Looking for a good counterargument against vector space decomposition. How do I see that I cannot write $\mathbb{R}^n = \bigcup_{\text{all possible }M} \operatorname{span}(M)$, where $M$ runs over the subsets with $n-1$ elements in it of the set of vectors $N=\{a_1,\ldots,a_n,\ldots,a_m\} \in \mathbb{R}^n$, where the total dimension of the span of all of them is $n$.
 A: Let $V$ be a vector space over an infinite field $F$, and $V_1, \ldots, V_n$ proper subspaces.  Then I claim $\bigcup _j V_j$ is not a vector space.
For each $k$ let $u_k$ be a vector not in $V_k$. We then inductively find
vectors $w_k$ not in $\bigcup_{j \le k} V_j$.  Namely, if $w_k \notin \bigcup_{j \le k} V_j$, and $u_{k+1} \notin V_{k+1}$, consider $f(t) = t u_{k+1} + (1-t) w_k$ for scalars $t$.  If this was in $V_j$ for two different values of $t$, say $t_1 \ne t_2$, then it would be in $V_j$ for all $t$, because
$$f(t) = \dfrac{t - t_1}{t_2 - t_1} f(t_2) + \dfrac{t - t_2}{t_1 - t_2} f(t_1)$$
This is not the case for any $j \in \{1,\ldots,k+1\}$ because $f(0) \notin V_{k+1}$ and $f(1) \notin V_j$ for $j \le k$.  So there are at most $k+1$
values of the scalar $t$ for which $f(t) \in \bigcup_{j \le k+1} V_j$, and
infinitely many for which it is not.  
A: This is a finite union of nowhere dense sets and so is nowhere dense in $\mathbb{R}^n$.
A: The solution I have for this problem is the same approach as Robert Israel's, except I think my method of finding my "escaping element" is slightly less elaborate. (However, I'm finding the method he used comes from a very nice picture, which seems a bit prettier than what I came up with!)
Here is how I solved it in the past:

Proposition If $F$ is an infinite ring, and $V$ is a vector space of dimension $n$, then for a fixed set of finitely many hyperplanes (subspaces of dimension $n-1$), you can always find a line avoiding the hyperplanes.

Proof: Let $V_i$ be as in the original post, and find $u_i\in V\setminus V_i$ for each $i$. We inductively find $w_k$ such that $w_k\notin\bigcup_{j\leq k}V_j$. The base case is easy since $u_1=w_1$ suffices.
Suppose that $w_k$ has already been found. We claim that $w_k+Fu_{k+1}$ cannot be confined in $\bigcup_{j\leq k+1}V_j$. Suppose that it were. Then by the pigeonhole principle and the infiniteness of $F$, there is a $V_j$ containing $w_k+\alpha u_{k+1}$ and $w_k+\beta u_{k+1}$ for distinct $\alpha,\beta\neq 0$.
Case 1: $j=k+1$.  In this case, $V_{k+1}$ contains $c-(w_k+\beta u_{k+1})=(\alpha-\beta)u_{k+1}$, but then $u_{k+1}\in V_{k+1}$, a contradiction.
Case 2: $j<k+1$. In this case, $V_j$ contains $w_k+\alpha u_{k+1}-\frac{\alpha}{\beta}(w_k+\beta u_{k+1})=(1-\frac{\alpha}{\beta})w_k$. But that means that $w_k\in V_j$, but this contradicts that $w_k\notin \bigcup_{j\leq k}V_j$.
Thus $w_k+Fu_{k+1}$ is not contained in $\bigcup_{j\leq k+1}V_j$, and we set $w_{k+1}=w_k+\gamma u_{k+1}$ by picking such a $\gamma$ to escape the union. By induction, this is true for any $k\leq n$. Once $w_n$ is found, it is clear that $\bigcup_{j\leq n}V_j\neq V$.

I also wanted to add an example of where things break down when the field is finite. It breaks with the smallest case you can imagine: consider $V=F_2\times F_2$. $V$ has four elements, and it has three hyperplanes which we can enumerate as $V_1=\langle(1,0) \rangle$, $V_2=\langle(0,1)\rangle$ and $V_3=\langle(1,1)\rangle$. You'll notice that $V=V_1\cup V_2\cup V_3$! After seeing that, I think you should be able to prove that for a finite field, the union of all (finitely many) hyperplanes of $V$ is equal to $V$.
