An interesting problem (closure, limit points)

[Buri]

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??

 

[Eynstone]

I reckon the Cantor set will do.

 

[Buri]

Yes it does. I realized this last night after looonngg hours of thinking lol And I've written up my proof that it does. Thanks a lot!