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I was glanced at this question here and it cause me to wonder the following:

Question: Is there a simple description of the subsets $L \subset [0,1]$ with the property that there exists a sequence in $[0,1] \setminus L$ whose limit points are exactly $L$?

A few things are pretty easy to see

  • Any finite nonempty $L$ has this property.
  • Some strange sets have this property. For instance, I think you can get the Cantor set from, say, an enumeration of the midpoints of the intervals deleted during the "take out the middle thirds" construction.

Anyway, I thought it might be a fun problem to find a precise description.

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Mike F
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    You should convince yourself that the set of all limit points of a sequence (or of any set, for that matter) is always closed. I think that every closed set can be thusly characterized. – Ben Grossmann Sep 30 '14 at 23:07
  • I agree that the set of limit points is closed, but the requirement that the sequence stay out of $L$ prevents you from getting any set containing an interval. – Mike F Sep 30 '14 at 23:09
  • @Omnomnomnom: I guess it could just be all closed sets $L \subset [0,1]$ with dense complement...? – Mike F Sep 30 '14 at 23:09
  • I forgot about the "staying out of $L$" requirement, this is interesting after all. I think that your characterization is correct. Another way of putting this is that $L$ must be a closed, measure-zero set (I might have to double check that, though). – Ben Grossmann Sep 30 '14 at 23:15
  • @Omnomnomnom: Actually those don't mean the same thing. For instance, if $q_1,q_2,\ldots$ are the rational points in $[0,1]$ and you delete an open interval of length $(1/2)^{n+1}$ around each $q_n$, then you are left with a closed set of positive measure whose complement is dense. – Mike F Sep 30 '14 at 23:20
  • I'm aware that sequences have measure zero. I'm saying that if a closed set (i.e. the set of limit points) has a dense complement, then it must have measure zero. – Ben Grossmann Sep 30 '14 at 23:23
  • @Omnomnomnom: Your last statement is precisely what I'm disagreeing with. Have a look here for more info. – Mike F Sep 30 '14 at 23:27
  • Omno... is right: you get exactly all closed sets. To obtain a given closed set as the set of limit points, just pick a countable dense subset. –  Sep 30 '14 at 23:47
  • @MikeF ah, you're right. – Ben Grossmann Sep 30 '14 at 23:58
  • @ChristianRemling The sequence must be disjoint from the set, so a countable subset does not work. –  Oct 24 '14 at 22:30

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