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?
Union of two vector subspaces equals the vector space
abstract-algebralinear algebravector-spaces
Related Solutions
If $|F|=q<\infty$, and $V$ is $m$-dimensional ($m<\infty$), then any proper subspace $U_i$ has at most $q^{m-1}-1$ non-zero elements. So to cover the $q^m-1$ non-zero vectors of $V\,$, the given $n\le q$ subspaces are not going to be enough, because $$n(q^{m-1}-1)\le q(q^{m-1}-1)<q^m-1.$$ So we need at least $|F|+1>n$ subspaces to get the job done.
If $m=\infty$, then we can extend all the subspaces to have codimension one (i.e. $\dim_F(V/U_i)=1$ for all $i$). In that case the intersection $U$ of all the $U_i$:s has finite codimension, and we can study $V/U$ instead of $V$ reducing the probelm to the previous case.
If $|F|=\infty, m<\infty$? Well, then we need some reinterpretation. The following argument shows that we need an infinite number of subspaces to cover $V$, and an uncountable number of subspaces to cover $\mathbf{R}^m$. Again, assume that all the subspaces have codimension one (w.l.o.g.), and that $m\geq 2$ (also w.l.o.g.). Identify $V$ with $F^m$, and consider the set $$ S=\{(1,t,t^2,\ldots,t^{m-1})\in V\mid t\in F\}. $$ Any $U_i$ is now a hyperplane and consists of zeros $(x_1,x_2,\ldots,x_m)$ of a single non-trivial homogeneous linear equation $$a_{i1}x_1+a_{i2}x_2+\cdots+a_{in}x_m=0.$$ Therefore the number of elements of the intersection $S\cap U_i$ is equal to the number of solutions $t\in F$ of $ a_{i1}+a_{i2}t+\cdots+a_{im}t^{m-1}=0$ and is thus $<m$, because a non-zero polynomial of degree $<m$ has less than $m$ solutions in a field. This shows that if $F$ is infinite, we need an infinite number of subspaces to cover all of $S$. Also, if $F$ is uncountable, then we need an uncountable number of subspaces to cover $S$. Obviously it is necessary to cover all of $S$ in order to cover all of $V$.
\begin{equation} U_1 \cup U_2 = V \iff ( \textbf{v} \in U_1 \cup U_2 \iff \textbf{v} \in V) \iff ((\textbf{v} \in U_1 \lor \textbf{v} \in U_2) \iff \textbf{v} \in V) \iff (\textbf{v} \in U_1 \iff \textbf{v} \in V) \lor (\textbf{v} \in U_2 \iff \textbf{v} \in V) \iff (U_1 = V) \lor (U_2=V) \end{equation}
Best Answer
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 $U_1\cup U_2 = V$.