Interesting problem? boundaries/limit points etc

Buri
Messages
271
Reaction score
0
I've been working on this problem and would like some help or any hints.

Give an example of a nonempty set A subset of R such that A = br(A) = Lim(A) = Cl(A), where br(A) denotes the boundary points of A, Lim(A) denotes the limit points of A and Cl(A) denotes the closure of A.

I've tried finding conditions that this A seems to have to satisfy and this is what I have so far. It appears that A cannot be dense in R. Otherwise, br(A) = R which won't equal A. And since Cl(A) = int(A) U br(A) then I should have that int(A) is empty. So it seems like A is going to be a set of discrete points. But at the same time, the fact that it isn't dense sort of confuses me because I must have that every point is a boundary point and hence any ball around it has points of A and R\A, so its rather weird.

Any help or ideas?

EDIT:

Hmm I just thought of something, maybe I could let A to be an convergent infinite sequence...once the terms begin to become arbitrarily close to each other...I'll have to look at this...

EDIT:

Wait does the Cantor set satisfy these properties?

Any ideas??
 
Physics news on Phys.org
What's so weird about a discrete point being in the ball centered at it?

What should be weirding you out is the limit points of such a discrete set.
 
ZioX said:
What's so weird about a discrete point being in the ball centered at it?.

I didn't say that...

ZioX said:
What should be weirding you out is the limit points of such a discrete set.

Hmm I think that's my problem
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top