$span\{S \cap T\} \subset span\{S\} \cap span\{T\}$ Here's what I'm working on:
Let $V$ be a vector space over some field $F$, and $S,T \subset V$. I have to prove that $span\{S \cap T\} \subset span\{S\} \cap span\{T\}$.
If I let: 
$S=\{s_1,s_2,...,s_n\}$, $T=\{t_1,t_2,...,t_m\}$, and $S\cap T=\{u_1,u_2,...,u_i\}$, then 
$span\{S \cap T \}=\{x \in V | x=c_1u_1+c_2u_2+...+c_iu_i, \forall u \in S \cap T, \forall x \in F\}$
I'm not too sure where to go from here.
Any help would be appreciated. Thanks.
 A: Quite obviously, if $X,Y$  are subsets of a vector space satisfying $X \subseteq Y,$ then
$$
\mathrm{span}(X) \subseteq \mathrm{span}(Y).
$$
Therefore,
$$
A \cap B \subseteq A
$$
implies
$$
\mathrm{span}(A \cap B) \subseteq \mathrm{span}(A).
$$
and then, by symmetry,
$$
\mathrm{span}(A \cap B) \subseteq \mathrm{span}(B),
$$
whence
$$
\mathrm{span}(A \cap B) \subseteq \mathrm{span}(A) \cap \mathrm{span}(B)
$$
for all subsets $A,B$ of $V.$ 
A: The way I'd do this is as follows:
Let $x\in\text{span}(S\cap T)$.
It follows that $x = \sum_i c_ie_i$ where $e_i\in S\cap T$.
As $e_i\in S$, we have that $x\in\text{Span}(S)$, and as $e_i\in T$, we have that $x\in\text{span}(T)$, so $x\in\text{span}(S)\cap\text{span}(T)$.
I find people doing this all too often, where they want to show that $A\subseteq B$ (as sets), and make some argument that doesn't start with "let $x\in A$".
To show that $A\subseteq B$, what you do is:


*

*"Let $x\in A$"

*Write what it means for $x$ to be in $A$.

*Try to use various definitions and results to show how this also means that $x\in B$.
As you see from the way I proved this, I:


*

*Started with $x\in\text{span}(S\cap T)$

*Wrote what this means

*Noticed this means that $x\in\text{span}(S)$, and $x\in\text{span}(T)$.

*Used the last point to show that $x\in\text{span}(S)\cap\text{span}(T)$
This is really all it takes, and required no appeal to the full basis of $S$ or $T$, as we don't care about that.
