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The word "or" in set theory

  1. Jan 6, 2016 #1
    1. The problem statement, all variables and given/known data
    So I was doing this problem in Munkres's Topology book:

    Determine whether the statement is true or false, If a double implication fails, determine whether one or the other of the possible implications holds:

    A ⊂ B or A ⊂ C ⇔ A ⊂ ( B ∪ C )

    2. Relevant equations
    -

    3. The attempt at a solution
    I know that the ⇒ direction is true because (x∈A→x∈B or x∈A→x∈C)⇒(x∈A→(x∈B or x∈C))
    For the other direction, I thought at first that it's true, but I checked some online answers and what I found is it's false. I thought it's true because what I had in mind is the word "or" stands for "either A is a subset of B, or A is a subset of C, or both". And if A is a subset of the union of B and C, then it's implied that A is a subset of at least one of them. What am I getting wrong?

    This is the answer that I found online:

    If the LHS statement is true then for each latex.png we have that latex.png and so latex.png , i.e. latex.png . The other direction is not correct though. Consider the sets latex.png then latex.png but it is not true that latex.png .
     
  2. jcsd
  3. Jan 6, 2016 #2

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    The example shows that this is wrong: "And if A is a subset of the union of B and C, then it's implied that A is a subset of at least one of them."
    A={1,3}⊂ B∪C={1,2,3,4}, but clearly A is not a subset of either B or C.

    venn.jpg
     
  4. Jan 6, 2016 #3

    Mark44

    Staff: Mentor

    Your problem isn't in misunderstanding of the word "or" (as implied by your thread title). The counterexample given below clearly shows that A can be contained in ##B \cup C## without A being a subset of either of the other sets.

    Note that in the example, one element of A is contained in B, and the other element of A is contained in C, but A is not a subset of either B or C.
     
  5. Jan 6, 2016 #4
    Oh, I don't how I missed that, feel stupid for asking haha.
    Thanks.
     
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