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Every closed subset of $\Bbb R$ has countably many isolated points.

Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!


My attempt:

Let $A \subseteq \Bbb R$ be a closed set and $I$ be the set of all isolated points of $A$.

If $a \in I$, then there exists $\delta > 0$ such that $x \neq a$ and $|x-a|<\delta$ implies $x \notin A$. Since $\Bbb Q$ is dense in $\Bbb R$, there exist $r,s \in \Bbb Q$ such that $a-\delta<r<a<s<a+\delta$.

As a result, if $a \in I$, then there exist $r,s \in \Bbb Q$ such that $a$ is the only element of $A$ in $(r,s)$.

Let $\langle r_n,s_n \rangle_{n \in \Bbb N}$ be an enumeration of the set $\{\langle r,s \rangle \mid r,s \in \Bbb Q\}$.

We define a mapping $\mathcal F:I \to \Bbb N$ by $$\mathcal F(a) = \min \{n \in \Bbb N \mid a \text{ is the only element of } A \text{ in } (r_n,s_n)\}$$

Clearly, $\mathcal F$ is injective and thus $|I| \le |\Bbb N| = \aleph_0$. Hence $I$ is countable.

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    $\begingroup$ I think real-analysis is a better tag because you're using two facts: 1- $\mathbb{R}$ is a metric space. 2- It has a dense subset $\mathbb{Q}$. Topology can also deal with spaces that are not metrizable which is way more abstract than $\mathbb{R}$. You can also add 'metric spaces' to your tags but I'm still not sure that the result holds in general metric spaces. (+1) $\endgroup$ – stressed out Feb 14 at 14:01
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    $\begingroup$ Also, your proof looks correct for $\mathbb{R}$. Well done. $\endgroup$ – stressed out Feb 14 at 14:05
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    $\begingroup$ I suggest you also edit the title and the first sentence: Every closed subset of $\mathbb R$ has countably many isolated points. $\endgroup$ – Lee Mosher Feb 14 at 14:06
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    $\begingroup$ And while I agree with @stressedout that your proof is correct, it is just a little bit too complicated. You don't need two points $r,s \in \mathbb Q \cap (a-\delta,a+\delta)$, you only need one point $r \in \mathbb Q \cap (a-\delta,a+\delta)$ to make the proof work. $\endgroup$ – Lee Mosher Feb 14 at 14:08
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    $\begingroup$ Thank you @stressedout! $\endgroup$ – MadnessFor MATH Feb 14 at 14:08
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Your proof looks correct for $\mathbb{R}$ with its standard topology. Although, as Lee Mosher pointed out, it could be simplified.

To see that your statement is something that depends on our knowledge of $\mathbb{R}$ with its usual metric, and it is not necessarily true in general topological spaces, or even metric spaces, consider the same set, the good old $\mathbb{R}$ with the discrete topology this time. That is, consider $\mathbb{R}$ equipped with the discrete metric:

$$d(x,y) = \begin{cases}0 & x=y \\ 1 & x\neq y\end{cases}$$

Then all points of $\mathbb{R}$ will become isolated points (take $0<\delta<1$), proving that your statement is not necessarily true about $\mathbb{R}$ with a topology other than the one coming from the Euclidean distance and it's a very specific statement in real analysis.

Also, your statement is true about separable metric spaces, i.e. any metric space that has a countable, dense subset. And the proof is similar to what you wrote taken Lee Mosher's comment into account.

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  • $\begingroup$ Is it that, instead of choosing $(r,s)$, we can accomplish the task by choosing $(r, 2a-r)$ where $r \in \Bbb Q \cap (a-\delta,a)$? $\endgroup$ – MadnessFor MATH Feb 14 at 14:16
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    $\begingroup$ @LeAnhDung I think what Lee Mosher meant is that you need only one point in each set to construct your function $\mathcal{F}$ and you can find it because $\mathbb{Q}\cap (a-\delta, a+\delta) \neq \emptyset$, because rationals are dense in real numbers. I think you don't even need to think about the order on real numbers for this to work, just the fact that rationals are dense suffices. Although, if you have found a relation between $r$ and $s$ depending on $a$, then that's good too. $\endgroup$ – stressed out Feb 14 at 14:20
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    $\begingroup$ Hi @stressedout, after looking back at my proof, i don't see the closeness of $A$ is used. So i think that the theorem generalizes to Every subset of $\Bbb R$ has countably many isolated points. $\endgroup$ – MadnessFor MATH Feb 17 at 9:40
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    $\begingroup$ @LeAnhDung Hi again. You're absolutely correct. What you have proved works for any subset of $\mathbb{R}$ and also implies that if $A$ is uncountable, then $A'$ must be uncountable too. Beautiful. $\endgroup$ – stressed out Feb 17 at 9:46
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    $\begingroup$ Thank you so much for your prompt reply ^^ $\endgroup$ – MadnessFor MATH Feb 17 at 9:47
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Thank to @Lee Mosher and @stressed out's suggestions, I presented a simpler proof here.


Let $A \subseteq \Bbb R$ be a closed set and $I$ be the set of all isolated points of $A$ .

If $a \in I$ , then there exists $\delta > 0$ such that $x \neq a$ and $|x-a|<\delta$ implies $x \notin A$ . Since $\Bbb Q$ is dense in $\Bbb R$ , there exist $r \in \Bbb Q$ such that $a-\delta<r<a<2a-r<a+\delta$ .

As a result, if $a \in I$ , then there exist $r \in \Bbb Q$ such that $a$ is the only element of $A$ in $(r,2a-r)$ .

Let $\langle r_n \rangle_{n \in \Bbb N}$ be an enumeration of $\Bbb Q$ .

We define a mapping $\mathcal F:I \to \Bbb N$ by $$\mathcal F(a) = \min \{n \in \Bbb N \mid a \text{ is the only element of } A \text{ in } (r_n,2a-r_n)\}$$

Clearly, $\mathcal F$ is injective and thus $|I| \le |\Bbb N| = \aleph_0$ . Hence $I$ is countable.

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    $\begingroup$ Yes. This is correct. You asked for more explanation about you need only one point in each set to construct your function F and you can find it because $\mathbb{Q}\cap(a-\delta,a+\delta) \neq \emptyset$. Well, the idea, as you know, is that you want to represent isolated points by elements of a countable set to find a function $\mathcal{F}$ from $I$ to $\mathbb{N}$. This is essentially what you did in your first solution. You considered all pairs $(r,s)$ which is a subset of $\mathbb{Q} \times \mathbb{Q}$, which is enumerable and then you considered an enumeration. To be continued... $\endgroup$ – stressed out Feb 14 at 16:54
  • $\begingroup$ Thank you so much @stressedout ^^ I got it. $\endgroup$ – MadnessFor MATH Feb 14 at 16:57
  • $\begingroup$ OK. I was writing the second part but I'm glad that you got it already. (+1) $\endgroup$ – stressed out Feb 14 at 16:58

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