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I think $S = \{r \in \Bbb Q: 0 \leq r \leq 1\}$ in $\Bbb R$ can be a set that satisfies the conditions. First, it is compact by the Heine-Borel theorem since it is closed and bounded. It is closed because it contains all its limit points. Every rational number is a limit point; if we take an open neighbourhood of radius s, we can find some rational number contained in the open ball. Finally, the rationals are countable.

Is this correct?

Thank you.

August
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  • No, it's not. As limit points you get the whole interval $[0,1]$. – Leo163 Nov 02 '14 at 08:57
  • Sorry, I don't think I understand. – August Nov 02 '14 at 08:59
  • If you can find a set in $[0,1]$ with only one limit point, then you can use scaling to define one in $[1, 1.5]$, $[1.5, 1.75]$.... –  Nov 02 '14 at 08:59
  • Your set $S$ is not closed. –  Nov 02 '14 at 09:00
  • In the example where the integers are a subset of $\Bbb R^2$, the set of integers is trivially closed because every point in the integers was not a limit point. So would the complement of the rationals, the irrationals, be similar; trivially closed? (Immediately before and after an irrational is a rational). So would it be open then? – August Nov 02 '14 at 09:10
  • See this. – David Mitra Nov 02 '14 at 09:12
  • Thanks, David. I saw that but wanted to try my own. :) I'll take a look again. – August Nov 02 '14 at 09:14

1 Answers1

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What about $\{\frac 1n:n\in Z^+\} \cup \{0\}$?

Paul
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