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Simple Set Theory Proofs

  1. May 27, 2015 #1

    B3NR4Y

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    1. The problem statement, all variables and given/known data
    1. Prove that if [itex] A \cap B = A [/itex] and [itex] A \cup B = A [/itex], then [itex] A = B [/itex]
    2. Show that in general [itex] (A-B) \cup B \neq A [/itex]
    3. Prove that [itex](A-B) \cap C = (A \cap C) - (B \cap C)[/itex]
    4. Prove that [itex] \cup_{\alpha} A_{\alpha} - \cup_{\alpha} B_{\alpha} \subset \cup_{\alpha} (A_{\alpha} - B_{\alpha})[/itex]
    2. Relevant equations
    None that I can think of.

    3. The attempt at a solution
    1. For the first one, it seems pretty clear to me going by the logic of the operations [itex] \cap [/itex] and [itex] \cup [/itex], because [itex] A \cup B [/itex] is both the sets joined, and if that equals A, B must be the empty set or A. The second part says [itex] A \cap B = A [/itex], the intersection of two sets is only what the two sets have in common, therefore B must be equivalent to A because in the previous part we found [itex] B=A [/itex] or [itex] B = \emptyset [/itex] and it cannot be the empty set because the empty set intersected with any set is the empty set. I came to the correct conclusion but I don't think my logic is sound.

    2. I can show by an example, if [itex] A = {1, 2, 3} [/itex] and [itex] B = {5, 6 ,7} [/itex], therefore [itex] A- B = {1 , 2, 3} [/itex] and [itex] (A-B) \cup B = {1,2,3,5,6,7} \neq A [/itex]. This is just one example and I know this doesn't prove it in general. But I'm not sure how to start.

    3. If [itex] x \in (A-B)\cap C [/itex] then [itex]x\in (A-B)[/itex] AND [itex]x\in C[/itex], since [itex]x \in (A-B), x\notin B[/itex] by the definition of subtraction of sets. Therefore [itex] x\in A\cap C[/itex] and [itex] x\notin B\cap C [/itex]. Therefore x is in the subtraction of [itex]A\cap C[/itex] and [itex]B\cap C[/itex]. Which proves the identity for any arbitrary element [itex]x[/itex]..... right? (the best way to end a proof is ".... right?")

    I am not sure where to start for 4.
     
  2. jcsd
  3. May 27, 2015 #2

    Zondrina

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    I believe your logic is quite sound for question 1. Near the end to clarify the argument, I would have said ##B## cannot be the empty set because the intersection of the empty set with anything is the empty set. We know ##A## intersect ##B## is equal to ##A## and not the empty set, so it must be the case ##A = B##.

    2. Your logic for question 2 is good as well. You have found a counter example; One counter example is enough to verify the claim in general.

    3. Right.

    4. I will think about this more after dinner.
     
  4. May 27, 2015 #3

    B3NR4Y

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    Okay, cool. I'm not as bad as I thought!
     
  5. May 27, 2015 #4

    Zondrina

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    4. Prove that: ##\cup_{\alpha} A_{\alpha} - \cup_{\alpha} B_{\alpha} \subset \cup_{\alpha} (A_{\alpha} - B_{\alpha})##.

    To gain some insight, lets write:

    $$\cup_{\alpha} A_{\alpha} - \cup_{\alpha} B_{\alpha} \subset \cup_{\alpha} (A_{\alpha} - B_{\alpha})$$
    $$(A_1 \cup \cdots \cup A_n) - (B_1 \cup \cdots \cup B_n) \subset (A_1 - B_1) \cup \cdots \cup (A_n - B_n)$$

    Can you show every element in the set on the left is contained in the set on the right? That is, ##\forall x \in (A_1 \cup \cdots \cup A_n) - (B_1 \cup \cdots \cup B_n)## can you show ##x \in (A_1 - B_1) \cup \cdots \cup (A_n - B_n)##?
     
  6. May 27, 2015 #5

    Ray Vickson

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  7. May 27, 2015 #6

    vela

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    If A = {1, 2, 3, 4} and B = {2, 3}, then ##A \cup B = \{1,2,3,4\}=A## but ##B \ne A## and ##B \ne \emptyset##.
     
  8. May 28, 2015 #7

    wabbit

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    Sorry but this is completely wrong. ##A\cap B =A## does not imply in any way that B should be A or the empty set.

    @B3NR4Y, try to think again about what ##A\cap B=A## means. First, do you see that this is equivalent to ##A\subset A\cap B## ?
     
  9. May 28, 2015 #8

    PeroK

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    I guess you mean that ##A\cap B=A## is equivalent to ##A\subset B##
     
  10. May 28, 2015 #9

    wabbit

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    No, I meant exactly what I said. I was trying to get OP think by himself by only spelling out a minimal step.
     
  11. May 28, 2015 #10

    HallsofIvy

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    I would try to be much more specific. To prove "A= B" you need to prove "A is a subset of B" and "B is a subset of A".

    To prove "A is a subset of B" start with "If x is an element of A" and use the properties of both A and B to conclude "x is an element of B".

    To prove " if [itex] A \cap B = A [/itex] and [itex] A \cup B = A [/itex], then [itex] A = B [/itex]"
    start, "if x is an element of A, then, because [itex]A \cap B= A[/itex] x is an element of B. Therefore A is a subset of B.
    if y is an element of B then [itex]A\cup B= A[/itex], y is an element of A. Therefore B is a subset of A. Therefore A= B."
     
  12. May 28, 2015 #11

    B3NR4Y

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    Okay so for 3, [itex]x\in (A\cap C) - (B\cap C) [/itex], which implies [itex]x\in A \cap C[/itex] but [itex]x\notin B\cap C[/itex] and from that it follows [itex]x \in A,C[/itex] and [itex]x\notin B[/itex], therefore [itex]\forall x\in (A\cap C) - (B\cap C), \, \, x\in (A-B)\cap C[/itex] which proves [itex](A\cap C) - (B\cap C)\subset (A-B)\cap C[/itex] and I've already proven that the converse is true, so this completes the proof?

    Okay I think I understand it better now, to prove two things are equivalent I need to prove the are subsets of each other, and for problems like 4, I only need to prove that any arbitrary "x" in the first part is also in the second part.

    I am working from "Introductory Real Analysis" by Kolmogorov and Fomin.
     
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