the set definitions of what I've already told you to be true.

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

Given any set, both intersection and union map P|A| x P|A| -> P|A|

Union an intersection each take two different sets. Not the same set, though they can be.

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

Ricky wrote:

Given any set, both intersection and union map P|A| x P|A| -> P|A|

Union an intersection each take two different sets. Not the same set, though they can be.

So you're saying any binary operator can only ever take the same element for it's operands? 2 + 3 = 5 last time I heard.

The mapping maps all elements of the set cart set to another element of the set, and they will forever be included in the power set of any given set, thusly.

Let A = {1,2,3}

P|A| = {{},{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}

Now take the union and intersection of all possible cartesian tuples of the power set, and the yield will ALWAYS be an element of the power set, hence P|A| x P|A| -> P|A|

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

So you're saying any binary operator can only ever take the same element for it's operands? 2 + 3 = 5 last time I heard.

No, but the elements must belong to the same set. In other words, the integers.

Let A = {1,2,3}

P|A| = {{},{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}

Now take the union and intersection of all possible cartesian tuples of the power set, and the yield will ALWAYS be an element of the power set, hence P|A| x P|A| -> P|A|

Yes, but this is restricting the definition of union because I can take:

B = {4, 5, 6}

And do A U B.

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

hence A and B are both elements of P|{1,2,3,4,5,6}| and no generality is lost, Z x Z -> Z necessarily for some Z = P|{1,2,3,4,5,6}|

Operators are associated with algebras to form more composite algebras, if I take a set, I can find an operator to form an algebra. Likewise if I find an operator and element I wish to perform, I can find a set containing them such that the operator will hold an algebra.

*Last edited by Sekky (2007-02-23 13:10:43)*

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

or better, why not just consider the power set of the universal set under both operators, and I believe that forms a boolean lattice.

Well, any set under both operators will be a boolean lattice, I believe, unless you can think of an axiom that fails.

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**Ricky****Moderator**- Registered: 2005-12-04
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hence A and B are both elements of P|{1,2,3,4,5,6}|

Ah, I see. So now we're defining union by taking the union of two sets? And you don't see the problem in that?

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

Ricky wrote:

hence A and B are both elements of P|{1,2,3,4,5,6}|

Ah, I see. So now we're defining union by taking the union of two sets? And you don't see the problem in that?

Union and Intersection are defined by P|A| x P|A| -> P|A|

Pick anything for A, and the operation closes to a lattice.

Obviously the requisit is that it is a power set of some set, to ensure that every element is in fact a set itself.

And no, I'm not defining it that way, you just happened to pick two that were disjoint, they're also contained in the power set of {1,2,3,4,5,6,7,fish,carpet}, but I exampled the simplest possible power set.

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**Set Theory**