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construct compact set of R with countable limit points

 
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Oct7-04, 07:43 PM   #1
 

construct compact set of R with countable limit points


Construct a compact set of real numbers whose limit points form a
countable set.
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Oct7-04, 07:48 PM   #2
 
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Shouldn't a single point be enough?
Oct7-04, 08:04 PM   #3
 
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Note:
I may have forgotten the precise definition of "the limit point".
You might instead look at a convergent sequence in R; that is a compact set, with one limit point.
Oct7-04, 10:07 PM   #4
 

construct compact set of R with countable limit points


for example {(0, 1/n) : n=1,2,3,......} is compact but the only limit point is 0. Still I need countable limit points.
Oct7-04, 10:11 PM   #5
 
Note: A single point has no limit point, since
a limit point of a set A is a point p such that for any neighborhood of p
(ie Ball(p,r) , where p is the origin and r=radius can take any value >0)
there exists a q≠p where q belongs in B(p,r) and q belongs to A.
Oct8-04, 03:55 AM   #6
 
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You can construct a set with one limit point. Now you can make one with two limit points, 3 limit points, indeed any number of limit points countable or otherwise.
Oct8-04, 04:58 AM   #7
 
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Quote by forget_f1
Note: A single point has no limit point, since
a limit point of a set A is a point p such that for any neighborhood of p
(ie Ball(p,r) , where p is the origin and r=radius can take any value >0)
there exists a q≠p where q belongs in B(p,r) and q belongs to A.
Yeah, I kind of remembered that a bit late...

Finite sets are countable.
Oct8-04, 06:51 AM   #8
 
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Quote by arildno
Yeah, I kind of remembered that a bit late...

Finite sets are countable.
But the original post probably meant "countably infinite".
Oct8-04, 09:35 AM   #9
 
Taking A={0, 1/n + 1/m | n,m >=1 in N}. Thus the limit points are 1/n which are countable.
Since the set is closed and bouned then it is compact. (theorem)
Or
It can prove by definition that A is compact, which is what I did since I forgot to use the theorem above which would have made life easier :)
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