Empty Set $\{\}$ is the Only Basis of the Zero Vector Space $\{0\}$ Question
Suppose we want to find a basis for the vector space $\{0\}$. 
I know that the answer is that the only basis is the empty set.
Is this answer a definition itself or it is a result of the definitions for linearly independent/dependent sets and Spanning/Generating sets? If it is a result then would you mind mentioning the definitions of bold items which based on them this answer can be deduced.

Useful Links
These are the links that I found useful for answering this question. It needs some elementary background form mathematical logic. You can learn it by spending a few hours on this Wikipedia page.
Link 1, Link 2, Link 3, Link 4, Link 5, Link 6

To answer the question of why $\text{Span}\{\}=\{0\}$ is true, I considered the following argument for myself.
I think all the things draw back to the operation of addition.
Addition is a map defined as follows
$$
\begin{align}
(\cdot+\cdot): & V \times V \to V \\
& (u,v) \mapsto (u+v)
\end{align}$$
with the commutative and associative properties
$$
\begin{align}
(u+v) &= (v+u) \\
((u+v)+w) &= (u+(v+w))
\end{align}
$$
so according to this definition, whenever we are talking about addition we should provide two inputs to get one output. 
From a programming point of view it is useful to have outputs in case when we have one or no inputs (see the detail of Plus function in wolfram language). It also turns out to be useful in proofs like the ones using induction. So what is the most useful definitions to make for such cases? Experience shows that these are
$$
\begin{align}
(u+\text{null}) &= u\\
(\text{null}+u) &= u \\
(\text{null}+\text{null}) &= 0
\end{align}
\tag{1}$$
where you can think of null meaning that no argument is provided!
Now the following definition can be easily interpreted in the special cases when we have a set with one element or no element.

Linear Combination and Span.
  A linear combination of a set $A=\{v_1,v_2,...,v_m\} \subseteq V$ is a vector $v$ defined by $v=\sum_{j=1}^{m}a_jv_j$. The set of all linear combinations of $A$ is called the span of $A$ denoted by $\text{Span}A$.

Now, if we make the convention that $m=0$ means $A=\{\}$ and when $m=1$ then $A=\{v_1\}$, according to $(1)$, we can interpret the definition as follows
$$
v=\sum_{j=1}^{m}a_jv_j:=s_m, \qquad
s_i =
\begin{cases}
0,  & i=0 \\
(a_1v_1+\text{null}),  & i=1 \\
(a_1v_1+a_2v_2),  & i=2 \\
(s_{i-1}+a_{i}v_{i}), & \text{otherwise} \\
\end{cases}
, \qquad 0 \le i \le m
$$
So, we can see that $\text{Span}\{\}=\{0\}$. Note that this is a result of our own convention for the addition operation and the definition of the span. I think that the vacuous truth argument has no advantage over this one! However, it keeps repeating in many other examples! So it is good to learn it once and for all!
