Removing countable hyperplanes from a ball in $\mathbb{R}^n$ I would like to claim that an open ball in $\mathbb{R}^n$ cannot be covered by a
countable collection of $(n-1)$-dimensional hyperplanes, that in fact excluding
the hyperplanes from the ball still leaves a set that contains an
uncountable number of points.
For $n=2$, this says removing a countable number of lines from a disk
still leaves an uncountable number of points of the disk.
What is the correct language to justify this?  It's a bit out of my expertise...
Thanks for suggestions!
 A: There are various ways to do this. The most intuitive is probably to show that any hyperplane has zero measure and use the fact that a countable union of sets of zero measure has zero measure. The simplest is probably to take a closed ball inside your open ball and apply the Baire category theorem to it. 
A: Here's an algebraic method that reduces the $n \geq 2$ case to the $n=1$ case, which is relatively easy.  Useless bonus: you can replace $\mathbb{R}$ with any uncountable field (for suitable notions of "ball").
For each hyperplane in the collection, we can take the unique normal line through the origin.  We get a pair of points on the unit hypersphere by taking the intersection with this line.  If we collect all of the points arising from the hyperplanes, the coordinates generate a field extension $K$ of $\mathbb{Q}$ that has at most countably infinite transcendence degree (and is therefore countable and not all of $\mathbb{R}$).
Choose a point $x$ on the unit hypersphere whose first $n-1$ coordinates are algebraically independent of $K$ and of each other, and consider the line $\ell$ passing through $x$ and the origin.  Each hyperplane can intersect $\ell$ in at most one point, since $x$ has nonzero inner product with the normal vector of any hyperplane in the collection.
