Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Supposing that, for example, we have two sets $A$ and $B$ where $\;A = \varnothing \;$ and $\,B = \{a,b\}$.

What is the result of the cross product of those sets?

My first intuition would be to say that the resulting set would be the empty set as well since you can't set up any ordered pairs of $B$.

Am I correct?

share|improve this question

marked as duplicate by Asaf Karagila, Brian M. Scott, Rahul, Thomas, Cameron Buie Mar 5 '13 at 1:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1  
This is just one amongst the many duplicates this question have on this site. If one is so inclined it is likely to find at least two more duplicates. –  Asaf Karagila Mar 5 '13 at 0:28
    
As a result the only relation on $A\times B$ is the so-called empty function with codomain $B$. –  peoplepower Mar 5 '13 at 0:38
2  
@RahulNarain: And in the comments, and related links, of that duplicate there are more duplicates... –  Asaf Karagila Mar 5 '13 at 0:39
    
@Asaf: Good point. Voting to close that one as well... –  Rahul Mar 5 '13 at 0:55
add comment

2 Answers 2

up vote 3 down vote accepted

Yes, you're correct: For all sets $B$, if $A = \varnothing$, then $$A \times B = \varnothing \times B = B\times A = B\times \varnothing = \varnothing$$ and for the reason you argue: there exist no ordered pairs in $\varnothing \times B$, by the definition of the Cartesian Product.

share|improve this answer
add comment

Yes, you are correct.

The cartesian product is defined as $$A \times B = \{(a,b) \mid a\in A, b\in B\}.$$

In the case that one or both of the sets $A$ and $B$ are empty, there is no single index pair $(a,b)$ such that $a\in A$ and $b\in B$, which means $A\times B = \emptyset$.

share|improve this answer
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.