# Construct compact set of R with countable limit points

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

arildno
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Shouldn't a single point be enough?

arildno
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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.

for example {(0, 1/n) : n=1,2,3,......} is compact but the only limit point is 0. Still I need countable limit points.

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.

matt grime
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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.

arildno
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forget_f1 said:
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.

Last edited:
HallsofIvy
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arildno said:
Yeah, I kind of remembered that a bit late... Finite sets are countable.

But the original post probably meant "countably infinite".

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 :)