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Math Help - Empty Cartesian product implies...?

  1. #1
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    Empty Cartesian product implies...?

    Suppose that A x B = null set, where A and B are sets. What can you conclude?
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  2. #2
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    By definition, A\times B=\{(x,y)\mid x\in A\mbox{ and }y\in B\}. Now, when does it happen that any set of the form \{f(x,y)\mid A(x,y)\} is empty? This happens when it is not the case that there exist x and y such that A(x,y), or, equivalently, when for all x and y, it is not the case that A(x,y). Can you state this more simply for this particular A?
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    I am not sure. Can one be a set and another the null? I am very confused on the null set.
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    Suppose that A x B = null set, where A and B are sets.
    Can one be a set and another the null?
    It is given that both A and B are sets, though one or both can be empty.

    The easiest way to find the answer is to consider several cases. Suppose both A and B are nonempty. Then there exists an x\in A and a y\in B. By the definition of A\times B, then (x,y)\in A\times B; therefore, A\times B\ne\emptyset.

    Suppose that A is empty, regardless of B. Then you can't find a single pair of x and y such that x\in A and y\in B. Therefore, A\times B=\emptyset. The same happens when B is empty.

    The same can be shown more formally. I'll write \exists for "there exists", \land for "and", \lor for "or" and \neg for negation. Then for any function f(x,y) and a predicate P(x,y), \{f(x,y)\mid P(x,y)\}\ne\emptyset iff \exists x\exists y.\,P(x,y). In our case, P(x,y) is x\in A\land y\in B, and f(x,y) is the function that returns a pair (x,y). So A\times B=\{(x,y)\mid x\in A\land y\in B\}\ne\emptyset iff \exists x\exists y.\,x\in A\land y\in B. This in turn is equivalent to (\exists x.\,x\in A)\land(\exists y.\,y\in B), or A\ne\emptyset\land B\ne\emptyset.

    Taking negation, A\times B=\emptyset iff \neg(A\ne\emptyset\land B\ne\emptyset) iff (\neg A\ne\emptyset)\lor (\neg B\ne\emptyset) iff A=\emptyset\lor B=\emptyset.
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