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In principals of mathematical analysis on page 32, Rudin defines (definition 2.18e-f): a point $p$ is an interior point of $E$ of there is a neighborhood $N$ of $p$ such that $N\subset E$, $E$ is open if every point of $E$ is an interior point of $E$

By this definition, shouldn't $\mathbb{Z}$ be open? Because $\forall n \in \mathbb{Z} \exists m \in \mathbb{Z}s.t. |n-m|>0$ and so $m$ is in a neighborhood of $n$ and in $\mathbb{Z}$


marked as duplicate by Dietrich Burde, user223391, Alex M., Bungo, Community Nov 14 '16 at 20:52

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  • $\begingroup$ in which topology ? If it's as a subset of $\mathbb R$, what is the $\partial \mathbb Z$ ? conclusion ? $\endgroup$ – Surb Nov 14 '16 at 19:54

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Assuming the usual, Euclidean topology on the reals: for any $\;m\in\Bbb Z\;$ and for any $\;\epsilon>0\;$ , the neighborhood $\;(m-\epsilon,\,m+\epsilon)\rlap{\;\,/}\subset\Bbb Z\;$ and thus $\;\Bbb Z\;$ cannot be open


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