If $S \subset \mathbb{R}$, can $x \in S$ be both an isolated point and an interior point of $S$?
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Well, if you are considering the topology of $\mathbb R$, then no. If you are considering the subspace topology of $S$, then yes. What do you want? – user251257 Sep 14 '15 at 03:37
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the topology of $\mathbb{R}$ – user270566 Sep 14 '15 at 03:40
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1just compare the definition of both concepts – user251257 Sep 14 '15 at 03:41
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Recall definitions. Interior: There is an open ball around x, all of whose points are in S. Isolated: There is an open ball around x, none of whose points are in S.
If it's in the interior, then any ball you draw around x will be containing points in S, and thus it's impossible to satisfy the definition of being isolated at the same time.
(Edit) This assumes that the space is dense everywhere, as for rationals, reals, and complex numbers. Thanks to the commenter below who points out that if a space is discrete then you can draw a ball around x empty of other points, which thus degenerately meets both requirements.
Daniel R. Collins
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8to be fair, you should say some thing about $\mathbb R$, as in a discrete space, every point is isolated and interior at the same time. – user251257 Sep 14 '15 at 04:10