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 $$\langle E\rangle:=\left\{\mathbf{a}\in M\Bigg|\exists(n\in\mathbb{Z}^+)\left[\exists(r:I_n\rightarrow R)\exists(\mathbf{u}:I_n\rightarrow E)\left[\mathbf{a}=\sum_{i=1}^n r(i)\mathbf{u}(i)\right]\right]\right\}.$$

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 little advice: don't spend to much of your youth (I assume you are young because these kind of questions come usually from beginning mathematicians) on such... empty :) subjects. Commented Apr 28, 2017 at 11:23
• Great piece of advice by JeanMarie. First of all, the theorem that any linear space has a basis (AC included, of course) would be false. Second, that "$\,E=\emptyset\iff\langle\,E\,\rangle=\emptyset\;$" is something I doubt you can prove from the axioms. And third and most important: who cares? This can be something interesting, perhaps, for someone trying to dwell deeply in fundaments, logic and etc.. not so much for "usual" mathematicians (no offense intended for anyone...) Commented Apr 28, 2017 at 11:29
• dimension of zero subspace is zero? Commented Apr 28, 2017 at 11:29
• @JeanMarie It becomes a problem because, I am compiling my own lecture notes. I want it to be rigorous as much as possible, so I won't be confused when I read it again next time. But anyway thanks! It's good to know that this kind of problem is not that much problematic. Commented Apr 28, 2017 at 11:31
• @DonAntonio Please, I'm a physicist, and it's well-known that this job does not care about math (well, even $\sin\theta=\theta$ for many of us). Put the joke aside, I just want to make sure that my book is consistent, not to have some contradiction in it. But thanks anyway. Commented Apr 28, 2017 at 11:40

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.