Finite Union of proper subspaces of $\mathbb C^2$ can equal to $\mathbb C^2$? My instructor for Linear Algebra gave us a problem to think about but am quite unsure on how to approach it:
Let $V_1, V_2, ... V_{100}$ be $100$ proper subspaces of the complex vector space $V=\mathbb C^2$. Can it be possible that $\bigcup _{i=1}^{100} V_i = \mathbb C^2$?
 A: Nope, there are two cases, all $V_i$ are the same, then the union is a subspace to, and a proper one. The second case is that not all $V_i$ are the same, then the union of all $V_i$ is not closed under addition and can't be a vector space. I used here that we have a 2 dimensional vectorspace over an infinite field, so we have an infinite number of subspaces.
Proof the following Lemma:
Let $U_1,U_2$ be subspaces of a vectorspace $V$, then 
\[ U_1 \cup U_2 \]
is a subspace iff $U_1\subseteq U_2$ or $U_1 \supseteq U_2$
A: If $V_i$ has as equation $l_i(z,w)=a_iz+b_iw=0$, then the non-zero polynomial $P(z,w)=\prod^{100}_{i=1} l_i(z,w)$  of degree $100$ vanishes on $\bigcup^{100}_{i=1} V_i$.
Since a non-zero polynomial cannot vanish on the whole of $\mathbb C^2$, we have $\bigcup^{100}_{i=1} V_i\subsetneq \mathbb C^2$
A: One has to be careful, because a finite vector space can  well be the union of finitely many proper subspaces. For instance the vector space of dimension two over the field with two elements is the union of three subspaces of dimension one.
Note first that each $V_i$ can be assumed to be one-dimensional. Add the subspace $V_{0} = \{ (0, a) : a \in \Bbb{C} \}$ to the $V_{i}$ for good measure, and suppose each $V_i \ne V_0$ for $i \ne 0$.
Then in each $V_i$ (with $i \ne 0$) there is a unique element of the form $(1, a_i)$. Choose an element $a \notin \{ a_i : i \in 1, \dots, 100 \}$. (Here of course we are using the fact that $\Bbb{C}$ is infinite.) Then $(1, a) \notin \bigcup_{i=0}^{100} V_i$, so $V \ne \bigcup_{i=0}^{100} V_i$.
So this works for any finite number of subspaces, and any infinite field.
