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 $\emptyset$ ?)
$\endgroup$5 Answers
$\begingroup$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.
$\endgroup$ 10 $\begingroup$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 $\emptyset \times \emptyset = \emptyset$.
$\endgroup$ $\begingroup$$\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.
$\endgroup$ $\begingroup$You can convince yourself that they're all empty sets by the fact that a Cartesian product $X_1 \times X_2 \times ... \times X_n$ is the empty set iff at least one $X_i$ is empty.
Proof. From the definition of the Cartesian product we have that an arbitrary object $a$ is in $X_1 \times X_2 \times ... \times X_n$ if and only if for every component $a_i$ of $a$ there is an element $x_i$ in $X_i$ such that $a_i = x_i$.
$\leftarrow$: If at least one $X_i$ is empty there is no such $x_i$.
$\rightarrow$: If $X_1 \times X_2 \times ... \times X_n$ is empty, there is no tuple $a$. $\Box$
In what follows we make use of the definitions found in the wiki article on binary relations.
Let $R$ be a binary relation, $R \subset X \times Y$.
The next two statements have direct proofs:
$\tag 1 R \ne \emptyset \; \text{ iff } \; \text{Domain(}R\text{)} \ne \emptyset$
$\tag 2 R \ne \emptyset \; \text{ iff } \; \text{Image(}R\text{)} \ne \emptyset$
Since it is always true that $\text{Domain(}R\text{)} \subset X$, whenever $X = \emptyset$ it has to follow that $R = \emptyset$.
Since it is always true that $\text{Image(}R\text{)} \subset Y$, whenever $Y = \emptyset$ it has to follow that $R = \emptyset$.
Setting $R$ equal to, say, $A \times \emptyset$, we see that $A \times \emptyset = \emptyset $.
$\endgroup$
