The question is:
Let $F$ be a field. $F$ is a vector space over itself which we'll represent as $V$. Find all the vector sub-spaces of $V$ and prove your answer.
After thinking about it, I'm pretty sure the only subspace of $V$ is $V$ itself. Since for every element we remove (Let's say $V_2 +V_3 = V_1 $) the elements $V_2$ or $V_3 $ can't be in the subspace either otherwise the closure axiom wouldn't hold, and the cycle continues.
But, after thinking a little longer, I have no idea how to prove it. I don't know how I can say that every element is the result of addition of two other elements. Also I have no idea how to prove this will result in no elements in the subspace.