# How do the sets $\emptyset\times B,\ A\ \times \emptyset, \ \emptyset \times \emptyset$ look like?

If we have a function $f:A \rightarrow B$, then one way to give meaning, I think, to this function, in terms of set theory, is to say, that $f$ is actually a binary relation $f=(A,B,G_f)$, where $G_f \subseteq A \times B$ is the graph of the function. Now my question is: what is $f$ if

$\bullet \ A=\emptyset, \ B\neq\emptyset$,?

$\bullet \ B=\emptyset, \ A\neq\emptyset$ ?

$\bullet \ B=\emptyset, \ A=\emptyset$ ?

(Another way to formulate this, I think, would be: How do the sets $\emptyset\times B,\ A\ \times \emptyset, \ \emptyset \times \emptyset$ look like ? Are they all $\emptyset$ ?)

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## 1 Answer

Yes, they're all empty sets. For example, $\emptyset \times A$ consists of all pairs of the form $(o,a)$ with $o \in \emptyset, a \in A$. But the empty set has no elements, hence $\emptyset \times A$ has no elements, hence $\emptyset \times A$ is the empty set. A similar argument works for the other two sets.

Here is how this problem can be interpreted in terms of cardinalities. For any sets $A,B$ the cardinality of $A \times B$ is the product of cardinalities of $A$ and $B$. Hence the cardinality of $\emptyset \times A$ is just $0 \cdot |A| = 0$ so $\emptyset \times A$ has $0$ elements, and hence $\emptyset \times A = \emptyset$. And a similar argument will work in the other two cases.

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Although I understand well what you mean, isn't writing $o\in \emptyset$ a bit weird, since the emptyset should not contain any other set ? – temo Jul 14 '11 at 9:07
Yes, you are correct. I just used the definition of Cartesian product to write down how an element of $\emptyset \times A$ "looks like". And since $\emptyset$ has no elements, the condition $o \in \emptyset$ is never satisfied, hence $\emptyset \times A$ is empty. Basically, I was trying to say "the empty set has no elements, hence $\emptyset \times A$ has no elements" rigorously. – algebra_fan Jul 14 '11 at 9:15
The first answer is spot on, but I don't see a way of formalizing the second that doesn't pass through the first, especially if $B$ is infinite (but even if $B$ is finite, there's something to show). – ccc Jul 14 '11 at 10:01
– algebra_fan Jul 14 '11 at 10:18
It's not that I think it's overkill.  I think it's begging the question argue that $\emptyset \times B$ is empty since $0 \times \kappa = 0$ for all $\kappa$.  Like I said, I think the first answer is spot on, and I'm sorry if this seems pedantic, but I think it's misleading to suggest that changing the language of the question to an equivalent statement in cardinal arithmetic (and implicitly appealing to an unjustified "$0$ times anything is $0$" intuition) is a suitable replacement for the actual argument. – ccc Jul 14 '11 at 10:37
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