Is the span of the empty set in a vector space equal to $\lbrace 0\rbrace$, or does it have no span? The "correct" answer in my opinion is the latter. See Example 3.10.3 of http://math.mit.edu/~rstan/ec/ec1.pdfhttps://web.archive.org/web/20201101123724id_/http://math.mit.edu/~rstan/ec/ec1.pdf for a reason.:
3.10.3 Example. We give one simple example of the use of equation (3.34). We wish to count the number of spanning subsets of $V_n$. For the purpose of this example, we say that the empty set $\emptyset$ spans no space, while the subset $\{0\}$ spans the zero-dimensional subspace $\{0\}$.* If $W \in B_n(q)$, then let $f(W)$ be the number of subsets of $V_n$ whose span is $W$, and let $g(W)$ be the number whose span is contained in $W$. Hence $g(W) = 2^{q^{\dim W}} − 1$, since $\emptyset$ has no span. Clearly $$ g(W) = \sum_{T \le W} f(T) $$ so by Möbius inversion in $B_n(q)$, $$ f(W) = \sum_{T \le W} g(T) \mu(T, W) $$ Putting $W = V_n$, there follows $$ \begin {aligned} f (V_n) &= \sum_{T \in B_n(q)} g(T) \mu(T, V_n) \\ &= \sum_{k=0}^n {\boldsymbol n \choose \boldsymbol k} (-1)^{n-k} q^{n - k \choose 2} ( 2^{q^k} - 1 ). \end {aligned} $$
* The standard convention is that the empty set spans $\{0\}$. If we wish to retain this convention, then we need to enlarge $B_n(q)$ by adding $\emptyset$ below $\{0\}$.
On the other hand, a reason (which I find unconvincing) for the span to be $\lbrace 0\rbrace$ is given by PBRMEASAP at https://www.physicsforums.com/archive/index.php/t-84017.html. This site has a discussion of whether the empty set is a vector space. The correct answer is that it isn't, because one of the axioms is the existence of an additive identity 0.
Update. I agree with the comments that the span of the empty set is $\lbrace 0\rbrace$. What I said above was foolish.
