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i have a problem about the definition of isolated point. In my notes, it says that a boundary point z of A may not be an accumulation point. To prove this, it says if z is an isolated point of A, then there is a ball B(z,r) such that B(z,r) intersection with A = {z} and this z is the isolated point. The only possibility i think is that if A is a discrete metrics then this situation is possible. Otherwise, that ball includes infinitely many points. Am i right here? What exactly is an isolated point? I would appreciate if someone could help. Thanks

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3 Answers 3

Consider the plane as a metric space and let $K$ be the closed unit disk along with the point $(0,2)$. Then $(0,2)$ is an isolated point of $K$. This is so because there is an $r > 0$ so that $$B_r((0,2))\cap K = \{(0,2)\}.$$

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thanks, so, am i right this onyl happens with the discrete metrics? –  user49065 Dec 3 '12 at 15:47

Definition of isolated point: Let $X$ be a metric speac, $x$ is an isolated point of $S\subset X$ if there exist an $\epsilon >0$ st $B_S(x,\epsilon)=\{x\} $. In another word, you can consider it as a point which has no neighborhood, just a single point. Infact it appears in lots of the metric space. One example is $\mathbb{N}$, all point of $\mathbb{N}$ are isolated points.

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It seems to me that the result your interested in is the following. Let $(X,d)$ be a metric space and $x \in X$. Then $x$ is an isolated point of $X$ if and only if there exists some $\varepsilon>0$ such that for every $y \in X\setminus\{x\}$ we have $d(x,y)>\varepsilon$. Proving this is mostly just moving definitions around.

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