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Simple sets question

  1. Aug 17, 2011 #1

    WannabeNewton

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    1. The problem statement, all variables and given/known data
    All the b's in f should be capitalized for the problem statement and attempt; I had it in the latex but it showed up lower case in the post I don't know why, my apologies =p.
    If [itex]f:X \mapsto Y [/itex] and [itex]A \subset X, B \subset X[/itex], is:
    (a) [itex]f[A \cap B] = f[A] \cap f [/itex] in general?
    (b) [itex]f[A - B] = f[A] - f [/itex] in general?
    3. The attempt at a solution
    My ability to write proofs is atrocious at best so bear with me please =D.
    For (a), let [itex]y\in f[A \cap B][/itex], then there is an [itex]x\in A \cap B[/itex] such that [itex](x, y) \in f[/itex]. Since [itex]x\in A [/itex] and [itex]x\in B [/itex], [itex]y\in f[A]\cap f [/itex] and [itex]f[A\cap B]\subset f[A]\cap f[/itex]. Now, let [itex]y\in f[A]\cap f[/itex]. For [itex](a, y)\in f, a\in A[/itex] and [itex](b, y)\in f, b\in B [/itex] [itex]a \neq b[/itex] in general so even if [itex]y\in f[A]\cap f[/itex], [itex]x\notin A\cap B[/itex] in general. Therefore, the statement (a) is not true in general. Is this enough?
    (b) I have more of a question with this one: if [itex]y\in f[A] [/itex] and [itex]y\notin f [/itex] does that necessarily mean [itex]x\in A - B[/itex]?
     
  2. jcsd
  3. Aug 17, 2011 #2

    CompuChip

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    The first part of a) looks OK.
    The second part is probably true, but it is not very clear.
    Note that if you suspect that a statement is false, in general it is simplest to try and find a counter-example. You have already shown that you cannot find an [itex]y \in f(A \cap B)[/itex] which is not in [itex]f[(A) \cap f(B)[/itex], so you know where to look for the counter-example.

    For b), the same advice. Can you think of a set A and a set B, such that x is not in A - B, but f(x) is in f(A) - f(B)?

    Both statements are true by the way, when you replace f by f-1 (defined by [itex]f^{-1}(y) = \{ x \in X \mid f(x) = y \}[/itex]). This should also give you the hint that you won't find any counterexamples for injective functions (for which f-1(y) is unique for all y).
     
    Last edited: Aug 18, 2011
  4. Aug 17, 2011 #3

    micromass

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    Note that it IS true that

    [tex]f(A\cup B)=f(A)\cup f(B)[/tex]

    so the union behaves nicely. Unlike the two operations in the OP.

    Just throwing that out there :smile:
     
  5. Aug 18, 2011 #4

    WannabeNewton

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    Sorry for responding so late but I'm on the move as it would seem =D.
    For (a) if [itex]f:x \mapsto x^2 [/itex] and [itex]A = [1 , 2, 3][/itex], [itex]B = [-5 , -4, -1][/itex] then [itex]f(A) = [1, 4, 9][/itex], [itex]f(B) = [25, 16, 1][/itex]. In this case, [itex]f(A \cap B) = \varnothing [/itex] so [itex]f(A)\cap f(B) \nsubseteq f(A \cap B)[/itex]. I guess that makes it more clear.
    For (b), if [itex]x\in A[/itex] and [itex]x\in B[/itex] then [itex]y\notin f(A) - f(B)[/itex] which can't be. So I think the statement is true? I'm still not sure because the question asks me to determine which is more well behaved, [itex]f[/itex] or [itex]f^{-1}[/itex], in terms of sets so I'm thinking something should be wrong with the statement (b) xD.
     
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