some properties of uncountable subsets of the real numbers

Let $S$ be an uncountable subset of $ℝ$.Let $𝒜:=\{(x,y):(x,y)\cap S\text{ is countable}\}$. For $ℝ$ is hereditarily Lindelöff, there is a countable subfamily ${𝒜}^{\prime }$ of $𝒜$ such that $\bigcup {𝒜}^{\prime }=\bigcup 𝒜$. For the reason that each of members of ${𝒜}^{\prime }$ has a countable intersection with $S$, we have that $(\bigcup {𝒜}^{\prime })\cap S$ is countable. As the open set $\bigcup {𝒜}^{\prime }$ can be expressed uniquely as the union of its components, and the components are countably many, we label the components as $\{(a_n,b_n):n\in \mathbb{N}\}$.

See that $(\bigcup {𝒜}^{\prime })\cap S$ is precisely the set of the elements of $S$ that are NOT the condensation points of $S$.

Now we’d propose to show that $\{a_n,b_n:n\in \mathbb{N}\}$ is precisely the set of the points which are unilateral condensation points of $S$.

Let $x$ be a unilateral (left, say) condensation point of $S$. So, there is some $r>0$ with $(x,x+r)\cap S$ countable. So, there is some $(a_n,b_n)$ such that $(x,x+r)\subseteq (a_n,b_n)$. See, if $x\in (a_n,b_n)$, then $x$ is NOT a condensation point, for $x$ has a neighbourhood $(a_n,b_n)$ which has a countable intersection with $S$. But $x$ is a condensation point; so, $x=a_n$. Similarly, if $x$ is a right condensation point, then $x=b_n$.

Conversely, each $a_n(b_n,\text{ resp})$ is a left (right, resp) condensation point. Because, for each $ϵ\in (0,b_n-a_n)$, we have $(a_n,a_n+ϵ)\cap S$ countable. And as no $a_n,b_n$ is in $\bigcup {𝒜}^{\prime }$, $a_n,b_n$ are condensation points.

So, $\bigcup {𝒜}^{\prime }$ is the set of non-condensation points - it is countable; and $\{a_n,b_n\}$ are precisely the unilateral condensation points. So, all the rest are bilateral condensation points. Now we see, all but a countable number of points of $S$ are the bilateral condensation points of $S$.

Call $T$ the set of all the bilateral condensation points that are IN $S$. Now, take two in $T$. As $x$ is a bilateral condensation point of $S$, $(x,y)\cap S$ is uncountable; and as $T$ misses atmost countably many points of $S$, $(x,y)\cap T$ is uncountable. So, $T$ is a subset of $S$ with in-between property.

We summarize the moral of the story: If $S$ is an uncountable subset of $ℝ$, then

1. The points of $S$ which are NOT condensation points of $S$, are at most countable.

2. The set of points in $S$ which are unilateral condensation points of $S$, is, again, countable.

3. The bilateral condensation points of $S$, that are in $S$, are uncountable; even, all but countably many points of $S$ are bilateral condensation points of $S$.

4. The set $T\subseteq S$ of all the bilateral condensation points of $S$ has got the property: if , then there is also $z\in T$ with .