# If $F \subseteq \mathbb{R}$ is closed, countable infinity. Prove that $F$ has a countless infinite number of isolated points.

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 $$a_i$$.

• I am very confused by this question. Do we have the same definition of 'isolated point'? – Servaes Feb 16 '19 at 23:35
• Let $F \subseteq \mathbb{R}$, a point $x \in F$ is isolated if exists $\delta>0$ such that $(a-\delta,a+\delta)\cap F = \{a\}$ – Juan Daniel Valdivia Fuentes Feb 16 '19 at 23:38
• 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. – Andrés E. Caicedo Feb 16 '19 at 23:44
• Thanks for answering! – Juan Daniel Valdivia Fuentes Feb 16 '19 at 23:47

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.