
#1
Oct704, 07:43 PM

P: 11

Construct a compact set of real numbers whose limit points form a
countable set. 



#2
Oct704, 07:48 PM

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PF Gold
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Shouldn't a single point be enough?




#3
Oct704, 08:04 PM

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PF Gold
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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. 



#4
Oct704, 10:07 PM

P: 11

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.




#5
Oct704, 10:11 PM

P: 11

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. 



#6
Oct804, 03:55 AM

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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.




#7
Oct804, 04:58 AM

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PF Gold
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Finite sets are countable. 



#8
Oct804, 06:51 AM

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Thanks
PF Gold
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#9
Oct804, 09:35 AM

P: 11

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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