A question on countability of isolated points of a subset of R

Solution 1:

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 $\phi$.

Solution 2:

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.

Solution 3:

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 $\mathbb{R}$.