## Uncountable Subset A of [0,1] with A – A empty.

### December 20, 2011

I’m going through a few books so that I can start doing lots and lots of problems to prepare for my quals. I’ll be posting some of the “cuter” problems.

Here’s one that, on the surface, looks strange. But ultimately, the solution is straightforward.

**Problem.** Find an uncountable subset such that has empty interior.

**Solution**. What’s our favorite empty interior set? That’s right, . It would be really nice if we could make equal to something like . So, what should we do?

Let’s make an equivalence class. Let’s say that iff . Each equivalence class must be countable (why?) and so we must have an uncountable number of equivalence classes. (Here’s where the AoC comes in.) Pick one element from each class and let this be our .

Note now that contains only elements which are irrational, and hence is a subset of , i.e., it is contained in the irrationals. This is not too hard to prove; suppose there was a rational number in ; then we’d have that there exists such that or, in other words, the difference of and is a rational. This would imply they are in the same equivalence class. Bam.

It suffices now to show that if some set has empty interior, and then has empty interior. This is straightforward. That completes the problem.

**Edit: **Thanks to Vipul for pointing out that I made an error in saying that is a subset of the rationals.

**Edit Edit: **I woke up in a cold sweat because I realized I had left something out here! The elements of aren’t JUST the irrationals, there is also 0 inside of it (since, in particular, for any ). This complicates things only a little bit; the last paragraph, in particular, doesn’t apply directly, but it reduces our workload: we only need to show that has empty interior. Of course, this is not very hard; if we supposed some open ball was inside of this, it would need to contain another rational number besides 0. That’s a contradiction.

Vitali Set?

Yep. It turns out to be an extremely natural set to use in this case. It’s unfortunate that an easier set doesn’t exist, but it’s often extremely difficult to visualize A – A.

Parenthetically, on the Analysis Qual that I took, one of the questions was: Let E be a subset of the reals and let the lebesgue measure of E be positive. Show that E – E contains a neighborhood of the origin. It’s a good question to think about, and if you want it to be a bit easier, pretend that E is open or closed to begin with.