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# prove? watch

1. I was wondering how do I prove this ?? Can I use the axiom of extensionality
{x∈ A' l P(x)}= {x∈ A l P(x)}
2. The axiom of extensionality states a set A is equal to set B if for all x , x an element of A is implied and implied by x is an element of B - i.e. memebership uniquely determines a set. I don't see why this would be needed [or what you are asking to prove].

But if A' means the complement of A then the two sets are definitely not equal [as P(x) must be satisfied by some element in your universe that is either in A or A' - it cannot be in both].
3. (Original post by DeanK22)
The axiom of extensionality states a set A is equal to set B if for all x , x an element of A is implied and implied by x is an element of B - i.e. memebership uniquely determines a set. I don't see why this would be needed [or what you are asking to prove].

But if A' means the complement of A then the two sets are definitely not equal [as P(x) must be satisfied by some element in your universe that is either in A or A' - it cannot be in both].
So are you saying they asked me to prove a false statement
4. (Original post by rbnphlp)
So are you saying they asked me to prove a false statement
I was saying that I don't know what your notation is - if you wrote the question out exactly then that what would be great. (Also I was assuming P(x) was well defined and actually some property that could be possessed by some element in your universe - if that is not the case then both sets would be empty and trivially the same).
5. (Original post by DeanK22)
I was saying that I don't know what your notation is - if you wrote the question out exactly then that what would be great. (Also I was assuming P(x) was well defined and actually some property that could be possessed by some element in your universe - if that is not the case then both sets would be empty and trivially the same).
this is what the textbook says

"If there is a set A ,such that for all x ,P(x) implies x ∈ A, then {x∈ A l P(x)} exists.and moreover does not depend on A. That means if set A' is another such set such that for all x, P(x) implies x∈A' , then
{x∈ A' l P(x)}= {x∈ A l P(x)}"

It then says prove it .
6. (Original post by rbnphlp)
this is what the textbook says

"If there is a set A ,such that for all x ,P(x) implies x ∈ A, then {x∈ A' l P(x)} exists.and moreover does not depend on A. That means if set A' is another such set such that for all x, P(x) implies x∈A' , then
{x∈ A' l P(x)}= {x∈ A l P(x)}"

It then says prove it .
Well that doesn't seem true. Take A = {1,2,3,4,...} and P(x) the property that x is even. Then { x in A ; P(x) } is the set {2,4,6...}. Take A' which is {4,5,6,7,8..}. Clearly you have problems as they are not equal.
7. I don't think A' means the compliment set in this case but just another set it would be better if it said B.

Yeah, I think you just use the axiom of extensionality.

P.S. Stupid notation.
8. (Original post by Simplicity)
I don't think A' means the compliment set in this case but just another set it would be better if it said B.

Yeah, I think you just use the axiom of extensionality.

P.S. Stupid notation.
Please state how the axiom of extensionality is used here...
9. Well, DeanK22

I will get back to you.

I just got owned.
10. (Original post by DeanK22)
Well that doesn't seem true. Take A = {1,2,3,4,...} and P(x) the property that x is even. Then { x in A ; P(x) } is the set {2,4,6...}. Take A' which is {4,5,6,7,8..}. Clearly you have problems as they are not equal.
sorry there is a mistake in the first line of my quote from the text book it should have been A instead of A' in {x∈ A' l P(x)}

Just edited
11. What book is this from - alot of things look like definitions to me.
12. (Original post by DeanK22)
What book is this from - alot of things look like definitions to me.
http://www.amazon.com/Introduction-R.../dp/0824779150
13. (Original post by rbnphlp)
"If there is a set A ,such that for all x ,P(x) implies x ∈ A, then {x∈ A l P(x)} exists.and moreover does not depend on A. That means if set A' is another such set such that for all x, P(x) implies x∈A' , then
{x∈ A' l P(x)}= {x∈ A l P(x)}"

It then says prove it .
The first coloured section is stating that if we take every element in our Universe of Discourse with some property, P(x) [for example P(x) could be x is yellow or something] then A contains every element that satisfies P(x). It is exactly identical to the case for A' - except instead of using the letter A you have used the letter A'. So it is similar to writing;

Spoiler:
Show
yeah you use extensionality in the argument to show that both sets are identical
14. (Original post by DeanK22)
The first coloured section is stating that if we take every element in our Universe of Discourse with some property, P(x) [for example P(x) could be x is yellow or something] then A contains every element that satisfies P(x). It is exactly identical to the case for A' - except instead of using the letter A you have used the letter A'. So it is similar to writing;

Spoiler:
Show
yeah you use extensionality in the argument to show that both sets are identical
thanks got it..
15. (Original post by DeanK22)
Well that doesn't seem true. Take A = {1,2,3,4,...} and P(x) the property that x is even. Then { x in A ; P(x) } is the set {2,4,6...}. Take A' which is {4,5,6,7,8..}. Clearly you have problems as they are not equal.
Sigh.

If P(x) is the statement "x is even", then P(x) implies x is in A', so A' must contain 2 (just like A). Then {x in A | P(x)} is the set of all even numbers in A, and {x in A' | P(x)} is the set of all even numbers in A', which happen to be the same set.
16. (Original post by generalebriety)
Sigh.

If P(x) is the statement "x is even", then P(x) implies x is in A', so A' must contain 2 (just like A). Then {x in A | P(x)} is the set of all even numbers in A, and {x in A' | P(x)} is the set of all even numbers in A', which happen to be the same set.
So is A' the same set as A , as in they both contain only positive integers?
17. (Original post by rbnphlp)
So is A' the same set as A , as in they both contain only positive integers?
No. DeanK22 is talking out of his arse, I'm afraid.
18. (Original post by generalebriety)
Sigh.

If P(x) is the statement "x is even", then P(x) implies x is in A', so A' must contain 2 (just like A). Then {x in A | P(x)} is the set of all even numbers in A, and {x in A' | P(x)} is the set of all even numbers in A', which happen to be the same set.
I even said this was wrong afterwards and explained the conditions on the sets - tbh why did you bother quoting this?
19. (Original post by DeanK22)
I even said this was wrong afterwards and explained the conditions on the sets - tbh why did you bother quoting this?
Because not very much that you've said in this thread has been correct.

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