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So the question is,

Give an example of a set of real numbers which has exactly 3 limit points and the set is closed.

I have been trying to solve it by myself, and I know one potential set can be $A$, such that, $A = \{k + \frac{1}{n}\ : n \in \mathbb{N} \}$ Where $K$ is the limit point of $A$.

The problem with this set is, this set is not closed, by the definition of a closed set, the set also need to contain it's limit points, but $k \notin A$

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    $\begingroup$ Hint: Look at the closure of that set, which just adds $k$. Now do it 3 times $\endgroup$
    – Alan
    Jul 2 at 10:44
  • $\begingroup$ What do you mean by closure of a set? Sorry I am very new in real analysis... $\endgroup$
    – Vincent
    Jul 2 at 10:47
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    $\begingroup$ The closure of a set is the smallest closed set it is a subset of, you get it by adding all the limit points $\endgroup$
    – Alan
    Jul 2 at 10:50

1 Answer 1

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Let $A(k)=\{k+\frac{1}{n}\mid n\in\mathbb{N}\}\cup \{k\}$, then $A(k)$ is closed and has one limit point, ie $k$. Now take $A(0)\cup A(1) \cup A(2)$. which is a finite union of closed sets, and hence closed. It contains three limit points, ie 0, 1, and 2, and it's easy to check it doesn't contain any others.

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  • $\begingroup$ I don't know why I did not think about adding k into the set in this way.... This is simple but genius. Thank you! $\endgroup$
    – Vincent
    Jul 2 at 10:49

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