Homework Help: Non-empty perfect set in R with no rational number

1. Mar 17, 2008

PingPong

1. The problem statement, all variables and given/known data
Is there a non-empty perfect set that contains no rational number?

2. Relevant equations
None

3. The attempt at a solution
I thought the answer was no, but my professor said that there is. My reasoning is as follows (please let me know if I'm wrong here):

If p is an irrational limit point of a perfect set P, then every open ball B(p;r) around the point such that B(p;r) that contains another point in P. But this ball contains rational numbers, so a rational number q is in B(p;r). Thus a ball of the same radius around q contains the point p, which is in P. So q is a limit point of P (because r was arbitrary). Since P must be closed, it contains all of its limit points, so q is in P.

Where'd I mess up? Thanks in advance!

2. Mar 18, 2008

StatusX

This doesn't follow - there may be different q for different r.

I agree with your professor. To get the example I have in mind, start by trying to construct an open set containing the rationals whose complement has positive measure.

3. Mar 18, 2008

PingPong

Ok, I can see where I messed up, but now I haven't got a clue about where to start. We haven't gone over anything with measure - the only tools I've got are those presented in Rudin's Principles of Mathematical Analysis, chapters 1-2.

Thanks again!

4. Mar 18, 2008

morphism

You can construct a "fat Cantor set" in [0,1] that misses all the rationals (or any other countable set of your choosing). These things I believe still have the property of being perfect.

5. Jun 3, 2010

a_Vatar

Instead of creating a new thread, I decided to just write in here.

If we define a set X={p in (0, 1), p in R} and let E = X \ Q.

Now, E non-empty(since Q is countable and X is not) and bounded, furthermore, no point of Q is in E. Now what remains to be shown is that if p in E, then p is a limit point of E. Clearly, every neighborhood of p contains another irrational(this point can made rigorous, I suppose) and since E does not contain any rational, the neighborhood doesn't either.
So E is the required set.

Is there a flaw in the (naive) arguement somewhere? The answers I've seen so far proceed with a construction of E by removing neigborhoods around enumaration of rational, etc.

6. Jun 6, 2010

a_Vatar

C'mon, guys. There must be something wrong with this naive construction, otherwise it would be used as an example(on planetmath for instance).

7. Sep 16, 2010

nstrehlke

There are several flaws in your argument a_Vatar. If the set E is, as you described, the set of all real numbers p in the interval (0, 1) such that p is not rational, then clearly 0 is a limit point of E. To see this, note that sqrt(2)/2 is contained in E, as is sqrt(2)/(2^n), for any positive integer n. By the archimedean property of R, there exists n such that sqrt(2)/(2^n)< r for any positive real number r. This shows therefore that there is a point of E in every neighborhood of 0, and so 0 is a limit point of E. Since 0 is not contained in E, E is not closed, and hence not perfect.

A more general argument can be made, in fact, which shows that for any interval I of R (open or closed), the intersection of I with the complement of the set Q of all rationals is not perfect. In order to prove this, suppose to the contrary that there exists such a set, say P, and suppose q is a rational number such that q is contained in I and q is not an endpoint of I (we are guaranteed such a q by the density of Q in R). Since P is perfect, q is not a limit point of P, hence there exists a neighborhood of q which is a subset of I and which contains no point of P. It follows that there exists a subinterval I* of I which contains only rational numbers. However, I* is clearly countable, which contradicts the fact that every interval of real numbers is uncountable.

8. Jul 18, 2012

finkeljo

I fail to see why $[e,\pi]\bigcap \mathbb{Q}^c$ is not a perfect set...

9. Jul 18, 2012

finkeljo

Oh gosh nevermind, it isnt closed.