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Possible Duplicate:
A vector space over $R$ is not a countable union of proper subspaces

This is a single step in a larger homework problem that I'm having difficulty with.


Consider a finite set of vectors $A$ in $\mathbb{R}^n$ of length $k$

$A$ has $m =\binom{k}{n-1}$ subsets of size $n-1$, designated by $A_1,A_2,...,A_m$

Let $S_j$ be the subspace spanned by $A_j$ (so $dim(S_j) \le n-1)$

Show that there must be an element in $\mathbb{R}^n$ that is not in any $S_j$ for all $j\in\{1...n\}$


This makes intuitive sense to me for $\mathbb{R}^2$ if thought of as the cartesian plane. Given any finite set of lines in $\mathbb{R}^2$,there must be a point in $\mathbb{R}^2$ that is not on any of those lines.

I can't think of a way to demonstrate this, even in $\mathbb{R}^2$.

  • 0
    In the plane you can write each space as $\{(r,\theta)\mid r\in\mathbb R^{\ge 0}, \theta\in[0,2\pi)\}$. Out of finitely many subspaces you can always find such $\theta$ which is in no subspace. Pick $r=1$, and you have your vector.2011-10-30
  • 0
    Rather simpler with a finite union of subspaces than a countable union, no?2011-10-31

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