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Math Help - Cartesian product

  1. #1
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    Cartesian product

    My book says the converse of the following implication is false:

    A, B, C, D are sets such that A\subseteq C and B \subseteq D, then A \times B \subseteq C \times D.

    Although P \Rightarrow Q \not \equiv Q \Rightarrow P, I don't understand why since the above implication still can be true under special circumstances.

    For exmple, When A=D and C=B or when A = B = \emptyset or C=D=\emptyset , I still can make it true.

    I think P \Rightarrow Q \not \equiv Q \Rightarrow Q is not always true. Am I missing something?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by novice View Post
    My book says the converse of the following implication is false:

    A, B, C, D are sets such that A\subseteq C and B \subseteq D, then A \times B \subseteq C \times D.

    Although P \Rightarrow Q \not \equiv Q \Rightarrow P, I don't understand why since the above implication still can be true under special circumstances.

    For exmple, When A=D and C=B or when A = B = \emptyset or C=D=\emptyset , I still can make it true.

    I think P \Rightarrow Q \not \equiv Q \Rightarrow Q is not always true. Am I missing something?
    Let (x,y)\in A\times B then x\in A\text{ and }y\in B\implies x\in C\text{ and }y\in D\implies (x,y)\in C\times D. Done.
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  3. #3
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    Quote Originally Posted by novice View Post
    My book says the converse of the following implication is false:
    A, B, C, D are sets such that A\subseteq C and B \subseteq D, then A \times B \subseteq C \times D.
    Consider: A = \{ a\} ,\;B = \emptyset ,\;C = \{ c\} ,\;D = \{ d\} .
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  4. #4
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    Quote Originally Posted by Plato View Post
    Consider: A = \{ a\} ,\;B = \emptyset ,\;C = \{ c\} ,\;D = \{ d\} .
    You are quite amazing. How could I be so blind?
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    Quote Originally Posted by Drexel28 View Post
    Let (x,y)\in A\times B then x\in A\text{ and }y\in B\implies x\in C\text{ and }y\in D\implies (x,y)\in C\times D. Done.
    The implication is not a problem, but the converse.

    Plato showed the condition for which when it's false. I thought of an empty set on each side of the inclusion, but never thought of having only one emptyset for the leftside while keeping the rightside nonempty.

    This is example or over focusing to the point that I loss the entire picture.
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  6. #6
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    Quote Originally Posted by Plato View Post
    Consider: A = \{ a\} ,\;B = \emptyset ,\;C = \{ c\} ,\;D = \{ d\} .
    Plato,

    Now, with the implication, if I declare that A, B, C, and D being nonempty sets and that A \not = B, I think it will be thought to make the converse false.

    Do you agree?
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