# Thread: Empty Cartesian product implies...?

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

Suppose that A x B = null set, where A and B are sets. What can you conclude?

2. By definition, $\displaystyle A\times B=\{(x,y)\mid x\in A\mbox{ and }y\in B\}$. Now, when does it happen that any set of the form $\displaystyle \{f(x,y)\mid A(x,y)\}$ 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?

3. I am not sure. Can one be a set and another the null? I am very confused on the null set.

4. 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 $\displaystyle x\in A$ and a $\displaystyle y\in B$. By the definition of $\displaystyle A\times B$, then $\displaystyle (x,y)\in A\times B$; therefore, $\displaystyle A\times B\ne\emptyset$.

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

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

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

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# what can one conclude if the Cartesian product of two sets is null

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