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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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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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    $\begingroup$ 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 ? $\endgroup$ – temo Jul 14 '11 at 9:07
  • $\begingroup$ 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. $\endgroup$ – algebra_fan Jul 14 '11 at 9:15
  • $\begingroup$ 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). $\endgroup$ – user83827 Jul 14 '11 at 10:01
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    $\begingroup$ @ccc: The first argument was the formal one. But it is always useful to have other views of the matter, either informal, or formalizable, but with some effort, as in this case. Anything that bolsters the intuition is useful. Think of the set $S$ of all ordered pairs $(a,b)$, where $a$ is a unicorn and $b$ is a giraffe. How many ordered pairs are in $S$? Clearly $0$, since there are no unicorns. Yes, I have dragged in biology, but most students will remember this argument. $\endgroup$ – André Nicolas Jul 14 '11 at 12:28
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    $\begingroup$ @temo, there is nothing weird in a stamente like «$x\in\emptyset$»: is is a statement like any other, and a false one. $\endgroup$ – Mariano Suárez-Álvarez Feb 15 '15 at 4:07
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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 $\emptyset \times \emptyset = \emptyset$.

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$\emptyset \times A$ 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.

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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 $a$. $\Box$

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    $\begingroup$ Your proof is a great proof of the fact that English is an immensely more efficient way to express proofs that whatever it is your notation is. If instead of those huge formulas you'd have explained in words what you meant! $\endgroup$ – Mariano Suárez-Álvarez Feb 15 '15 at 4:09
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    $\begingroup$ Sometimes, writing a proof in a "mathmatic language" is necessary to be sure that our intuition is not leading us astray as the mathematicians of the early 19th century were before an explicit formulation of the limit concept was found. $\endgroup$ – Bernard Massé Feb 15 '15 at 4:35
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    $\begingroup$ I agree to the both of you. @MarianoSuárez-Alvarez: Yes, it's annoying to go through the process of translating my notation into yours. But since I'm new to mathematical rigour I'm trying to stick to the few axioms and methods that I'm given. I know it lacks the light-footedness of a pro but I don't want to risk falling for my wrong intuition. That's why I tend towards Bernard's position. $\endgroup$ – Max Herrmann Feb 15 '15 at 10:13
  • $\begingroup$ Mathematical rigor has absolutely nothing to do with writing incomprehensible formulas. $\endgroup$ – Mariano Suárez-Álvarez Feb 15 '15 at 10:18
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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 $.

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