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To prove it use the following lemma : If $F$ is closed and $x \in F$ then $x$ is a isolated point of $F$ if and only if $F-\{x\}$ is closed.

In a book, a solution is as follows :

Let's suppose that $\{ a_1,a_2, \cdots ,a_n \}$ is the set of all isolated points of F. Then by the lemma : $F-\{a_1,a_2, \cdots ,a_n \}$ is closed. Then this set which is closed and infinite countable does not have isolated points, a contradiction since everything closed countable has some isolated point.

My question comes in this last part, the statement that the set $G=F-\{a_1,a_2, \cdots ,a_n \}$ does not have any isolated point does not seem to be true.

Why? , because although it is true that the set $G$ can not contain an isolated point of $F$, that does not mean that the same set by itself has an isolated point.

I would like to know if my doubt is true or not.

My prove is identical until that last part. I affirm that there is an isolated point of $G$ let's call it $a$. So exists $\delta_1>0$ such that $(a-\delta_1,a+\delta_1) \cap G =\{a\}$. Let´s put $\delta_2 = min_{1\leq i \leq n}\{ \vert a-a_i \vert \}$. Then for $\epsilon = min\{\delta_1,\delta_2\}$ we have to $(a-\epsilon,a+\epsilon)\cap F =\{a\}$ so we would get an isolated point of $F$ different from all the others $a_i$.

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  • $\begingroup$ I am very confused by this question. Do we have the same definition of 'isolated point'? $\endgroup$ – Servaes Feb 16 '19 at 23:35
  • $\begingroup$ Let $F \subseteq \mathbb{R}$, a point $x \in F$ is isolated if exists $\delta>0$ such that $(a-\delta,a+\delta)\cap F = \{a\}$ $\endgroup$ – Juan Daniel Valdivia Fuentes Feb 16 '19 at 23:38
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    $\begingroup$ Yes, you are correct that the argument you added at the end is needed. In fact, if $F$ is closed and countable, removing from $F$ its isolated points gives us a new closed set which, if nonempty, has again isolated points but none of them are isolated in $F$. The difference with the situation in your argument is of course that the number of isolated points we removed is infinite and thus the number corresponding to your $\delta_2$ could now be zero. $\endgroup$ – Andrés E. Caicedo Feb 16 '19 at 23:44
  • $\begingroup$ Thanks for answering! $\endgroup$ – Juan Daniel Valdivia Fuentes Feb 16 '19 at 23:47
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Your findings agree with the proof; you have proved that if $G$ has an isolated point then $F$ has an isolated point different from all the $a_i$. But by assumption (toward a contradiction), the $a_i$ are all the isolated points of $F$. Hence $G$ does not have an isolated point.

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