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 $n$.

This is a finite union of nowhere dense sets and so is nowhere dense in $\mathbb{R}^n$.
