Prove that $U⊆W_1$ or $ U⊆W_2$ if $U⊆W_1∪W_2$.

Question

Let $V$ be a vector space over a field $F$, and $U,W_1,W_2$ three subspaces of $V$, with

$$U⊆W_1∪W_2$$

Prove that $U⊆W_1$ or $U⊆W_2$.

Answer 1

Hint:

Suppose that it is not the case, that is, $u=w_1+w_2\in U$, with $w_1\in W_1\setminus W_2$ and $w_2\in W_2\setminus W_1$. Can you reach a contradiction?

By our choice $w_1\in W_1$, so $w_1=u-w_2\in W_1$. Now, $w_2\not\in W_1$ therefore $u\not\in W_1$. Similarly with $w_2$, using the same argument, we can show that $u\not\in W_2$, but then $u\not\in U$. This is a contradiction, so either $U\subseteq W_1$ or $U\subseteq W_2$

Answer 2

Let's assume it isn't the case. Then there are vectors $u_1\in U\setminus W_1$ and $u_2\in U\setminus W_2$. Both vectors are in $U$. But $U$ is a vector space so $u=u_1+u_2\in U$. But $u$ is not in $W_1$ and not in $W_2$. Why? Well, suppose it is in $W_1$. Then we get $u_1=u-u_2\in W_1$ which is a contradiction. In the same way prove that $u\notin W_2$. And that is a contradiction to $U\subset W_1\cup W_2$.