I am aware that the union of subspaces does not necessarily yield a subspace. However, I am confused about the following question:
(i) Let $U, U'$ be subspaces of a vector space $V$ (both not equal to $V$). Prove that the union of $U$ and $U'$ does not equal $V$.
(ii) Find an example of $V$ and $U,U',U''$ contained in $V$ (all not equal to $V$) such that the union of $U,U'$ and $U''$ is equal to $V$.
Thank you.
(i) Suppose $V = U \cup U'$. Since the inclusions aren't strict, there is at least one vector in $U'$ not in $U$, and viceversa. Call those $u' \in U'\setminus U$ and $u \in U\setminus U'$. Then, $u+u' \not \in U \cup U'$, but this is absurd since $u+u' \in V$.
(ii) If we take three subspaces, the above reasoning doesn't hold: for example, if we take $V=\mathbb{F}_2^2$ the union of the subspaces $(1,0)$ and $(0,1)$ and $(1,1)$ is the entire $V$.
(i) Let's assume $U,U',V$ as you said, but $U\cup U'=V$. Obviously we have neither $U\subseteq U'$ nor $U'\subseteq U$ as else we had $V=U\cup U'=U$, or $V=U\cup U'=U'$. So we find $v\in U\setminus U'$ and $v'\in U'\setminus U$. As $U\cup U'=V$ we have $v+v'\in U\cup U'$. If $v+v'\in U$ we have $v'=(v+v')-v\in U$, contradiction. If $v+v'\in U'$ we have $v=(v+v')-v'\in U'$, contradiction.
(ii) But let's assume the $\mathbb{F}_2$-space $\mathbb{F}_2^2$ and the three subspaces $U=\lbrace 0, (1,0)\rbrace$, $U'=\lbrace 0, (0,1)\rbrace$, $U''=\lbrace 0, (1,1)\rbrace$. Then $\mathbb{F}_2^2 = U\cup U'\cup U''$.
This is perfectly impossible if the base field is infinite, and is known as the avoidance lemma for subspaces:
If a subspace of a vector space over an infinite field is contained in a finite union of subspaces, it is contained in one of them.
You can find a proof in my answer to this question
Note the analogue for subgroups of a group is true only for the union of two groups.
