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Non-empty perfect set in R with no rational number

  • Thread starter PingPong
  • Start date
62
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1. Homework Statement
Is there a non-empty perfect set that contains no rational number?


2. Homework 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!
 

Answers and Replies

StatusX
Homework Helper
2,563
1
So q is a limit point of P (because r was arbitrary).
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.
 
62
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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!
 
morphism
Science Advisor
Homework Helper
2,013
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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.
 
18
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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.
 
18
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C'mon, guys. There must be something wrong with this naive construction, otherwise it would be used as an example(on planetmath for instance).
 
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.
 
10
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I fail to see why [itex][e,\pi]\bigcap \mathbb{Q}^c[/itex] is not a perfect set...
 
10
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Oh gosh nevermind, it isnt closed.
 

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