Proving $L(S)= \cap_{S ⊆ W}\space W$ Let , $S$ be a subset of a vector space , then how do we prove that $L(S)$ , the linear span of $S$ , is the intersection of all subspaces containing $S$ i.e. $L(S)= \cap_{S ⊆ W} \space W$ ? 
( I have been able to prove 
$L(S) ⊆ \cap_{S ⊆ W} \space W$ , 
but 
 having trouble with the other side , please help )
 A: Let $S$ be a set of vectors. Then $ \text{span}(S)$ is equal to the intersection of all the subspaces which contain $S$, which we denote $\cap W$. 
One direction is obvious, since each subspace which contain $S$ also contain all the linear combinations, $ \text{span}(S) \subset \cap W$. In ordered to show that $\cap W \subset \text{span}(S)$. We have to show that $\text{span}(S)$ is a subspace and also that contain $S$ but this last point is trivial. First we have to show that $0 \in \text{span}(S)$. Since $x\in \text{span}(S)$ iff $x= \sum_{v\in S}a_iv$ for $a_v \in \mathbb{F}$, so setting $a_v=0$ we get the zero. Now suppose that $x,y \in \text{span}(S)$ and $c\in \mathbb{F}$. So $x=\sum_{v\in S}a_vv,\, y=\sum_{v\in S}b_vv$. Thus 
$$cx+y=c\cdot \sum_{v\in S}a_vv+\sum_{v\in S}b_vv=\sum_{v\in S}ca_v+b_v \in \text{span}(S)$$
Hence is a subspace, $\cap W \subset \text{span}(S)$.
A: Note that $L(S)$ is a subspace containing $S$ that means $L(S)\in A$ with $A$ is the set of all subspaces containing $S$. Consequently, $\cap_{W\in A} \space W\subseteq L(S)$.
