## You are here

Homesome properties of uncountable subsets of the real numbers

## Primary tabs

# some properties of uncountable subsets of the real numbers

Let $S$ be an uncountable subset of $\mathbb{R}$. Let $\mathscr{A}:=\{(x,y):(x,y)\cap S\mbox{ is countable}\}$. For $\mathbb{R}$ is hereditarily LindelΓΆff, there is a countable subfamily $\mathscr{A}^{{\prime}}$ of $\mathscr{A}$ such that $\bigcup\mathscr{A}^{{\prime}}=\bigcup\mathscr{A}$. For the reason that each of members of $\mathscr{A}^{{\prime}}$ has a countable intersection with $S$, we have that $(\bigcup\mathscr{A}^{{\prime}})\cap S$ is countable. As the open set $\bigcup\mathscr{A}^{{\prime}}$ 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 $(\bigcup\mathscr{A}^{{\prime}})\cap S$ is precisely the set of the elements of $S$ that are NOT the condensation points of $S$.

Now weβd propose to show that $\{a_{n},b_{n}:n\in\mathbb{N}\}$ 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 $a_{n}(b_{n},\mbox{ resp})$ is a left (right, resp) condensation point. Because, for each $\epsilon\in(0,b_{n}-a_{n})$, we have $(a_{n},a_{n}+\epsilon)\cap S$ countable. And as no $a_{n},b_{n}$ is in $\bigcup\mathscr{A}^{{\prime}}$, $a_{n},b_{n}$ are condensation points.

So, $\bigcup\mathscr{A}^{{\prime}}$ 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 $x<y$ 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 $\mathbb{R}$, 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 $\exists x<y\in T$, then there is also $z\in T$ with $x<z<y$.

## Mathematics Subject Classification

54F65*no label found*54F05

*no label found*12J15

*no label found*54E35

*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff

## Recent Activity

## Info

## Corrections

spelling by Mathprof β

capitalization of title by Mathprof β

missing parentheses? by Mathprof β

confusing to me by Mathprof β