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As stated in the title. At the first glance I think the approach can be constructing an injection from $A$ to $\mathbb Q$, since obviously $\mathbb Q$ is a set that satisfies such condition. However I have no idea on how to get such injection. Any hints would be appreciated.

Caleb Stanford
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zeno tsang
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    Why are you mixing the terminology "countable" and "denumerable" in the same question? – Caleb Stanford Feb 05 '14 at 05:15
  • @Darrin That will not include $A$ necessarily. For instance, suppose that the value of $\delta$ you get for a rational number $q$ is $\frac12 |\sqrt{2} - q|$, and that $\sqrt{2} \in A$. – Caleb Stanford Feb 05 '14 at 05:28
  • Also @PedroTamaroff I'm not sure what you have in mind, but this seems to me in danger of the same problem as Darrin's solution. You need to use that this is true for all $x \in \mathbb{R}$; it is not sufficient that it holds for all $x \in \mathbb{Q}$. – Caleb Stanford Feb 05 '14 at 05:30
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    @Goos I've withdrawn the hint, using $\Bbb R$ is Lindelöf is the way to go. – Pedro Feb 05 '14 at 05:35
  • @Darrin see my example. I thought what you did at first too, but the problem is that $\bigcup_{p \in \mathbb{Q}} (p-\delta, p+\delta)$ does not necessarily cover $\mathbb{R}$ just because $\mathbb{Q}$ is dense. – Caleb Stanford Feb 05 '14 at 05:36
  • @Goos yes, I've withdrawn as well. I see what you mean now. – Darrin Feb 05 '14 at 05:37

3 Answers3

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For every $x\in \mathbb{R}$, let $\delta_x$ such that $(x-\delta_x,x+\delta_x)\cap A$ is countable. Set $B_x=(x-\delta_x,x+\delta_x)$. Since we have that $\mathbb{R}=\bigcup_{x\in \mathbb{R}}B_x$ and since $\mathbb{R}$ is separable, by this there's a countable subcover, that is, there's a sequence of points $(x_n)_{n\in\mathbb{N}}$ such that $\mathbb{R}=\bigcup_{n=1}^\infty B_{x_n}$. Can you conclude what you want from here?

Community
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Brandon
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    Your proof is neat and nice. However, it uses a sophisticated result in topology that $\mathbb{R}$ is Lindelof, while the book where this problem comes from only assume basic background of real analysis. So is there a proof looks more fundamental? Or I will pick this as the answer. – zeno tsang Feb 05 '14 at 05:53
  • @zenotsang Best to wait a day or so before picking the answer. Usually people will be working on an alternate solution but if you just accept an answer then they will give up. – Caleb Stanford Feb 05 '14 at 06:02
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Because $[n,n+1]$ is compact, an open cover $\cup_{x\in[n,n+1]}(x-\delta_x,x+\delta_x)$ has a finite sub cover. That is, there is a finite set $B_n$ where $[n,n+1] \subseteq \cup_{x \in B_n} (x-\delta_x, x+\delta_x)$. We have $B = \cup_{n \in Z} B_n$ a countable set and hence $R = \cup_{n\in Z} [n, n+1] \subseteq \cup_{x \in B} (x-\delta_x, x+\delta_x)$. Finally, $A = A \cap R = \cup_{x \in B} A \cap (x-\delta_x, x+\delta_x)$ is a countable union of countable sets by assumption. Hence it is countable.

abnry
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  • clear and good. It only uses Heine-Borel theorem in mathematical analysis, and in fact it follows that $\mathbb{R}$ is Lindelof. – zeno tsang Feb 06 '14 at 06:30
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An elementary argument: Let $$ S = \left\{ r \in \mathbb{R}^+ \; : \; (-r, r) \cap A \text{ is countable} \right\} $$

Clearly if $r \in S$ and $r_0 < r$, then $r_0 \in S$. Also, $S$ is nonempty, since we can find $\delta > 0$ such that $(-\delta, \delta) \cap A$ is countable.

We would like to show $\sup S = \infty$, because then we will have $n \in S$ for all positive integers $n$, and $$ A = \bigcup_{n = 1}^\infty (-n, n) \cap A $$ will be a countable union of countable sets, hence countable.

Suppose towards contradiction $\sup S < \infty$. Then let $R = \sup S$, and find $\delta_1$ and $\delta_2$ such that $(-R - \delta_1, -R+ \delta_1) \cap A$ and $(R - \delta_2, R + \delta_2) \cap A$ are countable. Then observe that $R - \delta_1, R - \delta_2 \in S$, so in fact $(-R - \delta_1, R + \delta_2) \cap A$ is countable. Letting $\delta = \min(\delta_1, \delta_2)$, it follows that $R + \delta \in S$, contradicting that $R$ was the supremum.

Alternatively, you can show that $S$ is open, closed, and nonempty. Since the only clopen sets in $\mathbb{R}^+$ are $\varnothing$ and $\mathbb{R}^+$ itself, $S = \mathbb{R}^+$.

Caleb Stanford
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