TITLE: For subspaces, if $N\subseteq M_1\cup\cdots\cup M_k$, then $N\subseteq M_i$ for some $i$?
QUESTION [6 upvotes]: I have a vector space $V$ over a field of characteristic $0$. If $M_1,\dots,M_k$ are proper subspaces of $V$, and $N$ is a subspace of $V$ such that $N\subseteq M_1\cup\cdots\cup M_k$, how can you tell $N\subseteq M_i$ for some $i$?
I was first trying to show it just in the case $N\subseteq M_1\cup M_2$, and hoping to extend it to finitely many $M_i$. If either of the $M_i$ contains the other, the claim follows, so I suppose neither $M_i$ contains the other. 
In hopes of a contradiction, I suppose $N\not\subseteq M_1$ and $N\not\subseteq M_2$, Picking $x_1\in N\setminus M_1$ and $x_2\in N\setminus M_2$, I'd have $x_1+x_2\in N\subseteq M_1\cup M_2$. The only thing I could conclude was that actually $x_1,x_2\notin M_1$ and $x_1,x_2\notin M_2$, which seems like a dead end. What's the right way to do this?

REPLY [4 votes]: Recall that a vector space over an infinite field is not a finite union of proper subspaces. Since your scalar field has characteristic $0$, it is necessarily infinite, as it contains an isomorphic copy of $\mathbb{Z}$ for instance.
If $N\subseteq M_1\cup\cdots\cup M_k$, then 
$$
N=(N\cap M_1)\cup\cdots\cup(N\cap M_k).
$$
Viewing $N$ as a vector space in its own right, with $N\cap M_i$ subspaces of $N$, we see they cannot all be proper. So for some $i$, $N\cap M_i=N$, hence $N\subseteq M_i$.