Difference between $\operatorname{Span}(e_i)_{i\in I}$ and $\overline{\operatorname{Span}(e_i)_{i\in I}}$? Let $(e_i)_{i\in I}$ a family of linear independent vector and $I$ not finite (take for example $I=\mathbb N$. 
For me, $$\operatorname{Span}(e_i)_{i\in I}=\left\{\sum_{i\in I}a_ie_i\mid a_i\in\mathbb R\right\}.$$
So what is the difference between $\operatorname{Span}(e_i)_{i\in I}$ and $\overline{\operatorname{Span}(e_i)_{i\in I}}$ ? Indeed, if $x\in \overline{\operatorname{Span}(e_i)_{i\in I}}$, then there is a sequence of $\sum_{i=1}^n a_ie_i$ s.t. $$x=\lim_{n\to \infty }\sum_{i=1}^n a_ie_i=\sum_{i\in I}a_ie_i$$
and thus $x\in \operatorname{Span}(e_i)_{i\in I}$. So isn't it the same space ? 
 A: The definition of $\operatorname{span} (e_i)_{i\in I}$ does not allow infinite linear combinations. The closure of this set contains such linear combinations.
A: If we define the span as has been done by John in the question; that is, if we define
$$ \DeclareMathOperator{\span}{Span} \span_{\text{John}}(e_i)_{i\in I} := \left\{ \sum_{i\in I} a_i e_i \ \middle|\ a_i \in \mathbb{R} \right\}, $$
then
$$ \span_{\text{John}}(e_i) = \overline{\span_{\text{John}}(e_i)}. $$
Observe that I have called this $\span_{\text{John}}$ because it is distinct from the usual definition of the span of a set of vectors, which is given by
$$ \span(e_i)_{i\in I} := \left\{ \sum_{n=1}^{N} a_{n} e_{i_n} \ \middle|\ a_{n} \in \mathbb{R}, i_n \in I, N \in \mathbb{N} \right\}. $$
Note the distinction: the usual span consists only of finite linear combinations, while John's span include infinite "linear combinations" as well.
If the index set $I$ happens to be finite (i.e. it has finite cardinality), then
$$ \span_{\text{John}}(e_i)_{i\in I}
= \overline{\span_{\text{John}}(e_i)_{i\in I}}
= \span(e_i)_{i\in I}
= \overline{\span(e_i)_{i\in I}}. $$
However, if $I$ has infinite cardinality, then this identity needn't hold—in particular, the span and its closure (using the usual definition) needn't agree.  For example, let $\ell^2$ be the set of all sequences in $\mathbb{R}$ which are bounded with respect to the $\ell^2$-norm, i.e.
$$ \ell^2
:= \left\{ (a_1, a_2, a_3, \dotsc) \ \middle|\ a_n \in \mathbb{R} \forall n, \sum_{n=1}^{\infty} |a_n|^2 < \infty \right\}. $$
If for each $i \in \mathbb{N}$, we define $e_i$ to be the sequence having a $1$ in the $i$-th place, then
$$ \ell^2 = \overline{\span(e_i)_{i\in\mathbb{N}}} \ne \span(e_i)_{i\in\mathbb{N}}. $$
I think that it is clear from the definition that each element of $\ell^2$ can be written as an infinite linear combination of the $e_i$ (for a sequence $(a_n)$, take the coefficient of $e_i$ to be $a_i$).  From this, a few technical arguments will show that
$$ \ell^2 = \overline{\span(e_i)}. $$
On the other hand, the sequence
$$ \left( 1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \dotsc \right) $$
is an element of $\ell^2$, but cannot be written as a finite linear combination of the $e_i$, and is therefore not contained in $\span(e_i)$.
It is perhaps worth noting that the closure of the span is important in the study of Banach spaces, i.e. in functional analysis.  As such, it is tempting to call this set the "analytic span" of a set of vectors, rather than the "algebraic span".  I am not sure that this is standard terminology, but you would likely be understood if you used it.  It may also be worth reading up on the distinction between a Schauder basis and a Hamel basis (see also Wikipedia).
A: If the dimension is finite, then they are the same. However, in general they are not. For instance, in $C^0[0,1]$, the space of continuous, real-valued functions on $[0,1]$, there are functions (such as $e^x$) that is not in the span of the set of polynomials. However, it is in the closure of that span. In fact, all of $C^0[0,1]$ is in the closure of the set of polynomials.
