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I tried doing that, but I didn't get anything at all.

Could you provide me with some hints?

What I'm sure of Is that, such a set doesn't contain any interval and it's infinite so I think it's a set like cantor's one or something like that.

FNH
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    You’re trying to make it too complicated. Start by finding a set with exactly one limit point; Seth’s hint would be a good place to begin. – Brian M. Scott Oct 10 '14 at 22:11

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Hint: Consider the set $\{ 1/n : n\in\mathbb{N}\}$. Can you use this set to solve the problem?

Seth
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  • I think that is a good hint! – MJD Oct 10 '14 at 22:11
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    If we let, $H={\frac{1}{n}:n\in \mathbb{N}} \cup {1+\frac{1}{n}:n\in \mathbb{N}} \cup {2+\frac{1}{n}:n\in \mathbb{N}}$, then $H$ has exactly three limit points, which are $0,1$ and $2$, right? – FNH Oct 10 '14 at 22:18
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    @MathsLover yup, you got it. – Seth Oct 10 '14 at 22:18
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    @MathsLover Note that, as you said, the set does not contain any interval, and yet it is much simpler than a Cantor set. – MJD Oct 10 '14 at 22:27