Results 1 to 4 of 4

Thread: Empty Cartesian product implies...?

  1. #1
    Junior Member
    Joined
    Oct 2009
    Posts
    59

    Empty Cartesian product implies...?

    Suppose that A x B = null set, where A and B are sets. What can you conclude?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,577
    Thanks
    790
    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?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2009
    Posts
    59
    I am not sure. Can one be a set and another the null? I am very confused on the null set.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,577
    Thanks
    790
    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 $\displaystyle A=\emptyset\lor B=\emptyset$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Cartesian product.
    Posted in the Discrete Math Forum
    Replies: 9
    Last Post: Nov 4th 2010, 01:59 PM
  2. Replies: 5
    Last Post: Apr 12th 2010, 02:16 PM
  3. Cartesian product
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Mar 9th 2010, 02:13 AM
  4. Cartesian product
    Posted in the Discrete Math Forum
    Replies: 7
    Last Post: Aug 27th 2009, 08:51 AM
  5. Empty Product Limit
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Jun 17th 2009, 11:58 AM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum