# Properties of sets under operations help

1. Feb 5, 2016

### Kilo Vectors

Hi

So I am learning about sets and I wanted to know if these definitions was correct, specifically the properties of sets under operations, and I had a question. please help.

The closure property: A set has closure under an operation if the result of combining ANY TWO elements under that operation is another element of the set.

The identity property: A set has identity under an operation if the result of combining ANY OTHER element with one element, results in the first element itself. The element is known as the identity element under that operation.

The inverse property: A set has an inverse property under an operation if the result of combining ANY OTHER
ELEMENT
with that element results in the identity element of that set for that operation.

If the closure property applies to any two, can it also to all? if the set is infinite?

2. Feb 5, 2016

### Staff: Mentor

This is basically true what you said. Whether the set is finite or infinite does not play a role here.

You obviously talk about a binary operation $\cdot$ on the set $S \times S$.

Closure under $\cdot$ means: For all pairs $(s,t) ∈ S \times S$ it holds that $\cdot(s,t) = s \cdot t ∈ S$.
So it holds for any two. Anyway which two. What do you mean by all?

A left identity element $e_1∈S$ means: There is an element $e_1∈S$ such that for all $s∈S$ it holds $\cdot(e_1,s) = e_1 \cdot s = s$.
A right identity element $e_2∈S$ means: There is an element $e_2∈S$ such that for all $s∈S$ it holds $\cdot(s,e_2) = s \cdot e_2 = s$.
Usually one shows in a first step that $e_1 = e_2 = e$ (for short) by using an additional property not mentioned yet.

A left inverse element means: For all $s∈S$ there is an element $s_1∈S$ such that $\cdot(s_1,s) = s_1 \cdot s = e$.
A right inverse element means: For all $s∈S$ there is an element $s_2∈S$ such that $\cdot(s,s_2) = s \cdot s_2 = e$.
Here, too, one usually shows that $s_1 = s_2 = s^{-1}$ (for short) by using this additional property, called associativity.

Associativity means for all $r,s,t ∈ S$ it holds $r \cdot (s \cdot t) = (r \cdot s) \cdot t$.

3. Feb 5, 2016

### Kilo Vectors

Sorry I didnt make myself clear, well for all I meant say there is:
1. a finite set of elements say the set of natural numbers from 1 to 6..
2. the set of all natural numbers.

For set number one, as far as I know we cannot assign closure under addition? as 1 + 6 is not a part of the set.

When I said all I meant that for the set of all natural numbers, when we have 50 elements operated under addition, the result will still be a part of the set....instead of operation between two elements, we can have 3, 4 elements and still get the result as being an element part of the set. But not in finite sets?

Basically (im new to mathematics) what I am saying in the set {1,2,3,4,5,6} exists closure only between certain elements? but not between any pair of two elements. So, does this mean it does not have closure property under addition? Finite sets of numbers do not seem to have this property in general. Because 1 + 6 is 7, not a part of the set. but in the set number 2, any two elements have closure under addition.

So, in finite sets of numbers, the property of closure under certain operation does not exist?

4. Feb 6, 2016

### Staff: Mentor

Well, it gets closed under addition, if you wrap it around: Identify 7 with 1, 8 with 2 and so on and it is closed. You have to decide whether you consider this set as a subset of the natural numbers or as a set in its own right. The first isn't closed under the inherited operation, on the second there can be defined another operation which is closed. If you consider, e.g. a set with five elements $\{0,1,2,3,4\}$ you can even define not only addition but also a multiplication on it, with subtraction and division all inclusive.

Sorry, I don't get you. See my example above.
You should forget about the meaning of the symbols you are used to. If you are talking on just sets, then it doesn't play a role how you note their elements. You can define addition, subtraction, multiplication and division on the set $\{Alice , Bob\}$, all of which are closed, have neutral (identity) and inverse elements. The computer on which you read these lines operates on this principle. Only that we usually call $Alice =0$ and $Bob=1$. Such a structure can be defined on every set with a prime (finite) number of elements. If you drop the requirement of being prime you can still define closed binary operations, just that you cannot divide all elements anymore.

5. Feb 6, 2016

### MrAnchovy

Just to confirm:
1. The set {1,2,3,4,5,6} is not closed under addition. You have proved this by counter-example (6 + 1 = 7 which is not an element of the set).
2. The (infinite) set of all integers is closed under addition.
3. The only finite set of integers which is closed under (normal integer) addition is {0}. We can prove that {0} is closed under addition by exhaustion: 0 + 0 = 0 is the only sum we need to inspect.
4. Finite sets of integers can be closed under other operations, such as the set {0, 1, 2... 11} under addition modulo 12 where 11 + 1 = 0 (similar to the way we tell the time where 12 o'clock + 1 hour = 1 o'clock).
[Edited: see post #7]

Last edited: Feb 6, 2016
6. Feb 6, 2016

### Staff: Mentor

This is only true if you restrict the term 'addition' to the inherited addition from the integers or natural numbers, i.e. regard those sets as subsets without changing the operation. The addition in finite rings like $ℤ_6$ is also called just 'addition'.

7. Feb 6, 2016

### MrAnchovy

I have qualified the term "addition" in my point 3.

8. Feb 6, 2016

### Staff: Mentor

If you use a common term for a special binary operation defined on many, many different sets of all kind, as addition is, in a such restricted manner you will have to make this clear in advance and certainly not implicitly.

9. Feb 7, 2016

### Kilo Vectors

Sorry I didn't make myself clear Mr Fresh sir. I was just asking if under finite sets, some of the properties do not apply. Pretty stupid question but I feel my book isn't that good..Is there any good books you can recommend? Maybe it is me too, I am just not able to understand easily. Thank you for your answer, really appreciate it, I did not know you could do define it like that, my level of maths is barely even scratching the surface.

thank you for your help sir.

