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Sets and Logic

  1. Sep 13, 2014 #1
    Hi ,

    Can anyone please give me an idea to disprove the following with counter example:
    A , B & C be sets. If A X C = B X C , the A = B .

    I tried giving random numbers in venn diagram but didn't work. And, using subset way to prove equal but still couldn't solve it.
     
  2. jcsd
  3. Sep 13, 2014 #2
    Are you convinced that the statement is false?
     
  4. Sep 14, 2014 #3
    Let ##Y## be some set (say ##Y=\{0,1\}## or ##Y=\mathbb R##), and consider the special case where ##C=Y^\infty##, the set of sequences in ##Y##.

    Does ##A\times Y^\infty = B\times Y^\infty## imply ##A=B##? Let's think about the case of ##A=Y##.
     
  5. Sep 14, 2014 #4
    Use the modus tollens. ##A \times C = B \times C \Rightarrow A = B## is the same as ##A \neq B \Rightarrow A \times C \neq B \times C##. Can you find a counterexample to the latter statement?
     
    Last edited: Sep 14, 2014
  6. Sep 15, 2014 #5
    Thanks. I tried keeping set C as an empty set. It worked. Cheers :)
     
  7. Oct 6, 2014 #6
    I'm new to Sets so if this doesn't apply my apolagizes.

    AxC=BxC this statement will only be valid if C is invertable. So the Det(C) cannot be equal to zero. If you use a random number generator there is a possibility that the Det(C) will equal zero.

    AxC=BxC
    (AxC=BxC)xC^-1
    A=B

    if C is not invertable then A=B is not valid
     
  8. Oct 8, 2014 #7

    FactChecker

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    Your doubts are correct, but you still might be close.

    In general, a set just sits there and does nothing. You can't talk about it being invertible. To talk about an inverse, there has to be more structure to C than just being a general set. Some sets are ordered pairs that represent functions. For a set like that, you can talk about the inverse of the function that the set defines.

    Since the OP was asking for a counterexample, you could define a specific C that is a function and use that to prove it is a counterexample. You would have to be specific about C and use that. But the definition of the Cartesian product, A x C doesn't give you much to work with.

    Consider the example that @klamgade came up with, C = [itex]\emptyset[/itex]. It is the fact that the operation "x[itex]\emptyset[/itex]" is not invertible that makes it work as a counterexample. It would not be correct to say that [itex]\emptyset[/itex] is not invertible.
     
  9. Oct 8, 2014 #8

    PeroK

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    The best way to show this is false, perhaps, is to prove it's true! As follows:

    ##Let \ \ c \in C##
    ##For \ \ a \in A##
    ##(a, c) \in A \times C##
    ##(a, c) \in B \times C##
    ##a \in B##
    ##\therefore \ A \subseteq B##

    By the same method you can show ##B \subseteq A## hence ##A = B##

    So, the statement is true as long as C is non-empty.
     
  10. Oct 8, 2014 #9
    I don't buy that argument. I don't think it follows from ##(a,c)\in A\times C## and ##c\in C## that ##a\in A##. Let ##C= \mathbb R^\infty## and ##A=\mathbb R##. Let ##c=(1,1,1,...)\in C## and ##a = (0,0)\in \mathbb R^2.## Then ##a\notin A##, but ##(a,c)=(0,d)##, where ##d=(0,1,1,1,...)\in C##, so that ##(a,c)\in A\times C##.
     
  11. Oct 8, 2014 #10

    PeroK

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    You're missing a subtlety of the definition of the Cartesian Product.

    Let ##A = \mathbb{R} \ \ and \ \ B = \mathbb{R}^2##

    Then a typical element of ##A \times B## would be:

    ##(x, (y, z))##

    Which is not the same as:

    ##((x, y), z) \ \ or \ \ (x, y, z)##

    Which are elements of ##B \times A## and ##\mathbb{R}^3## respectively.

    In any case:

    ##(a, b) \in A \times B \ \ iff \ \ a \in A \ \ and \ \ b \in B##
     
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