Can a Nonempty Perfect Set in R1 Exclude All Rational Numbers?

[boombaby]

Homework Statement

Is there a nonempty perfect set in R1 which contains no rational number?
(ex18,chap2,Principles of Mathematics 3rd edition)

Homework Equations

The Attempt at a Solution

I tried to prove that there is no such set (because I cannot construct one...) by trying to show that some points in it must be isolated. but a set of \sqrt{2}+1/n (closed but not perfect) prevents me to complete the proof.

Any hint would be appreciated!

 

[boombaby]

oh, that set contains sqrt{2} as a member

 

[Dick]

That isn't the problem with sqrt(2)+1/n. You want a set with no rational members. Your set works perfectly in that respect. But it's not a perfect set. Not all of it's points are limit points. sqrt(2) is. sqrt(2)+1 isn't. It's an isolated point. I was reclusing myself from this question because I looked up an answer while looking up the definition of a perfect set. Pick an interval with irrational endpoints and use a cantor set type construction. Remove all of the rationals but be careful to preserve the perfect set quality.

 

[boombaby]

Thanks a lot. The idea of Cantor set helps me a lot.
Let E be the interval [sqrt(2),2*sqrt(2)] and use the similar construction. every member of E should take the form of sqrt(2)+sqrt(2)*k/(3^m), where k is nonnegative integer and m is positive integer and hence is irrational.

 

[matt grime]

That's not a cantor set. That set is countable, for example.

 

[boombaby]

I'm a little confused.
I know Cantor set is uncountable and I know how to prove it . But it really confuses me because it seems that cantor set contains only the end points of those intervals and therefore should take the form k/(3^m), which forms a countable set. Hence cantor set is countable (which is wrong). since cantor set is not countable, some point cannot be written in k/(3^m) - and I've no idea of what kind of point it should be. This is important because if I don't get it, I can not confirm that set in my original question contains only irrational numbers.
Or maybe I should construct it in this way: take the end points of each intervals as members? But I'm not so clear about it yet...
Any help? Thanks

 

[Dick]

The usual cantor set consist of ternary decimals consisting of only 0's and 2's. Since you systematically subtract the points with a 1 in them. This set contains both rationals, like 0.0202020202... and irrationals. You have to find a way to alter this construction in such a way as to contain only irrationals. This is not an easy exercise and I'm assuming you wouldn't have been given it if you weren't capable of handling it. You can assume an enumeration of the rationals over an interval. Just try and subtract them all. To answer your question of finding an irrational in the ternary cantor set, how about 0.2020020002000020000020... If the ternary expansion doesn't repeat, it's not rational. Now is it?

 

[boombaby]

to be frank, I should say I encounter more problems as I look closer in it...
sorry that I'm not yet clear about the construction but I'm still working on it...
Any way, thanks for all your helps! It helps me a lot

 

[boombaby]

I got the answer. well, it's a somewhat direct instruction from my friend. The construction is easier than I expected...Thanks again