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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?

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    $\begingroup$ Note that if one of $U_1$ or $U_2$ is contained in the other, then the claim is obvious. If not, then consider $h_1 \in U_2 \setminus U_1$ and $h_2 \in U_1 \setminus U_2$, and debate about where $h_1 + h_2$ lies. $\endgroup$ – астон вілла олоф мэллбэрг Sep 19 '18 at 6:48
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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 $U_1\cup U_2 = V$.

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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 $u_1+u_2\notin U_1\cup U_2$.

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