Prove that intersection with each sub vector space is contained with the intersection of the sum of those vector spaces. I'm trying to find a way to prove the following:
Let there be $U, W_1, W_2$ sub vector spaces of linear space V
Prove that: $(U\cap W_1 ) + (U\cap W_2 ) \subseteq U \cap (W_1+W_2 )$
Logically, I can think see it, as the sum of $W_1 +W_2 $ is larger, but I can't think of the steps to take to prove it.
Help will be appreciated.
 A: To show that $(U \cap W_1) + (U \cap W_2) \subseteq W \cap (W_1 + W_2)$, we will take an arbitrary element of $(U \cap W_1) + (U \cap W_2)$ and show that it is in the set $U \cap(W_1+W_2)$.
Let $x \in (U \cap W_1) + (U \cap W_2)$. By definition, there exist vectors $ x_1 \in (U \cap W_1)$ and $x_2 \in (U \cap W_2)$ such that $x=x_1+x_2$. Observe the following:

*

*Because $x_1 \in U \cap W_1$, then $x_1 \in U$ and $x_1 \in W_1$.

*Because $x_2 \in U \cap W_2$, then $x_2 \in U$ and $x_2 \in W_2$.

$U$, $W_1$ and $W_2$ are vector spaces and thus are closed under vector addition. So,

*

*$x_1 \in U$, $x_2 \in U$, therefore $x=x_1+x_2 \in U$

*$x_1 \in W_1$, $\textbf{0}_\textbf{V} \in W_2$, therefore $x_1=x_1+\textbf{0}_\textbf{V} \in (W_1+W_2)$

*$x_2 \in W_2$, $\textbf{0}_\textbf{V} \in W_1$, therefore $x_2=x_2+\textbf{0}_\textbf{V}\in (W_1 + W_2)$

*$x_1 \in (W_1 + W_2)$, $x_2 \in (W_1 + W_2)$, therefore $x = x_1 + x_2 \in (W_1 + W_2)$

*$x \in (W_1 + W_2)$, $x \in U$, therefore $x \in U \cap (W_1+W_2)$
This concludes the proof.
