For any point $x \in \mathbb{R}$ and a set $S$, $x$ is called an isolated point if $x \in S$ and $x \notin S'$, where the set $S'$ denotes the set of accumulation points. But what confuses me is that, if $x \in S$, then every neighborhood of $x$ has a nonempty intersection with $S$, but if $x \notin S'$, then for some $\epsilon > 0$, the deleted neighborhood $N^*(x;\epsilon)$ satisfies $N^*(x;\epsilon) \cap S = \varnothing$. To me, this seems like a contradiction, because the first claims that every neighborhood of $x$ contains at least one point in $S$, but the second one states that there is at least one deleted neighborhood of $x$ whose members do not belong in $S$.
What am I overlooking?
$\endgroup$ 22 Answers
$\begingroup$The more insightful definition of an isolated point of $S$ is:
There is some $\varepsilon>0$ such that $N(x; \varepsilon) \cap S = \{x\}$.
This implies that $N^\ast(x; \varepsilon) \cap S = \emptyset$, indeed, so $x \notin S'$. But clearly the definition implies $x \in S$ as well.
Indeed every neighbourhood of $x$ contains a point of $S$ (namely $x$ itself!), and for small neighbourhoods (smaller radius than $\varepsilon$) this is the only point this neighbourhood has in common with $S$. In the deleted neighbourhood we delete this $x$ itself, so then there are no more points in common with $S$ in the deleted neighbourhood (for those small ones).
$\endgroup$ $\begingroup$The first claim is wrong as $\{x\}$ may be a neighbourhood of $x$.
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