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Let $A = \{ (x,y) \in \mathbb{R}^2 : x \in \mathbb{Q}, y = \frac{1}{n}, n \in \mathbb{N} \} $. I want to find its interior and its closure. My claim is that $int A = \varnothing $. To see this, we can take $X =(x,y) \in A $. Then we see that $B(X, \sqrt{2} )$ lies outside $A$. Hence, every point $A$ has no neighborhood $N$ that is contained in $A$. Is this correct? Also, im claiming that $\overline{A} = A $. How can I show this? Thanks

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    $\begingroup$ Please note that you can't pick a radius when showing the interior is empty. One has to deal with all possible open balls. Also, w.r.t. the closure, consider the point $(\sqrt{2},\frac{1}{2})$. How 'far' is it from $A$? $\endgroup$ Feb 8, 2014 at 16:48

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If $(x,y)$ were an interior point of $A$, then there would be would be an $\varepsilon>0$ such that $(x-ε,x+ε)\times(y-ε,y+ε)\subset A$. But $A=\Bbb Q×\{1/n\mid n\in\Bbb N\}$, so we would have $(x-ε,x+ε)⊂\Bbb Q$, which is not possible.

To determine $\overline A$, first show that $\overline{B\times C}=\overline B×\overline C$ for $B,C⊂\Bbb R$, then show that $\overline{\{1/n\mid n\in\Bbb N\}}=\{1/n\mid n\in\Bbb N\}\cup\{0\}$ and $\overline{\Bbb Q}=\Bbb R$

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  • $\begingroup$ why is it not possible that $(x-ε,x+ε)⊂\Bbb Q$ ? $\endgroup$
    – user124140
    Feb 9, 2014 at 3:43
  • $\begingroup$ The interval is uncountable but $\Bbb Q$ is countable. Remember: All open intervals in $\Bbb R$ are bijective, even homeomorphic(!) with the Euclidean topology, including $(-\infty,\infty)$. $\endgroup$ Feb 10, 2014 at 20:10

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