Non-measure-theoretic proof there is a basis of ${\mathbb R}^n$ disjoint from a countable union $\cup_j W_j$ of proper subspaces? Using measure theory arguments, it is trivial to prove there is a basis of ${\mathbb R}^n$ that is disjoint from a given countable union $\cup_j W_j$ of proper subspaces. Is there a more elementary proof that's not too hard, e.g. just using the fact that the reals are uncountable?  
 A: Corrected version: You can do it by induction on $n$.
Without loss of generality assume that each $W_k$ has dimension $n-1$ and let $x_k\in\Bbb R^n$ be a unit vector orthogonal to $W_k$; $C=\{x_k:k\in\Bbb N\}$ is a countable subset of $S^n$, so there is an $x\in S^n\setminus C$. Let $W$ be the subspace of $\Bbb R^n$ orthogonal to $x$; then $W_k\cap W$ is a proper subspace of $W$ for each $k\in\Bbb N$. By the induction hypothesis $W$ has a basis disjoint from $W\cap\bigcup_{k\in\Bbb N}W_k$, say $B$. Now repeat the argument to find a subspace $V$ of $\Bbb R^n$ of dimension $n-1$ such that $V$ has a basis disjoint from $V\cap\left(W\cup\bigcup_{k\in\Bbb N}W_k\right)$, and pick $y\in V\setminus\left(W\cup\bigcup_{k\in\Bbb N}W_k\right)$; $B\cup\{y\}$ is a basis for $\Bbb R^n$ disjoint from $\bigcup_{k\in\Bbb N}W_k$.
A: There's a geometric proof for the case of $\mathbb{R}^2$. Every proper subspace is a line, so it intersects the unit circle at two points. But the unit circle is uncountable, so it cannot be covered if you only use countably many lines.
Here is a a nice papaer by Pete L. Clark which covers the general case, and proves the claim using cardinality arguments only: http://alpha.math.uga.edu/~pete/coveringnumbersv2.pdf
A: Basically the same proof as Brian M. Scott, we just need to additionally prove that ${\mathbb R}^n$ is not the countable union $\cup_j W_j$ of proper subspaces. Same induction idea. For $n=1$ this is trivial. For $n \geq 2$, we assume all subspaces have dimension $n-1$ and we choose $x$ to be a non-zero vector whose direction is not parallel to any of the normal vectors of the proper subspaces $W_j$. Then let $W$ be the subspace orthogonal to $x$, and note that if $\cup_j W_j = {\mathbb R}^n$ then $W = \cup_j (W \cap W_j)$ which would mean ${\mathbb R}^{n-1}$ is a countable union of proper subspaces, contradicting the inductive hypothesis.
A: Perhaps you mean that $\mathbb{R}^n$ is the not the union of a countable collection of proper subspaces? For that, you can use the fact that a proper subspace is nowhere dense, and use Baire Category.   
