Why is there no function with a nonempty domain and an empty range?

The standard set-theoretic way to define functions is that:

1. The cartesian product of the sets $A$ and $B$, written as $A \times B$, is the set of all ordered pairs where the first element of the pair is in $A$ and the second in $B$: $$A \times B = \{(a,b): a \in A, b \in B\}.$$

   (Representing these ordered pairs as sets, and showing that the cartesian product of two sets is indeed a set under the axioms of set theory, are details that we may safely skip here.)

2. A relation $R$ between the sets $A$ and $B$ is any subset of their cartesian product: $R \subset A \times B$.

   (Often, by convention, we write $a \mathop R b$ as a shorthand for $(a,b) \in R$; this is particularly common when the symbol chosen for the relation is not a letter like $R$, but something abstract like $\sim$ or $\odot$.)

3. A function $f$ from $A$ to $B$ is a relation between $A$ and $B$ (i.e. a subset of their cartesian product) satisfying the following two extra conditions:

   • existence of images: for all $a \in A$, there is a $b \in B$ such that $(a,b) \in f$.

   • uniqueness of images: if $(a,b) \in f$ and $(a,b') \in f$, then $b = b'$.

   If the relation $f$ is a function, then, for each $a \in A$, there exists exactly one $b \in B$ satisfying $(a,b) \in f$. We call this $b$ the image of $a$ under $f$, written $f(a)$, so that: $$f(a) = b \iff (a,b) \in f.$$

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So, what about when $B = \varnothing$? In that case, for any $A$, the cartesian product $A \times B = \varnothing$, since there exist no pairs $(a,b)$ such that $a \in A$ and $b \in B$. (The same, of course, is also true whenever $A = \varnothing$.)

Since the only subset of $\varnothing$ is $\varnothing$, the only relation between $A$ and $B$ is the empty relation $\varnothing$. The question, then, is: is the empty relation a function from $A$ to $B$?

• If $A \ne \varnothing$, no, it is not, because there exists at least one $a \in A$, but there can be no $b$ such that $(a,b) \in \varnothing$.

• If $A = \varnothing$, yes, it is. In this case, both the existence and uniqueness conditions are vacuously true, since there is no $a \in A$ for which they could fail.

Thus, there is a (single) function from the empty set to any set (including the empty set itself), but there is no function from a non-empty set to the empty set.

You messed up because $A\times\varnothing$ is the empty set. So only if $A$ is empty as well such function exists.

A function $f\colon A\to B$ is a subset of $A\times B$ such that

1. for every $a\in A$, there exists $b\in B$ such that $(a,b)\in f$;

2. for every $b,b'\in B$, if there exists $a\in A$ with $(a,b)\in f$ and $(a,b')\in f$, then $b=b'$.

The first condition expresses the fact that every element in $A$ has an image in $B$. The second condition expresses the fact that such image is uniquely determined by the element in the domain. Thus, if $a\in A$, we can write $f(a)$ to denote the unique element in $B$ such that $(a,f(a))\in f$.

Now, if $B=\emptyset$, we have $f\subset A\times\emptyset=\emptyset$, so $f$ is empty. Thus every element of $A$ fails to have an image in $B$ under $f$, and so a function $f\colon A\to\emptyset$ can exist only when also $A$ is empty.

On the contrary, if $A=\emptyset$, condition 1 cannot fail for any element of $\emptyset$, which has no elements; also condition 2 cannot fail, so there is indeed a function $f\colon \emptyset\to B$. It's unique, because $f\subseteq \emptyset\times B=\emptyset$, so $f=\emptyset$.