Does every element of the empty list posses every property? Suppose we have a list of elements $v_1, v_2, \ldots, v_n$.
Then, as I've understood, setting $n=0$ above results in the empty list $v_1, v_2, \ldots, v_0$ of no elements (please correct me if I'm wrong ).
Now does every element of the empty list posses every property $P(x)$ ?
I know that $x \in \emptyset \Rightarrow P(x)$ is true for every $x$. However, a list is not set ?
As an example:
$v_1, v_2, \ldots v_n \in V$ is linearly independent if $c_1 v_1 + c_2 v_2 + \ldots + c_n v_n = 0 \Rightarrow c_i=0$.
Are $v_1, v_2, \ldots, v_0$ linearly independent then ?
I know this might be a "stupid" question, but I've been wondering ...
 A: Your definition of linear independence can be written as
$$\sum_{i=1}^n c_i v_i=0\implies c_i=0\text{ for all }i$$
There's almost no difference in considering $v_i$'s as a tuple or as a set (considering them as a tuple simplifies certain arguments).
It should be clear the empty tuple $()$ or the empty set $\varnothing$ of vectors satisfies the definition, as the empty sum is $0$ and all of the (empty set) of scalars are equal to $0$.
This is useful because then all vector spaces have a basis (the trivial vector space $\{0\}$ can only have empty bases).
A: Think of your list as a function $f:I\to S$ for index set $I\subset \mathbb{N}$ and arbitrary set $S$.
An empty list corresponds to $I=\emptyset$. It would then be vacuously true that $\forall x\in I: P(f(x))$, i.e. every element of the list has property P.
A: Yes, every element in the empty list possesses every property you can think of. Even the property $P(x)$ saying that $x$ is not an element of the empty list. So it is correct to say that every element on the empty list is not on the empty list. 
And yes, the empty list of vectors is linearly independent in every vector space, and a basis in every $0$ dimensional vector spaces, i.e., every vector space consisting of just the $0$ vector.  
