Fork me on GitHub
Math for the people, by the people.

User login

perfect set

Type of Math Object: 
Major Section: 

Mathematics Subject Classification

54A99 no label found


Is there a nonempty subset of IR which contains no rational number? (This is a problem from the second chapter, named "Basic Topology", of Walter Rudin's "Principles of Mathematical Analysis" ")

It's okay to just call it R.

Try {sqrt(2)} - it's a nonempty subset of R and contains no rational points. It happens to have only one sequence in it, namely {sqrt(2),sqrt(2),sqrt(2),...} which converges to sqrt(2). So the set of limit points of this set is {sqrt(2)}. So {sqrt(2)} is perfect.

This set has no limit points at all. A nonempty perfect set actually has to be uncountable.

Try taking a set somewhat like the Cantor set and translating it by a well-chosen irrational number. I think that might do it.

Oops :)

Sorry about that - I was using a different definition than is here (essentially a point in the closure). Yes, the condition that it should actually be in the closure of the set with the point removed makes things more difficult.

It might be possible to translate a Cantor set by some irrational x to get rid of the rational points, but yes, you'd have to be careful about not allowing x added to any irrational point in the Cantor set to be rational. That sort of thing might get tricky to work with. I'll have to look at it a bit more.

Rudin 1.1 shows us that for p $\in \mathbb{Q}$ and $x$ irrational, $p + x$ is irrational.

Subscribe to Comments for "perfect set"