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 $\emptyset$ ?)

Solution 1:

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.

Solution 2:

A slightly more formal rephrase of the other answer is to calculate which elements are in $A \times \emptyset$: for any $p$, \begin{align} & p \in A \times \emptyset \\ \equiv & \;\;\;\;\;\text{"definition of $\times$ on sets"} \\ & \textrm{isPair}(p) \;\land\; \textrm{fst}(p) \in A \;\land\; \textrm{snd}(p) \in \emptyset \\ \equiv & \;\;\;\;\;\text{"$x \in \emptyset \equiv \textrm{false}$ for any $x$; simplify"} \\ & \textrm{false} \\ \equiv & \;\;\;\;\;\text{"$x \in \emptyset \equiv \textrm{false}$ for any $x$"} \\ & p \in \emptyset \\ \end{align} and therefore, by set extensionality, $A \times \emptyset = \emptyset$. A very similar proof of course goes for $\emptyset \times B = \emptyset$, and obviously either of these implies $\emptyset \times \emptyset = \emptyset$.

Solution 3:

$\emptyset \times A$ is empty set. Let's prove by contradiction:

Suppose $\emptyset \times A$ is non-empty, then there exists an ordered pair $(x,y)$ such that $(x,y) \in \emptyset \times A$, so we have $x \in \emptyset$ and $y \in A$, which is a false statement because there is no object belongs to empty set. Thus, our hypothesis is false. In other words, $\emptyset \times A$ is empty set.

By the way, I've found out that it's not a bad idea to consider contradiction when dealing with the empty set.