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This is exercise $2.13$ in Rudin.

Can't we simply define such set as $[a, b]$, with all members being rational? It is bounded, and closed (proof is straightforward), and the limit points are all members of the set since $\mathbb{Q}$ is dense on an interval; and since rational numbers are countable then all the limit points are. Am I missing something? Thanks in advance.

Asaf Karagila
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G Ch
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2 Answers2

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You have stated why this doesn't work. The limit points comprise the entire interval $[a,b]$, which is not a countable set.

MPW
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  • But the interval contains only rational numbers, and even if their number is infinite in this interval, they are countable (as $Q$ is). – G Ch Feb 12 '15 at 17:53
  • "limit points" of a set can include things not in the set. Besides, your set is not compact, either. – GEdgar Feb 12 '15 at 17:56
  • Presumably, we are working in the reals, not the rationals. In the rationals, every subset is countable. You should be looking for this set to be closed and bounded in $\mathbb R$. – MPW Feb 12 '15 at 17:58
  • Got it, sorry, silly question. Thanks for the answer. – G Ch Feb 12 '15 at 17:59
  • Would the set of 1/n (n integers) and 0 do? In that case, 0 is the only limit point. – G Ch Feb 12 '15 at 18:12
  • Technically, yes, as would any convergent sequence together with its limit. However, I suspect the question intends there to actually be a countably infinite set of limit points. But that's easy, right? Just take the countable compact set you constructed, and for each point in that set, throw in a sequence converging to that point. So, I think $${\tfrac1k + \tfrac1n:k,n\in\mathbb N}\cup{0}$$ would work. – MPW Feb 12 '15 at 19:41
  • (addendum) You may want to ensure that the sequences don't overlap, to avoid introducing possible additional limit points; so perhaps adding the restriction that $n>k$ is a good idea. – MPW Feb 12 '15 at 20:02
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if you want limit points should be finite then {1/n| n $\in \mathbb{N} $}$\cup$ {0} in case of limit points countably infinite:- for each n$\in \mathbb{N}$ define $M_n$ = { 1/n +1/r | r$\in \mathbb{N}$ and 1/r < 1/(n-1) - 1/(n)} $\cup$ {1/n}, define $ M = \bigcup_{n>1} M_n$. claim: the required set is $M \cup {0}$... this set is compact since closed and bounded and set of limits points are {1/n| n $\in \mathbb{N} $}$\cup$ {0} which is countable

Anubhav Mukherjee
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