10. Feb 7, 2016

### HallsofIvy

Staff Emeritus
I would drop the word "another". That implies that the result of the operation cannot be one of the first two elements and that is not true.

Your wording is awkward. A set has 'identity under the operation' if there exist a (unique) element such that "combining" that element with any other results in that "other" element.

I doubt you are saying what you intend here! You are saying that there exist a unique element that, combined with any other, gives the identity- and that is just not true. A set has "inverses" if, for any element of the set there exist an element such that combining them gives the identity. Different elements will have different inverses. And, again, you should not have the word "other". It is possible for an element to be its own inverse.

It doesn't matter if the set is finite or infinite, the same definitions apply.

11. Feb 7, 2016

### Staff: Mentor

If you have a set of 6 elements, say $\{1,2,3,4,5,6\}$ you can define a binary operation $ω$, e.g.
$1ω1=1$ , $1ω2=2$ , $1ω3=3$ , $1ω4=4$ , $1ω5=5$ , $1ω6=6$ ,
$2ω1=2$ , $2ω2=3$ , $2ω3=4$ , $2ω4=5$ , $2ω5=6$ , $2ω6=1$ ,
$3ω1=3$ , $3ω2=4$ , $...$
$...$
Now we can as well rename the whole stuff:
Instead of $\{1,2,3,4,5,6\}$ we name our elements $\{0,1,2,3,4,5\}$, and instead of $ω$ we will write $+$ and call it addition.
This does not change anything despite the symbols we chose to write down our elements and our operation.
It's similar to change the language we use to talk: different symbols same content.
So our Cayley table has now the form:
$0+0=0$ , $0+1=1$ , $0+2=2$ , $0+3=3$ , $0+4=4$ , $0+5=5$ ,
$1+0=1$ , $1+1=2$ , $1+2=3$ , $1+3=4$ , $1+4=5$ , $1+5=0$ ,
$2+0=2$ , $2+1=3$ , $...$
$...$
This definition now is a very common one. It is what you get, when you calculate with integers, beside that you do not take the integers itself but their remainders if divided by 6. E.g. 14+31=45, but we wanted to take their remainders instead. So 14:6=2 remaining 2; 31:6=5 remaining 1 and 45:6=7 remaining 3, i.e. instead of 14+31=45 we write 2+1=3, which is exactly what our Cayley table tells us.
This is an example of an addition that works well on a finite set, a set with 6 elements. It is even the same addition as on the integers only that we restricted ourselves to the remainders of those integers which we get by a division by 6. That is what I meant by wrap it around: 7,8,9, ... become the remainders 1,2,3, ....
We have even an identity element, we usually call it neutral element: 0.
And we have inverse elements: e.g. 1+5=0 so 1 is inverse to 5 and vice versa.
Only, and only if you insist to strictly regard your finite set as subset of the integers with the addition on the integers, only then it is not closed under addition.
However, if you start with $\{1,2,3,4,5,6\}$ as a finite set, then there is no 7. 7 simply does not exist. And therefore an equation 1+6=7 does not exist either. You cannot write it down! If you want to write it, you have to tell us about integers first and finite subsets next, i.e. you completely change your context!

Last edited: Feb 7, 2016
12. Feb 7, 2016

### HallsofIvy

Staff Emeritus

13. Feb 7, 2016

### Staff: Mentor

14. Feb 7, 2016

### Staff: Mentor

Closure is not a property of elements of the set. It is a property of the whole set with the specified operation.
There are finite sets with an operation that are closed. fresh 42 showed an example of one.

If you want the addition structure of the natural numbers, there is just one closed finite set: {0} with 0+0=0. All other sets have to be infinite.

15. Feb 7, 2016

### Kilo Vectors

Hello sir

1. Closure .. Set x ={A, B,C} A*B = C (closure) though it can also be under another operation, as according to you

2. Identity ...set x = {A,B,C}... (A*B = B) AND (A*C= C) AND (A*A=A) ...so the identity element combined with all elements results in the element it is combined with.

3. Inverse...each element of the set can have a distinct inverse element..regardless of whether this element is a part of the set or not.

if it is combined with that element which is in the set under the same operation as above (*) and it gives the identity, then this is set has the inverse property for that operation. But only if the inverse element is a part of the set.

x is any real number
Inverse propert of real numbers = x + -x = 0
Identity property of addition in real numbers = 0 + x = x

So it also possible for a set to have an identity property, but no inverse..

as for the set of all positive numbers and 0..there exists an identity property under addition, but NO INVERSE as negative numbers are not part of the set.

Last edited: Feb 7, 2016
16. Feb 7, 2016

### Staff: Mentor

What are A*A, B*B, C*A, A#A and so on? You have to specify the result of all possible combinations.
To have A as identity element, you would need A*C=C.

17. Feb 7, 2016

### Kilo Vectors

Hello Sir, those are just general operations..or supposed to represent that..

18. Feb 7, 2016

### Staff: Mentor

The set of natural numbers including 0 has no inverse for addition.
The set of all (e.g. real) matrices has no inverse elements for multiplication for all of its elements.

19. Feb 7, 2016

### Kilo Vectors

yes sir, unfortunately I haven't covered matrices yet..I will reply to your posts above as soon as I can understand them well enough. Thank you for your answers though, they have helped me understand..I will update my notes..btw can you recommend any good book for a beginner?

20. Feb 7, 2016

### Staff: Mentor

Yes, but they are not fully specified yet. In particular, not sufficiently to study the properties you want to study.
It has to be a part of the set, otherwise not even the operation is defined.

You don't need matrices, all those things can be studied with sets like {A,B,C,D}.