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Just going over some old notes and I realized I always took this for granted without actually fleshing out exactly why it is true.

The set of all r is closed in $\mathbb{Q}$, because the set of all r is just all of the rationals in that interval, and obviously contains all of its limits. If you take its complement, the irrational numbers between $- \sqrt{2}$ and $\sqrt{2}$, what precisely can we say about this to conclude that the original set is clopen? i.e. why precisely is the complement open?

Thanks.

r123454321
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2 Answers2

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$\Bbb Q$ inherits ist topology from $\Bbb R$. So if a set $U$ is open in $\Bbb R$, then $\Bbb Q \cap U$ is open in $\Bbb Q$, and the same thing goes for closed sets.

Now, the funny thing is that your set is both the intersection of a closed interval in $\Bbb R$ with $\Bbb Q$ (namely $[-\sqrt 2, \sqrt 2] \cap \Bbb Q$), and it's the intersection of an open interval with $\Bbb Q$ (namely $(-\sqrt 2, \sqrt 2)\cap \Bbb Q$). Thus, from the topology it ingerits from $\Bbb R$, your set is both an open set and a closed set. That is why it's a clopen set.

Arthur
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  • Thanks. Can you explain precisely how your reasoning changes if we consider my set as a subset of $\mathbb{R}$? I know the set does not contain all of its limits in $\mathbb{R}$, so it cannot possibly be closed. To conclude it is not open, do I simply take any neighborhood of any point in the set, and note that for any such neighborhood, not every point in the neighborhood lies within the set? – r123454321 Oct 11 '13 at 08:04
  • Well, it depends which way you "expand" the set to $\Bbb R$. As I pointed out, there are several sets in $\Bbb R$ that becomes your interval when you intersect with $\Bbb Q$. From your exact definition, the set ${r \in \mathbb{R}: - \sqrt{2} < r < \sqrt{2}}$ is open. The set ${r \in \mathbb{R}: - \sqrt{2} \leq r \leq \sqrt{2}}$ is closed. The set ${r \in \mathbb{R}: - \sqrt{2} < r \leq \sqrt{2}}$ is neither, but that doesn't prove anything. – Arthur Oct 11 '13 at 08:12
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Let $A=\{r\in \mathbb Q: -\sqrt{2}<r <\sqrt{2}\}$. Since, $A=(-\sqrt{2},\sqrt{2})\cap \mathbb Q$, therefore, $A$ is open. Again $A=[-\sqrt{2},\sqrt{2}]\cap \mathbb{Q}$. Thus $A$ is closed in $\mathbb Q$.

Anupam
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