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If an interval $I$ is a non-empty subset of $\Bbb R^n$ with the property that $a,b \in I$ and $a<c<b$, then $c \in I$.

Is it possible I union the integers? Where $I$ is a subset of the real numbers?

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  • $\begingroup$ According to your definition, $\{x\}$ is an inerval, for any $x$. Then $\Bbb Z=\bigcup_{x\in\Bbb Z}\{x\}$ writes the set of integers as the union of intervals $\endgroup$ – Hagen von Eitzen Sep 15 '16 at 4:53
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One can verify that this property is equivalent to being a connected subset of $\Bbb R$, for instance the proof appears in Rudin. It's easy to verify that $I$ cannot be a union of integers then.

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