If a subset $S$ of a vector space $V$ is a subspace of $V$, then is $\langle S \rangle = S$? I'm reading here on page 22 of Axler, Linear Algebra Done Right, where the following is stated:

A $\bf{linear}$ $\bf{combination}$ of a list $(v_1,\dots,v_m)$ of vectors in $V$ is a vector of the form
$\hspace{5cm}$ $a_1v_1+\cdots+a_mv_m,$
where $a_1,\dots,a_m \in F$. The set of all linear combinations of $(v_1,\dots,v_m)$ is called the $\bf{span}$ of $(v_1,\dots,v_m)$, denoted $span(v_1,\dots,v_m)$. In other words,
$\hspace{5cm}$ $span(v_1,\dots,v_m)=\{a_1v_1+\cdots +a_mv_m:a_1,\dots,a_m \in F\}$.

Anyway, if $\langle S \rangle$, that is, the span of $S$ is equal to $S$, then 
$\hspace{5cm}$ $\{a_1v_1+\cdots +a_mv_m:a_1,\dots,a_m \in F\}=\{v_1,\dots, v_m\}$,
correct?

OK, so now I'm reading in Halmos's Finite-Dimensional Vector Spaces, and I feel that the theorem, Theorem 2, on page 17 suffices to prove the above problem. What do you think?
$\hspace{1.8cm}$
$\hspace{1.8cm}$
Ok, this seems so unnecessarily complicated. In Hoffman's Linear Algebra on page 35 a good definition is given for subspace:
Theorem 1. A non-empty subset $\text{W}$ of $\text{V}$ is a subspace of $\text{V}$ if and only if for each pair of vectors $\alpha,\beta$ in $\text{W}$ and each scalar $\text{c}$ in $\text{F}$ the vector $c\alpha+\beta$ is again in $\text{W}$
I mean, if $\alpha$ and $\beta$ are in $W$, then—if we say $W$ is a subspace with a basis $\{w_1,\dots,w_n\}$—we have that
\begin{eqnarray}
c\alpha + \beta = (ca_1+b_1)w_1+\cdots +(ca_n+b_n)w_n,
\end{eqnarray}
for $\alpha = a_1w_1+\cdots+a_nw_n$ and $\beta = b_1w_1+\cdots+b_nw_n$. The RHS here is clearly the very definition of $\text{Span}(W)$ for $k_i=ca_i+b_i$ with $1\leq i \leq n$.
 A: The definition of $\displaystyle{\operatorname{span}(S)}$  is more general (not only for finite $S$):

If $S\subseteq V$ then $\operatorname{span}(S)=\{a_1s_1+a_2s_2+\ldots+a_ns_n:a_i\in\mathbb F, s_i\in S, n\in\mathbb N\}$

Now it is not hard to show that $\displaystyle{\operatorname{span}(S)=\bigcap W}$ where the intersection is taken over all linear subspaces of $V$ that contain $S$. To prove this show that $\displaystyle{\operatorname{span}(S)}$ is a subspace of $V$ containing $S$ and that every such subspace must contain $\displaystyle{\operatorname{span}(S)}$.
Therefore, if $S$ is a subspace of 
$V, \ \displaystyle{\operatorname{span}(S)=\bigcap_{S\subseteq W\leq V} W} \subseteq S$.
 
Since we always have $\operatorname{span}(S)\supseteq S$ (as intersection of subspaces containing $S$),
it follows that $\displaystyle{\operatorname{span}(S)=S}$. 
A: Yes because a vector space is closed with sum and scalar product! Every $a_1 v_1 +a_2 v_ 2 +\cdots+a_m v_m$  must be equal with one of $\{v_1 ,v_2 ,\dots,v_m\}$   else $S$ would not be a subspace.
update: for you dear tranco:
If   $\, \exists\,\, a_1 v_1 +a_2 v_ 2 +\cdots \cdots +a_m v_m \notin S$ , it disturb that S is sub space! so span(S)$\subset$ S and it is clear that S$\subset$ span(S) so S=span(S)
