some properties of uncountable subsets of the real numbers
Let $S$ be an uncountable subset of $\mathrm{\beta \x84\x9d}$. Let $\mathrm{\pi \x9d\x92\x9c}:=\{(x,y):(x,y)\beta \x88\copyright S\beta \x81\u2019\text{\Beta is countable}\}$. For $\mathrm{\beta \x84\x9d}$ is hereditarily LindelΓΆff, there is a countable^{} subfamily ${\mathrm{\pi \x9d\x92\x9c}}^{\beta \x80\xb2}$ of $\mathrm{\pi \x9d\x92\x9c}$ such that $\beta \x8b\x83{\mathrm{\pi \x9d\x92\x9c}}^{\beta \x80\xb2}=\beta \x8b\x83\mathrm{\pi \x9d\x92\x9c}$. For the reason that each of members of ${\mathrm{\pi \x9d\x92\x9c}}^{\beta \x80\xb2}$ has a countable intersection^{} with $S$, we have that $(\beta \x8b\x83{\mathrm{\pi \x9d\x92\x9c}}^{\beta \x80\xb2})\beta \x88\copyright S$ is countable. As the open set $\beta \x8b\x83{\mathrm{\pi \x9d\x92\x9c}}^{\beta \x80\xb2}$ can be expressed uniquely as the union of its components, and the components are countably many, we label the components as $\{({a}_{n},{b}_{n}):n\beta \x88\x88\mathrm{\beta \x84\x95}\}$.
See that $(\beta \x8b\x83{\mathrm{\pi \x9d\x92\x9c}}^{\beta \x80\xb2})\beta \x88\copyright S$ is precisely the set of the elements of $S$ that are NOT the condensation points of $S$.
Now weβd propose to show that $\{{a}_{n},{b}_{n}:n\beta \x88\x88\mathrm{\beta \x84\x95}\}$ 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 $(x,x+r)\beta \x88\copyright S$ countable. So, there is some $({a}_{n},{b}_{n})$ such that $(x,x+r)\beta \x8a\x86({a}_{n},{b}_{n})$. See, if $x\beta \x88\x88({a}_{n},{b}_{n})$, 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 ${a}_{n}\beta \x81\u2019({b}_{n},\text{\Beta resp})$ is a left (right, resp) condensation point. Because, for each $\mathrm{{\rm O}\u0385}\beta \x88\x88(0,{b}_{n}{a}_{n})$, we have $({a}_{n},{a}_{n}+\mathrm{{\rm O}\u0385})\beta \x88\copyright S$ countable. And as no ${a}_{n},{b}_{n}$ is in $\beta \x8b\x83{\mathrm{\pi \x9d\x92\x9c}}^{\beta \x80\xb2}$, ${a}_{n},{b}_{n}$ are condensation points.
So, $\beta \x8b\x83{\mathrm{\pi \x9d\x92\x9c}}^{\beta \x80\xb2}$ is the set of noncondensation 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 $S$, $(x,y)\beta \x88\copyright S$ is uncountable; and as $T$ misses atmost countably many points of $S$, $(x,y)\beta \x88\copyright T$ is uncountable. So, $T$ is a subset of $S$ with inbetween property.
We summarize the moral of the story: If $S$ is an uncountable subset of $\mathrm{\beta \x84\x9d}$, 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 $T\beta \x8a\x86S$ of all the bilateral condensation points of $S$ has got the property: if $$, then there is also $z\beta \x88\x88T$ with $$.
Title  some properties of uncountable subsets of the real numbers 

Canonical name  SomePropertiesOfUncountableSubsetsOfTheRealNumbers 
Date of creation  20130322 16:40:42 
Last modified on  20130322 16:40:42 
Owner  sauravbhaumik (15615) 
Last modified by  sauravbhaumik (15615) 
Numerical id  21 
Author  sauravbhaumik (15615) 
Entry type  Topic 
Classification  msc 54F65 
Classification  msc 54F05 
Classification  msc 12J15 
Classification  msc 54E35 