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Is there a set $A$ such that for every $a,r>0$ we have that $(0,r)\cap A$ is non-countable but $(a,a+r)\cap A$ is countable?

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No. Because $$(0,r) \cap A = \left( \bigcup_{n>0} \left(\frac{1}{n}, r\right) \right) \cap A = \bigcup_{n>0} \left(\left(\frac{1}{n}, r\right) \cap A \right)$$

But a countable union of countable set is countable, hence if every $(\frac{1}{n}, r) \cap A$ is countable, $(0,r)\cap A$ is countable

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    $\begingroup$ It's nice to know that this answer uses the axiom of choice, and it is in fact consistent to have such $A$ in some models where the axiom of choice fails (very badly, though). $\endgroup$
    – Asaf Karagila
    Jun 13, 2015 at 7:43
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    $\begingroup$ @Harry: I think the proof of "countable union of countable sets is countable" uses it. $\endgroup$
    – Regret
    Jun 13, 2015 at 13:17
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    $\begingroup$ @Carlos: If I've understood correctly, the problem is with assigning the indices. Your set $A$ is constructed with them in place. $\endgroup$
    – Regret
    Jun 13, 2015 at 19:32
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    $\begingroup$ But there exist, by definition of the countability of $A_i$, a bijection between $\mathbb{Z}^+$ and $A_i$, so I don't think you need AC to assign the indices, as you have an explicit way to do this. $a_{i,j} = \varphi_{A_i}(j)$, and $\varphi_{A_i}$ exist by definition $\endgroup$
    – Tryss
    Jun 13, 2015 at 19:43
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    $\begingroup$ @Carlos: If you have a countable set, then there are $2^{\aleph_0}$ ways to put it in bijection with $\Bbb N$. These ways are completely arbitrary in the most case. So if you have infinitely many countable sets, how can you find a uniform way to match each one with $\Bbb N$? You can't. And there are models of set theory where the axiom of choice fails, and countable unions of countable sets are in fact uncountable. $\endgroup$
    – Asaf Karagila
    Jun 13, 2015 at 21:24

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