Show $U_1 \cup U_2=V \implies U_1=V$ or $U_2=V$ Let $V$ be a vectorspace over the field $K$ and $U_1, U_2$ subspaces of $V$.
Show $U_1 \cup U_2 = V \implies U_1=V$ or $U_2=V$
my thoughts:
Let $x_1 \in U_1$ and $x_2 \in U_2$, then $x_1+x_2 \in U_1 \cup U_2$. But this would mean
$x_1+x_2 \in U_2$ or $x_1+x_2 \in U_1$. This would mean $U_1=V$ or $ U_2=V,$ since $x_1$ and $x_2$ are any elements from $U_1 \cup U_2 = V.$
Could someone give me a feedback if its correct?
 A: The original proof is problematic because you go from the statement "for every $x_1,x_2,x_1 + x_2 \in U_1$ or $x_1 + x_2 \in U_2$" to the statement " $x_1 + x_2 \in U_1$ for every $x_1,x_2,$ or $x_1+x_2 \in U_2$ for every $x_1,x_2$" without justification.
Here is an alternative way to construct the proof that you might find helpful.  It suffices to show that $U_1 \cup U_2 = V$ and $V \neq U_1$, then it must hold that $V = U_2$. To that end, suppose (for the purpose of contradication) that there exists an element $x \in U_1 \setminus U_2$.
A: Completely wrong.
Proof Let $U_1\neq V$, then there is $v_0\in V\smallsetminus U_1$. But $U_2\cup U_1=V$ gives $v_0\in U_2\dots (1)$.
Let $v\in V$.
If $v\notin U_1$ then $v\in U_2$.
Otherwise if $v\in U_1$ then $v+v_0\notin U_1$ (since $v_0\notin U_1$). Thus $v+v_0\in U_2$. Hence from $(1)$ this implies $v\in U_2$.
Thus $V\subseteq U_2\implies V=U_2$. Hence it is proved.
A: If $x_1\in U_1$ but $x_2\notin U_1$ then we know $x_1+x_2\notin U_1$, 
similarly if $x_1\notin U_2$ but $x_2\in U_2$ then we know $x_1+x_2\notin U_2$,
So if $x_1,x_2$ satisfying above condition then $x_1+x_2\notin U_1\cup U_2$

So if $U_1,U_2$ both are proper subspaces/subgroups then we will always find $x_1,x_2$ satisfying above condition, and $x_1+x_2\notin U_1\cup U_2$. (neglecting the trivial case when either $U_1$ or $U_2$ is subgroup of the other and both are proper subgroups)

If $U_1\cup U_2=V$ then there is no such $x_1,x_2$ such that $x_1\in U_1,x_1\notin U_2,x_2\notin U_1,x_2\in U_2$) . With both $U_1,U_2$ being proper subgroups/subspaces,this is not possible; so at least one of them is $V$.
