# Is there a set non-countable around zero but countable elsewhere?

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?

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
• 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). Jun 13, 2015 at 7:43
• @Carlos: If I've understood correctly, the problem is with assigning the indices. Your set $A$ is constructed with them in place. Jun 13, 2015 at 19:32
• 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 Jun 13, 2015 at 19:43
• @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. Jun 13, 2015 at 21:24