What theorem would be false if the span of empty set is not $\lbrace \mathbf{0}\rbrace$ I know that many books define that the span of empty set is $\lbrace\mathbf{0}\rbrace$.
However, I came across this definition of spanning set from many places (like Roman Steven's Advanced Linear Algebra or even Proofwiki:

Let $M$ be a left $R$-module, let $E\subseteq M$, and let $I_n$ be the set $\lbrace x\in\mathbb{N}\mid 1\leq x\leq n\rbrace$ for each natural number $n$. The set span $E$
   (or $\langle E\rangle$) is defined by \begin{equation}  \langle
 E\rangle:=\left\{\mathbf{a}\in
 M\Bigg|\exists(n\in\mathbb{Z}^+)\left[\exists(r:I_n\rightarrow
 R)\exists(\mathbf{u}:I_n\rightarrow E)\left[\mathbf{a}=\sum_{i=1}^n
 r(i)\mathbf{u}(i)\right]\right]\right\}. \end{equation}

In other words, the set $\langle E\rangle$ is the set of all possible linear combinations of elements of $E$.
My problem is, according to the definition, we must have this:

A set $E$ is empty if and only if $\langle E\rangle$ is empty.

Why? I will only prove the forward case: Suppose that $E$ is empty, and we assume for the sake of contradiction that $\langle E\rangle$ is not empty. Since $\langle E\rangle$ is not empty, there must exist $\mathbf{a}\in\langle E\rangle$. This implies that there must exist a positive integer $n$ such that $\mathbf{a}=\sum_{i=1}^n r(i)\mathbf{u}(i)$ for some function $r:I_n\rightarrow R$ and some function $\mathbf{u}:I_n\rightarrow E$. Since $I_n$ is not empty, it follows that $E$ must not be empty, which is a contradiction to the assumption that $E$ is empty. QED.
So, according the definition, the span of empty set is EMPTY!
It's like we have two conventions here. The first one is that the span of empty set is the zero vector space while the second one is that the span of empty set is empty.
My problem is, which convention should I stick with (and no, not according to my prof, he doesn't care anyway, but I care). So, I'm considering the consequence of having $\langle\emptyset\rangle=\emptyset$. Well, it has the nice property that $E$ is empty iff $\langle E\rangle$ is empty. But if $\emptyset$ does not span the zero vector space, then the zero vector space would not have a basis, and that's pretty big drawback. So, it would not be true anymore that every vector space has a basis.
What other theorems would be false from this basic definition? If there are many drawbacks, should I discard this definition and the book itself?
 A: I saw some good advices for you in the comments. But I also understand that certain (seeming) irregularities can be irritating. It shouldn't. Consider the rule that the linear span $<E>$ of a set $E$ of vectors must be a linear space. It would be an exception if $<\emptyset>$ would be $\emptyset$ because that is not a linear space (it fails to have a "zero vector"). Just as numbers have their "zero" and sets have their "zero set", linear spaces have their "zero linear space" which is {0}. You should follow the definition that $<E>$ is the smallest vector space that includes $E$. There is no exception for $E = \emptyset$. The pragmatic computation of $<E>$ goes through your definition. For $\emptyset$, no computation is needed.
Perhaps you can get at ease with one more example. As a physicist, you are acquainted with the exponent notation $x^y$ of numbers, "$x$ to the $y$". If $y$ is a positive integer (say: $5$) then $x^y$ represents $x*x*x*x*x$. If $y = -5$ then $x^y$ stands for $1/x^5$ ($x \neq 0$). But what is $x^0$? Isn't that a multiplication with a list of zero $x$'s? Well, by the calculus rules for exponents, $x^5*x^{-5} = x^{5-5} = x^0$ on the one hand and $x^5/x^5 = 1$ on the other hand. Convention: $x^0 = 1$, the empty product of $x$'s is $1$ and an exception is avoided (though you must avoid $x = 0$ here). What about the proposal $x^0 = 0$? Just as zero is the neutral element of addition, $1$ is the neutral element of multiplication. Guess what's the sum of an empty list of numbers...? Right.
