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Math Help - Set Proof check

  1. #1
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    Set Proof check

    I need to prove each side is a subset of the other.

    Let F be a function. Show that F(A U B) =   F(A)  U   F(B)


    Let (x,y) be  \in F(A U B). Then there exists a point F(X) and a point F(y) such that x or y  \in   F(A)  or  F(B) . Since (x,y)  \in F(A U B) it follows (x,y)  \in   F(A)  U   F(B) .


    Am i on the right track?
    I did the other half but I did not want to waste time typing it if this was wrong, since I used a similar method.
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  2. #2
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    Set Proof

    Hello p00ndawg
    Quote Originally Posted by p00ndawg View Post
    I need to prove each side is a subset of the other.

    Let F be a function. Show that F(A U B) =   F(A)  U   F(B)


    Let (x,y) be  \in F(A U B). Then there exists a point F(X) and a point F(y) such that x or y  \in   F(A)  or  F(B) . Since (x,y)  \in F(A U B) it follows (x,y)  \in   F(A)  U   F(B) .


    Am i on the right track?
    I did the other half but I did not want to waste time typing it if this was wrong, since I used a similar method.
    You're kind of on the right track, but you're not using the notation correctly. So let me start you off.

    As you say, we need to prove that each set is a subset of the other, in order to prove they're equal. So let's prove first that F(A \cup B) \subseteq (F(A) \cup F(B)). We do this by choosing an element in the first set, and showing that it's also in the second.

    So let's suppose that y \in F(A \cup B). That means that there's an x \in (A \cup B) for which y = F(x).

    Now [x \in (A \cup B)] \Rightarrow [(x \in A) \lor (x\in B)]

    But [x \in A]\Rightarrow [y \in F(A)] and [x \in B]\Rightarrow [y \in F(B)]

    So [x \in (A \cup B)]\Rightarrow [(y \in F(A)) \lor (y \in F(B))]

    \Rightarrow y \in (F(A) \cup F(B))

    So we've just proved that [y \in F(A \cup B)] \Rightarrow [y \in (F(A) \cup F(B))]. In other words, that F(A \cup B) \subseteq (F(A) \cup F(B))

    OK? Now you've got to prove it the other way round. Start with y \in (F(A) \cup F(B)) and y = F(x) ...

    Grandad
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  3. #3
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    ok lets see if this is right...

    Suppose Y \in F(A) \cup F(B)  and y = F(x).
    Since y \in F(A) \cup F(B) , Y \in F(A) or  Y \in F(B) .
    If Y \in F(A) , then x \in A , and if y \in F(B)  then  x \in B .
    Because y \in F(A) \cup F(B) , it follows that  x \in A or  x \in B .

    Thus, F(A) \cup F(B) \subseteq F(A \cup B) .
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  4. #4
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    Quote Originally Posted by p00ndawg View Post
    Suppose Y \in F(A) \cup F(B)  and y = F(x).
    Since y \in F(A) \cup F(B) , Y \in F(A) or  Y \in F(B) .
    If Y \in F(A) , then x \in A , and if y \in F(B)  then  x \in B .
    Because y \in F(A) \cup F(B) , it follows that  x \in A or  x \in B .

    Thus, F(A) \cup F(B) \subseteq F(A \cup B) .
    I think that you have a great deal of work to do on understanding the notation involved in this question.
    There is no way to prove something if you cannot use the correct notation.
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  5. #5
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    Quote Originally Posted by Plato View Post
    I think that you have a great deal of work to do on understanding the notation involved in this question.
    There is no way to prove something if you cannot use the correct notation.
    what do you mean by notation?
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  6. #6
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    Hello p00ndawg
    Quote Originally Posted by p00ndawg View Post
    ok lets see if this is right...

    Suppose Y \in F(A) \cup F(B)  and y = F(x).
    Since y \in F(A) \cup F(B) , Y \in F(A) or  Y \in F(B) .
    If Y \in F(A) , then x \in A , and if y \in F(B)  then  x \in B .
    Because y \in F(A) \cup F(B) , it follows that  x \in A or  x \in B .

    Thus, F(A) \cup F(B) \subseteq F(A \cup B) .
    This is essentially correct, although not very elegant.

    Quote Originally Posted by Plato View Post
    I think that you have a great deal of work to do on understanding the notation involved in this question.
    There is no way to prove something if you cannot use the correct notation.
    Perhaps Plato would give us his version of the proof?

    Grandad
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  7. #7
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    Suppose Y \in F(A) \cup F(B) \Rightarrow x \in Aor x \in B for  y = F(x).
    If y \in F(A) \cup F(B) , then x \in A or y  \in F(A), if  x \in B then y \in F(B).
    Because x \in A or x \in B , then y \in F(A) or  y \in F(B) .
    It follows that F(A) \cup F(B) \subseteq F(A \cup B) .

    is this in anyway better?
    Last edited by p00ndawg; October 8th 2009 at 04:33 PM.
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