Question

Show that the following is a group:
\( (P(X),\bigtriangleup )\) Where X is a non empty set and \( \bigtriangleup \) is the symmetric difference operation \( A \bigtriangleup B= (A-B) \cup (B-A) \)

Answer 1

P(X) is the power set

Answer 2

ya sooo....

Answer 3

idk lol I dont even know the elements

Answer 4

do u like swiss cheez

Answer 5

@Herp_Derp can u help me?

Answer 6

Ok.
The elements of P are exactly the subsets of X, including the empty set (this is the definition of a powerset).

We need to prove: closure, associativity, existence of an identity, and existence of an inverse.

First we prove closure.

The symmetric difference of two sets is equivalent to the union minus the intersection. The union of two subsets A and B of X contains elements z, with each z being an element of at least A or B. Since each z is an element of a subset of X, each z is (by the definition of a subset) in X. Thus the union of A and B is a subset of X (by the definition of a subset, again). The symmetric difference is in a similar way a subset of X. Therefore it is in the powerset of X (the set of all subsets of X). Therefore P is closed under the symmetric difference operation.

Answer 7

so hwo wld we write that as a table. I think i gotta show it as a table

Answer 8

X is any nonempty set, so you cannot draw a general table. You can, however, consider an example set and draw a table for it, but it would be a large, complicated table and it would not be sufficient to prove the general case as posed in the problem.

Answer 9

thats what i figured
Thanks @Herp_Derp I really appreciate your help

Answer 10

Let \(R=A\Delta B\). \((A\Delta B)\Delta C\equiv R\Delta C\) means that for any \(x\in (R\Delta C)\), x is either in R, or x is in C, but not both. If x is in R, then x is either in A or B, but not both. So either x is in A, or x is in B, or x is in C, but if x is in C, then x is not in either A or B, but is possibly in both A and B. If x is in C and x is in A and x is in B, x is in the intersection of A,B,C. So either x is in exactly one of A,B,C; or x is in all three. In a similar way we can conclude that all x in \(A\Delta (B \Delta C)\) satisfy these same properties. Therefore \(A\Delta (B\Delta C)=(A\Delta B)\Delta C\). Therefore the operation is associative.

Answer 11

Wow Thanks @Herp_Derp I really appreciate this :)))))))))))))))))))

Answer 12

I have proved the most difficult parts for you. Can you do the other two?

Answer 13

well the identity is the emptyset

Answer 14

Good!

Answer 15

hahahah i dont even know how but i just know that from reading it somewhere

Answer 16

Ha! Well, think about it. What is the union of any set and the empty set, minus the intersection of any set and the empty set?

Answer 17

yaaaaaaaaaaaaaa ok got that

Answer 18

hmmm just thinking abt the inverse

Answer 19

its where the ordered pairs r reversed

Answer 20

like (x,y) becomes (y,x)

Answer 21

whooppps in this case its where it equals the empty set?

Answer 22

this is going to be nice and ugly
did you show it was associative?

Answer 23

ya just stuck with the inverse

Answer 24

oh yeah look @Herp_Derp did it above, nice work!!

Answer 25

Ya that waas reallllyy nice of himmmm

Answer 26

closer is no big deal
identity is empty set
so what is the inverse of \(A\)?

Answer 27

how would you get the symmetric difference to be the empty set? is another way of asking the question

Answer 28

the emptyset obviously

Answer 29

well 2 disjoint sets will give you the inverse

Answer 30

no i don't think so but i could be wrong

Answer 31

*will give u the emptyset

Answer 32

wait let me reread this

Answer 33

unless i miss my guess, doesn't symmetric difference mean union minus the intersection? in plain english i mean

Answer 34

when A=B?

Answer 35

yes i think that is right
every element is its own inverse

Answer 36

oh thats coolllll
Thankssssss guysssssss

Answer 37

@Herp_Derp did all the hard work

Answer 38

it looks scarier than it is

Answer 39

don't be scared by these, there are so few things to check and you have to work from the definition, so it is usually either obvious or false
i bet with some time you would have been able to do this on your own

Answer 40

hahahah Hope so. Thanks @Herp_Derp and @satellite73
That was awesomeeeee THANKS :)))))

Answer 41

Crap, I'm retarded. The symmetric difference of two disjoint sets is the union of the two sets, which is not the empty set. The symmetric difference of a set and itself, on the other hand, is always the empty set. So every element of P is its own inverse.

You should probably delete your posts saying that the inverse is a disjoint set, so we don't get laughed at when the Math Gods see this.

Answer 42

LOL Its fine @Herp_Derp I really appreicate ur help. It was awesome

Answer 43

I needed a little brush up on my set theory anyways :)