For subspaces, if $N\subseteq M_1\cup\cdots\cup M_k$, then $N\subseteq M_i$ for some $i$? 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? 
 A: First show that $M_1\cup M_2$ is a subspace of $V$ if and only if $M_1\subseteq M_2$ or $M_2\subseteq M_1$.
This extends by induction to any finite number of subspaces: if $M=M_1\cup\ldots\cup M_k$ is a subspace of $V$ then there is some $i$ such that $M=M_i$.
Next show that if $N\subseteq M_1\cup\ldots\cup M_k$ then $N_i=N\cap M_i$ is a subspace of $V$, and therefore $N=N_1\cup\ldots\cup N_k$ and it is a subspace, so $N=N_i$ for some $i$, so $N\subseteq M_i$ as wanted.
A: 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 $N\subseteq M_i$.
A: In this problem, you can try pointing out that if $N$ is not a subset of $M_1$, then it must be a subset of $M_2$. I'll give you some hints on this:


*

*Since $N \not \subset M_1$, then there exists some $x \in N \  \backslash \ M_1$.

*You should notice that, since $N = M_1 \cup M_2$ (i.e, every element in $N$ must belong to either $M_1$, or $M_2$), and we have that, $x \in N \  \backslash \ M_1$, so $x \in M_2$.

*Now, can you try showing that for all $n \in N$, $n$ must belong to $M_2$? Hint: Take their sum, and consider 2 cases. ;)
To show it holds not only for $k = 2$, but for any $k \in \mathbb{N} \ \backslash \ \{ 0 \}$, you can try using Proof by Induction. Case $k = 1$ is trivial, and $k = 2$ has already been done, can you continue from here? :)
