The question is to prove that with respect to the euclidean metric on the Real numbers prove that if A is any subset of R, isoA is countable and hence deduce that if A is uncountable the A' is uncountable.

I am still not that comfortable with proving things like countability so any answers or help is appreciated



3 Answers 3


Let $A \subset \mathbb{R}$ and $\text{iso}(A)$ the set of isolated points of $A$. For any $x \in \text{iso}(A)$, there exist two rationals $a<b$ such that $x \in (a,b)$ and $(a,b) \cap A= \{x\}$. Let $\phi(x)=(a,b) \in \mathbb{Q}^2$.

Then $\phi : \text{iso}(A) \to \mathbb{Q}^2$ is injective and $\text{iso}(A)$ is at most countable.

For your second question, write $A= \text{iso}(A) \coprod A'$ and use the fact that the union of two countable sets is countable.

Remark: The axiom of choice is not needed to construct $\phi$.

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    $\begingroup$ Thanks a lot! One small doubt by the definition of the isolated points would it be (a,b)∩A=∅ or (a,b)∩A={x}? $\endgroup$
    – user68099
    May 27, 2013 at 5:36
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    $\begingroup$ $(a,b) \cap A= \{x\}$ of course. Thank you for you comment, I edited my answer. $\endgroup$
    – Seirios
    May 27, 2013 at 5:44
  • $\begingroup$ how do you formally prove that $\phi$ is injective? $\endgroup$
    – user68099
    Jul 2, 2013 at 17:30
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    $\begingroup$ Because the balls may overlap, so $\phi$ is not necessarily injective: you may pick out the same rational twice. $\endgroup$
    – Seirios
    Jun 3, 2014 at 18:06
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    $\begingroup$ @Seirios Oops, you are right. Thank you for your reply. $\endgroup$ Jun 4, 2014 at 6:18

Seirios has given a fine answer directed at your specific exercises; mine is just additional commentary that may add some insight or give you another way of thinking about the problems.

Note that one can prove the second result without explicitly proving the first. Suppose that $A$ is an uncountable subset of $\Bbb R$. Let $\mathscr{B}$ be the set of open intervals with rational endpoints; $\Bbb Q$ is countable, so $\mathscr{B}$ is countable as well. If $x\in\Bbb R\setminus A'$, there is some $B_x\in\mathscr{B}$ such that $x\in B_x$ and

$$B_x\cap A=\begin{cases} \varnothing,&\text{if }x\notin A\\ \{x\},&\text{if }x\in A\;. \end{cases}$$

(The cases can be combined into the single statement that $B_x\cap A\subseteq\{x\}$, but I thought that it might be clearer to separate the two possibilities.)

Let $\mathscr{B}_0=\{B_x:x\in\Bbb R\setminus A'\}$, and let $U=\bigcup\mathscr{B}_0$, and let $A_0=A\setminus U$. $\mathscr{B}_0\subseteq\mathscr{B}$, so $\mathscr{B}_0$ is countable, and each member of $\mathscr{B}_0$ contains at most one point of $A$, so $U\cap A$ is countable. If $A_0$ were countable, then $A=A_0\cup(U\cap A)$ would be the union of two countable sets and would therefore be countable, so $A_0$ must be uncountable. Thus, if we can show that $A_0\subseteq A'$, it will follow immediately that $A'$ is uncountable.

Suppose that $x\in A_0$. Then $x\notin U$, so $x\notin B_y$ for any $y\in\Bbb R\setminus A'$. In particular, $x\notin\Bbb R\setminus A'$, i.e., $x\in A'$, and we’re done: $A'$ is uncountable.

Your first result is hidden inside the proof above: it’s essentially the observation that $U\cap A$ is countable.

Almost the same argument actually proves the following stronger result:

Let $A$ be an uncountable subset of $\Bbb R$. Then there is an uncountable $A_0\subseteq A$ such that if $x\in A_0$, and $U$ is open neighborhood of $x$, then $U\cap A_0$ is uncountable (and of course then $U\cap A$ is also uncountable). In other words, $A$ has uncountably many points that are very far indeed from being isolated: their open nbhds not only have infinite intersection with $A$, but actually have uncountable intersection with $A$.

The details are in this answer.

  • $\begingroup$ Great explanation! Was really helpful and this result be extended to other metric spaces as well? $\endgroup$
    – user68099
    May 27, 2013 at 5:48
  • $\begingroup$ @user68099: Glad you found it useful. It extends essentially unchanged to separable metric spaces, since they are precisely the second countable metric spaces. $\endgroup$ May 27, 2013 at 7:15

Let A $ \subset \mathbb{R} $ . Now iso(A) $\subset A$. Let a $\in$ iso(A). We can choose a nbd N(a) of a such that N(a) $\cap$ A= { a } and N(a) $\cap$ N(b) = $\emptyset$ if b $\in$ iso(A) and b$\neq$a. Since $\mathbb{Q }$ is dense in R, N(a) contains a rational number, say q. $\forall a\in$ iso(A), choosing a rational number in such way we can construct a mapping f:iso(A)$\mapsto \mathbb{Q}$. Since f is injective, cardinality of iso(A) is $\leq$ cardinality of $\mathbb{Q}$. As $\mathbb{Q}$ is countable, iso(A) is also countable. Let A be uncountable. Then A-iso(A) is also uncountable. Now A-iso(A) $\subseteq A' $. Therefore $A'$ is also uncountable. Hence every uncountable subset of $\mathbb{R}$ has uncountable number of limit points in $\mathbb{R}$.

  • $\begingroup$ its \forall not \all $\endgroup$ Dec 17, 2020 at 0:56
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    $\begingroup$ Thank you Aryadeva, I'm new to Stackexchange. $\endgroup$ Dec 17, 2020 at 1:56
  • $\begingroup$ You're very welcome Saikai Prime $\endgroup$ Dec 17, 2020 at 1:57

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