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Proving Two Sets Are Equivalent

  1. Aug 23, 2014 #1
    To prove that two sets are in fact the same, do I actually have to prove that the two are subsets of each other; or could I prove that they are equivalent by some other means, such as invoking the definitions of the sets?

    For instance, the I am trying to show that the binary set operator ##+## is commutative; that is,

    ## A + B = B + A##,

    where

    ##A + B = (A-B) \cup (B-A)##.

    By using the commutative property of the ##\cup## operator, I was able to prove the equality:

    ##A + B = (A-B) \cup (B-A) = (B-A) \cup (A-B) = B+A##

    Is this a valid proof?
     
  2. jcsd
  3. Aug 23, 2014 #2

    micromass

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    Yes, as long as you've already proven that union is commutative.
     
  4. Aug 23, 2014 #3
    °Yes, I did prove that the union is commutative.

    I have another question; it comes from the same problem. I am asked to determine whether ##A + \emptyset = A## is a true statement. Here is my proof:

    ##A + \emptyset = (A - \emptyset) \cup (\emptyset - A)##

    Before we proceed, let us determine the nature of ##(A - \emptyset)## and ##(\emptyset - A)##.

    ##(A - \emptyset) = \{ x~| x \in A \wedge x \notin \emptyset \}##. The statement ##x \notin x \emptyset ## is always true by definition. Therefore,

    ##(A- \emptyset) = \{x~| x \in A \wedge T \} = \{x~| x \in A\} = A##.

    Here is the portion of my proof that I am unsure of:

    ##\emptyset - A = \{x~| \underbrace{x \in \emptyset} \wedge x \notin A \}##. The underlined portion is always a false statement, as the empty set never houses any elements. As such,

    ## \emptyset - A = \{x~| F \wedge \underbrace{x \notin A}\}##. The truth value of the underlined portion is irrelevant to the truth value of the entire statement. Thus,

    ## \emptyset - A = \{x~| F \}##...

    I would interpret this as being the empty set, but I do not have any basis for such an inference. How might I justly proceed from this last step in my proof?

    EDIT:

    What if I wrote ##\emptyset - A = \{x~|\forall xp(x) = F\}##, where ##p(x)## is the condition that must be satisfied in order for the element ##x## to be a member. What ##p(x)=F## states is, that every ##x## makes the statement false, meaning that it can contain no elements.

    Would this be sufficient reasoning?
     
    Last edited: Aug 23, 2014
  5. Aug 24, 2014 #4

    HallsofIvy

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    B- A is, by definition, "All members of B that are not in A". If B is empty, B- A is the empty set no matter what A is.
     
  6. Aug 24, 2014 #5

    vela

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    I think you're fine simply saying that ##\emptyset-A = \emptyset##, but if you're still worried, you can show that ##\emptyset \subset \emptyset-A## and ##\emptyset-A \subset \emptyset##.
     
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