Alternative definition for span and proving it is equivalent to the most common one This is a question related to something that I asked here about this alternative definition of span. User hardmath has helped me a lot! Therefore, I can't still understand how to prove the equivalence between the definition:

Span:
  Suppose a vector space $(V,+,\cdot)$, and 
$$S = \{u_1,\cdots,u_n\}$$
(and $S$ is a subset of $V$, not a subspace)
$$[S]=:\cap_{w\subset V, w\supseteq S} W$$
In other words, $[S]$ is, by definition, the intersection of all $W$,
  such that $W$ is a subspace of $V$ and $W$ contains $S$.

And the definition:

Span:
  $$S = \{u_1,\cdots,u_n\}$$
  The span of $S$, denote by $[S]$ is, then:
  $$W = \{\alpha_1u_1+\cdots+\alpha_nu_n|\alpha_i\in\mathbb R\}$$

I am not a lazy person and I'm not asking for anyone to simply solve it for me, if it will bother you. Any hint is wellcome. So, how do I go from the first definition, to the second?
Couldn't find it on wikipedia, neither in my book. 
 A: You must take the intersection in the first definition over subspaces, not all subsets; indeed the intersection over all subsets will just be $S$ itself.
Let's look at one direction of containment. For convenience I will write $S_1$ for the first definition and $S_2$ for the second. (This is not standard notation by any means.) I'll show that $S_1 \subset S_2$. This follows because $S_2$ is one such subspace $W$, so it is one of the sets in the intersection defining $S_1$, so $S_1 \subset S_2$.
Now try to prove $S_2 \subset S_1$. 
A: Let $T= \{W \subseteq V| W \textrm{ is vector space and } S\subseteq W\}$
and Let $$U_1 = \bigcap_{W \in T} W$$
$$U_2= \{\alpha_1u_1+\cdots+\alpha_nu_n|\alpha_i\in\mathbb R\}$$
So $U_1$ is as in your first definition, and $U_2$ is as your second definition.
We want to show that $U_1=U_2$
Step 1: Notice that $U_2$ is vector space(Why!) and $S\subseteq U_2$(Why!). Hence $U_2 \in T$ and thus $\bigcap_{W \in T}W \subseteq U_2$, which implies that $U_1 \subseteq U_2$.
Step 2: We will show that $U_2 \subseteq U_1$. So let $u\in U_2$ be an arbitrary element. Then $u$ can be written as linear combination of the elements of $S$. Lets say $$u= \lambda_1u_1+ ... +\lambda_nu_n$$
Now, Let $W$ be any element in $T$, Then $S \subseteq W$. This means $u_1,u_2,...,u_n$ are elements of $W$. But this implies that $\lambda_1u_1+ ... +\lambda_nu_n \in W$ because any linear combination of elements from a vector space $is$ also element from the same vector space. Hence $u \in W$. This shows that $u$ is element in every vector space $W$ in $T$, thus $u \in \bigcap_{W \in T}W $, i.e. $u \in U_1$. But $u$ was arbitrary element in $U_2$, so we get $U_2 \subseteq U_1$.
