# 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!

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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.

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Why pass to a closed ball? The open ball is completely metrizable or locally compact as well. Nice, anyway. There are in fact many results that can be proved both with Baire and with measure techniques. The upshot of both arguments is: a countable union of small things remains small. – t.b. May 26 '11 at 13:04
@Theo: ah, I see. – Qiaochu Yuan May 26 '11 at 13:08
Thanks, Qiaochu & Theo! You've given me the tools I need. Many thanks! – Joseph O'Rourke May 26 '11 at 13:12

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.

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I'll bite: if $K$ is an arbitrary uncountable field, what is a suitable notion of "ball" in $K^n$? – Pete L. Clark May 27 '11 at 6:15
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. – Scott Carnahan May 29 '11 at 7:21