Why is $S$ specified to be nonempty in Axler’s definition of function spaces $\mathbf F^S$?

Question

In Example 1.24 of S. Axler’s Linear Algebra Done Right, the following statement is made:

If $S$ is a nonempty set, then $\mathbf F^S$ […] is a vector space over $\mathbf F$.

(Here, $\mathbf F^S$ is the set of functions from $S$ to $\mathbf F$, and $\mathbf F$ is either $\mathbb R$ or $\mathbb C$.)

However, this statement is true even if $S=\varnothing$. (Then there is only one function $S \to \mathbf F$, and $\mathbf F^S$ is trivial.) Why would Axler have written nonempty? Is he simply distracting from “weird” corner cases for didactic purposes, or is there some deeper motivation?

Answer 1

In Example 1.24 of Linear Algebra Done Right (3rd edition), I considered $\mathbf{F}^S$ only for nonempty sets $S$ because it is unusual to think about a function whose domain is the empty set. I did not want to distract the reader with that issue when there are more important things for the reader to think about in the context of learning about $\mathbf{F}^S$.

Answer 2

The reason ought to be your first, that is avoiding the corner case.

There is no reason I can think of for not considering $F^{\emptyset}$ as a (trivial) vector space. Indeed, to truly exclude it would be inconvenient.

But, in the context I can see the point to not ask for the argument. It is one of those things that could trip up somebody. Generally, I try to avoid such questions, too. (Or if not, I hint at the fact there is a corner case.)