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    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)}
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    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].
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    (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:confused:
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    (Original post by rbnphlp)
    So are you saying they asked me to prove a false statement:confused:
    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).
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    (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 .
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    (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.
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    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.
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    (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...
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    Well, DeanK22

    I will get back to you. :confused:

    I just got owned. :o:
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    (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
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    What book is this from - alot of things look like definitions to me.
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    (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
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    (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;

     \int_a^b f(t)dt = \int_a^b f(x)dx

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    yeah you use extensionality in the argument to show that both sets are identical
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    (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;

     \int_a^b f(t)dt = \int_a^b f(x)dx

    Spoiler:
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    yeah you use extensionality in the argument to show that both sets are identical
    thanks got it..
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    (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.
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    (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?
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    (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.
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    (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?
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    (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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