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Show that if f: A → B and E, F are subsets of A, then f(E ∪ F) = f(E)

  1. Aug 10, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that if f: A → B and E, F are subsets of A, then f(E ∪ F) = f(E) ∪ f(F).

    2. Relevant equations



    3. The attempt at a solution

    My attempt:

    Suppose x is an element of E. Then f(x) is an element of f(E), which means f(x) is a subset of f(E).
    But x is in E implies x is in E or F so that x is in E ∪ F.
    Thus f(E ∪ F) = f(x) is a subset of f(E), and is also a subset of f(E) ∪ f(F) by or introduction.

    *Similar argument applies when we suppose x is an element of F*

    Now I'm having trouble with the converse part of the proof; the only idea I've come up with is to use a similar strategy as in the first proof and suppose x is in E so that f(x) is in f(E). Then you can show that since x is in E or F that the left side of the equation is contained in the right.

    I'm a total newbie at proofs. please someone help me out :3
     
  2. jcsd
  3. Aug 10, 2013 #2
    Just some formalities: In the first line you say [itex]f(x)[/itex] is a subset of [itex]f(E)[/itex]. The More correct thing to say is that for [itex]x\in E[/itex], [itex]f(x)\in f(E)[/itex], i.e. f(x) is an element in f(E). But what you really want to say is this: Let [itex]x\in E[/itex] so that [itex] f(x)\in f(E)[/itex]. Then continue as you did: since [itex] x\in E, x\in E\cup F[/itex], and thus
    [itex]f(x)\in f(E\cup F)[/itex]. This shows [itex]f(E)\subset f(E\cup F)[/itex]. The same argument applied to [itex]x\in F[/itex] will give [itex]f(F)\subset f(E\cup F)[/itex] which together yield
    [tex]f(E)\cup f(F)\subset f(E\cup F).[/tex]
    Saying [itex]f(E\cup F)=f(x)[/itex] and that last line you wrote don't make sense to me. The idea is to start with an element from one side of the equation and then show it must also belong to the set on the other side of the equation : If you want to prove A=B, you need to show that if x belongs to A, it also belongs to B ( A is a subset of B), and ten you need to show if x belongs to B it also belongs to A (B is a subset of A). Together (A is a subset of B) and (B is a subset of A) yield B=A. Does this make sense to you? you need to keep track of where you start and make sure you understand what is going on!
    This side of the proof we want to show that [itex]f(E\cup F)\subset f(E)\cup f(F).[/itex] to do this you need to start with something like : Let [itex]x\in E\cup F[/itex] so that [itex]f(x)\in f(E\cup F)[/itex]. Then you brake it in to two cases, either x is in E or x is in F (or both). Try rewriting your argument and post what you get.
     
  4. Aug 11, 2013 #3
    Thank you for your reply :3 I think I get it now; here's my new proof:

    Let x be an element of E.
    Then f(x) is an element of f(E).
    But if x is an element of E then x is an element of E ∪ F
    Thus f(x) is an element of f(E ∪ F).
    Therefore f(E) is a subset of f(E ∪ F).

    Now let x be an element of F.
    Then f(x) is an element of f(F).
    But if x is an element of F then x is an element of E ∪ F.
    Thus f(x) is an element of f(E ∪ F).
    Therefore f(F) is a subset of f(E ∪ F).

    Together these two cases imply that f(E) ∪ f(F) is a subset of f(E ∪ F).

    Conversely, let x be an element of E ∪ F so that f(x) is an element of f(E ∪ F).

    Case 1: x is an element of E. Then f(x) is an element of f(E).
    By or introduction, f(x) is also an element of f(E) ∪ f(F).
    Therefore if x is an element of E then f(E ∪ F) is a subset of f(E) ∪ f(F).

    Case 2: x is an element of F. then f(x) is an element f(F).
    By or introduction, f(x) is also an element of f(E) ∪ f(F).
    Therefore if x is an element of F, f(E ∪ F) is a subset of f(E) ∪ f(F).

    These two cases together imply that f(E ∪ F) is a subset of f(E) ∪ f(F).

    Since f(E U F) is a subset of f(E) ∪ f(F), and vice versa, these two sets are equal.

    Hence f(E ∪ F) = f(E) ∪ f(F).
    qed

    The other thought I had for the 2nd part of the proof would be to consider 3 cases: x is in A/B, B/A and A ∩ B.
    But as I understand it my proof should work.

    Thanks again for your help :3
     
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