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I got stuck on this problem and can't find any hint to solve this. Hope some one can help me. I really appreciate.

Give an example of a collection of sets $A$ that is not locally finite, such that the collection $B = \{\bar{X} | X \in A\}$ is locally finite.

Note: Every element in $B$ must be unique, so maybe there exist 2 distinct sets $A_{1}, A_{2} \in A$, but have $\bar{A_{1}} = \bar{A_{2}}$. So that's why this problem would be right even though $A \subset \bar{A}$.

Thanks everybody.

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    $\begingroup$ And the note should be a big hint. Suppose all $X \in A$ are dense. Then $B$ is finite. $\endgroup$ – Daniel Fischer Sep 12 '13 at 15:52
  • $\begingroup$ By "bar" do you mean complement or closure? $\endgroup$ – Mikhail Katz Sep 12 '13 at 15:54
  • $\begingroup$ Oh right, but can you specify a topology X and an infinite collection A such that every elements of A is dense... $\endgroup$ – le duc quang Sep 12 '13 at 15:55
  • $\begingroup$ @user72694:oh, I mean the closure, for sure! $\endgroup$ – le duc quang Sep 12 '13 at 15:56
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Just take some non-empty open set $U$ such that $\overline{U} \setminus U$ is infinite. Actually $U$ does not have to be open, just non-empty and such that $\overline{U} \setminus U$ is infinite. This includes the correct example of dense subset of $\mathbb{R}$ other people mentioned.

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  • $\begingroup$ I guess what you mean then we will take $A$ as a collection of subsets defined by adjoining $A$ with one element in $\bar{U} - U$.Then $B$ is just collection contains $\bar{A}$.I take all $U$ as the ball $B(0,1)$(without the unit circle) plus one point on unit circle. Is it ok? $\endgroup$ – le duc quang Sep 12 '13 at 16:03
  • $\begingroup$ @leducquang: Yes, that's it. $\endgroup$ – user87690 Sep 12 '13 at 16:04
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Sorry, typo, now corrected. All elements of A are dense in R.

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  • $\begingroup$ Got what you mean, but I think we can let it easier by choosing collection of sets of of the form $Q - \{1/n\}$... $\endgroup$ – le duc quang Sep 12 '13 at 16:23
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Let Q be the rationals in R, and A be the collection of sets of the form {a+√p, a ∈ Q}, p ∈ N.

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  • $\begingroup$ I don't get what you mean. Then in your case, what is $A$? $\endgroup$ – le duc quang Sep 12 '13 at 16:07

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