Let $V$ be a vector space over $F$ and $U_1$ , $U_2$ be subspaces of $V$.
The claim is that if (the union) $U_1 \cup U_2 = V$, then $U_1 = V$ or $U_2 = V$, or both.
How would I go about trying to prove this claim?
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community
The opposite of the statement is $U_1 \neq V$ and $U_2\neq V$
Suppose the opposite is true, i.e. both $U_1, U_2$ are proper subsets of V, then $\exists v_1, v_2 \in V$ such that $v_1\notin U_2$ and $v_2\notin U_1$. Further, $U_1 \cup U2 = V $ implies $v_1\in U_1$ and $v_2\in U_2$.
Now since V is a vector space, $v_1, v_2 \in V$ implies $v_1+v_2 \in V$. But $v_1+v_2 \notin U_1$ since $v_2 \notin U_1$. Similarly, $v_1+v_2 \notin U_2$. So $v_1+v_2 \notin U_1\cup U_2$, contradicting $U_1\cup U_2 = V$.
Do it by contraposition: assume $U_1\neq V$ and $U_2\neq V$.
If $U_1\subseteq U_2$ or $U_2\subseteq U_1$ then you're done.
Assume $U_1\not\subset U_2$ and $U_2\not\subset U_1$. Take $u_1\in U_1\backslash U_2$ and $u_2\in U_2\backslash U_1$. Prove that $u_1+u_2\notin U_1\cup U_2$.