Set of functions from the empty set to other set

Question

so am a bit unsure about the whole propriety of the empty set function So the only function from the empty set to the empty set is the empty function and the only function from the empty set to a finite set is the empty function Does that mean the only function from a finite set to the empty set is the empty function?

Answer 1

No. If $B$ is a nonempty finite set, then there is no function from $B$ to $\emptyset$.

I don't know what definition of "function" you have. For me, it's that there's a set $D$ called the domain ($B$ in this case), another set $C$ called the codomain ($\emptyset$ in this case), and a third set, $R$, the 'rule', which is a subset of $D \times C$ with certain properties. In our case, that makes $R$ a subset of $B \times \emptyset = \emptyset$, so that would mean that $R$ must be the empty set.

One of the properties of $R$ that's required for $(D, C, R)$ to be a function is that for all $x \in D$, there's an element $r \in R$ where $r = (x, y)$ for some $y \in C$.

In our situation, consider any element $b \in B$. There's supposed to be an element $r \in R$ with that property. Unfortunately, $R = \emptyset$, so no such element $r$ can exist, i.e., there's no triple $(B, \emptyset, \emptyset)$ that matches the definition of "function" when $B$ is nonempty.