# Is it wrong to be irked when an author presents this linear algebra proposition without stating the subset is non-empty?

The following is given as a proposition in a book by Thomas Garrity: "A subset $U$ of a vector space $V$ is a subspace of $V$ if $U$ is closed under addition and scalar multiplication."

It would seem to me that the author should state that the subset is non-empty. The empty set is a subset of $V$ and it is vacuously true that if $u, v \in \emptyset$, then $\alpha{}u + \beta{}v \in \emptyset$ for all $\alpha, \beta \in \mathbb{F}$. But $\mathbf{0} \not\in \emptyset$, so $\emptyset$ is not a vector space.

Would this not be a legitimate concern, or am I being pedantic here?

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Non-empty should have been specified. – André Nicolas Aug 10 '14 at 20:00
In a context where addition is considered only as a binary operation, so "closed under addition" means that $a,b\in U$ implies $a+b\in U$, then "nonempty" needs to be included. In a context where "addition" refers to sums of any finite number of terms, rather than just two, "closed under addition" would mean that $U$ contains the sum of each finite subset of $U$. Then you don't need to say "nonempty" because $0$ is the sum of the empty set (which is a finite subset of any $U$). – Andreas Blass Aug 10 '14 at 21:10
It should be corrected; otherwise you're going to constantly have to write, "Suppose M is a non-empty subspace of X." And that would definitely become a source of irritation. :) – TrialAndError Aug 11 '14 at 0:13