# Vector space of functions on an empty domain?

I know that these sorts of pathological cases are irrelevant, but I want to "practice" as it were and so want to understand the following. Hoffman and Kunze give the following example of a vector space:

The space of functions from a set to a field. Let $$F$$ be any field and let $$S$$ be any non-empty set. Let $$V$$ be the set of all functions from the set $$S$$ into $$F$$.

They give the usual definitions of addition and scalar multiplication. At any rate, my question is about why they give the nonempty requirement. If $$S$$ is empty, then $$V$$ itself is not empty but rather contains the empty set (the empty set can be defined as a function on an empty set), $$V = \{ \emptyset \}$$.

If I define $$\emptyset$$ as my neutral element, then why on earth can't this be a vector space?

• I see @badjohn So, is it fair to say that we could define a vector space as I have above, but we choose not to?
– EE18
Commented Sep 7, 2023 at 14:43
• It is a vector space. It's just that some people don't like functions with empty domains. The example does not exclude this case, nor does it say that if you take $S=\varnothing$ then it is not a vector space. Commented Sep 7, 2023 at 14:59
• @ArturoMagidin I was just being cautious. "At least some" includes the case "all". Commented Sep 7, 2023 at 15:11
• Sorry, it was over-cautious. Now, with your answer, it serves no purpose and I will delete it. Commented Sep 7, 2023 at 15:19

I suspect that is what is at issue here. The author does not want to consider functions with empty domain, so they exclude $$S=\varnothing$$. The wording of the example does not categorically state that you will not get a vector space when $$S$$ is empty: it just does not treat that case and is silent about it.
You are correct that a one element set, say $$\{\star\}$$, is a vector space (over any field): the trivial vector space with operations $$\star+\star = \star$$ and $$\alpha\cdot\star = \star$$ for any scalar $$\alpha$$. Perfectly fine vector space. I know of no one who would exclude this vector space (it would introduce all sorts of exception statements in standard theorems: for example, it would no longer be true that an intersection of subspaces of $$V$$ is again a subspace of $$V$$, we would have to add a clause stating that the intersection consists of more than the zero vector).