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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 $\mathbb{R}^2$.

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user149100
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  • 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. – Asaf Karagila Oct 30 '11 at 23:19
  • Rather simpler with a finite union of subspaces than a countable union, no? – Robert Israel Oct 31 '11 at 02:29

1 Answers1

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Hint: You can prove, by mathematical induction, the stronger statement that $\mathbb{R}^n$ is strictly larger than any finite union of its proper subspaces. The case $n=1$ is trivial. Suppose $n>1$ and $S_1,\ldots,S_m$ are proper subspaces of $\mathbb{R}^n$. WLOG, assume that each $S_i$ has dimension $(n-1)$ and unit normal $v_i$.

  1. Argue that there is some unit vector $u$ such that $u\not=\pm v_i$ for all $i$.
  2. Denote by $P$ the hyperplane orthogonal to $u$. Determine $\dim(P)$ and $\dim(P\cap S_i)$.
  3. Show that there exists $v\in P$ such that $v\notin P\cap S_i$ for all $i$.
  4. Hence $v\notin S_i$ for all $i$.

In general, as noted by Gerry Myerson, $\mathbb{R}^n$ is not a countable union of proper subspaces.

user1551
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    Instead of algebraic manipulations one could also invoke the Baire category theorem: if $\mathbb{R}^n = \bigcup_{k=1}^{\infty} V_k$ then one of the $V_k$ must have non-empty interior, hence $V_k = \mathbb{R}^{n}$. – t.b. Oct 31 '11 at 13:27
  • Very nice argument! Yet perhaps a bit too advanced for the OP. In view of the problem's setting, I guess there might be some special, constructive proof. – user1551 Oct 31 '11 at 13:46
  • I only looked at Arturo's answer in the thread linked to above and didn't read the question there properly. I just noticed that the measure version of that argument was given in that question. I thought it's a nice, quick and dirty way of getting the same in one line and didn't intend this to be helpful for the OP. – t.b. Oct 31 '11 at 13:56