Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a set. What is $X\times \emptyset$ supposed to mean? Is it just the empty set?

share|cite|improve this question

marked as duplicate by MathOverview, Claude Leibovici, TooTone, Sami Ben Romdhane, Michael Hoppe Mar 1 '14 at 10:39

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.

up vote 8 down vote accepted

And more can be said: a cartesian product is empty if and only if one of the two factors is empty.

share|cite|improve this answer
My formal proof: – Dan Christensen Jul 31 '12 at 19:25
A side note: This proof also works fine for the Cartesian product of finitely many sets. For arbitrary products, this statement is precisely the axiom of choice. – Nate Eldredge Jul 31 '12 at 22:44

$X \times \emptyset = \emptyset$. In fact, if not we have $x \in X $ and $y \in \emptyset$ such that $(x,y) \in X \times \emptyset$. But $y \in \emptyset$ is impossible.

share|cite|improve this answer

Recall the definition of $A\times B$: $z\in A\times B$ if and only if $z=\langle a,b\rangle$ where $a\in A$ and $b\in B$. That is $A\times B$ is the set of all ordered pairs whose first coordinate is in $A$ and second in $B$.

If $B$ is empty then there are no ordered pairs $\langle a,b\rangle$ such that $b\in\varnothing$, therefore $A\times\varnothing=\varnothing$. Similarly $\varnothing\times B=\varnothing$.

share|cite|improve this answer

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