$span\{S \cap T\} \subset span\{S\} \cap span\{T\}$

Question

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.

Answer 1

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 $V.$

Answer 2

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:

1. "Let $x\in A$"

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

3. 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:

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

2. Wrote what this means

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

4. 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.