Solution
= Solution
\b[False]. In $\mathbb R^2$ let $U=\operatorname{span}(1,1)$, $W=\operatorname{span}(1,0)$ and $X=\operatorname{span}(0,1)$. Then $W+X=\mathbb R^2$, so $U\cap(W+X)=U$, but $U\cap W=U\cap X=\{0\}$ and their sum is zero. Thus intersection need not distribute over the <sum of vector subspaces>. The right side is always contained in the left; this example makes the inclusion strict.