4
$\begingroup$

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!

2 Answers 2

11

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.

  • 0
    Thanks, Qiaochu & Theo! You've given me the tools I need. Many thanks!2011-05-26
1

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.

  • 0
    If the field is normed, then you have a notion of unit ball. For an arbitrary field, I'd just take all of $K^n$ as the ball of infinite radius. The theorem in question doesn't really imply anything special about balls.2011-05-29