# nonempty perfect subset of $\mathbb{R}$ that contains no rational number, a

We will construct a nonempty perfect set contained in $\mathbb{R}$ that contains no rational number.

We will begin with a closed interval, and then, imitating the construction of Cantor set, we will inductively delete each rational number in it together with an open interval. We will do it in such a way that the end points of the open intervals will never be deleted afterwards.

Let $E_{0}=[b_{0},a_{0}]$ for some irrational numbers $a_{0}$ and $b_{0}$, with $b_{0} . Let $\{q_{1},q_{2},q_{3},\ldots\}$ be an enumeration of the rational numbers in $[b_{0},a_{0}]$. For each $q_{i}$, we will define an open interval $(a_{i},b_{i})$ and delete it.

Let $a_{1}$ and $b_{1}$ be two irrational numbers such that $b_{0}. Define $E_{1}=E_{0}\backslash(a_{1},b_{1})$.

Having defined $E_{1},E_{2},\ldots,E_{n}$, $a_{1},a_{2},\ldots,a_{n}$ and $b_{1},b_{2},\ldots,b_{n}$, let’s define $a_{n+1}$ and $b_{n+1}$:

If $\displaystyle q_{n+1}\in\bigcup_{k=1}^{n}(a_{k},b_{k})$ then there exists an $i\leq n$ such that $q_{n+1}\in(a_{i},b_{i})$. Let $a_{n+1}=a_{i}$ and $b_{n+1}=b_{i}$.

Otherwise let $a_{n+1}$ and $b_{n+1}$ be two irrational numbers such that $b_{0}, and which satisfy:

 $\displaystyle q_{n+1}-a_{n+1}<\min_{i=0,1,2,\ldots,n}\{|q_{n+1}-b_{i}|\}$

and

 $\displaystyle b_{n+1}-q_{n+1}<\min_{i=0,1,2,\ldots,n}\{|a_{i}-q_{n+1}|\}.$

Now define $E_{n+1}=E_{n}\backslash(a_{n+1},b_{n+1})$. Note that by our choice of $a_{n+1}$ and $b_{n+1}$ any of the previous end points are not removed from $E_{n}$.

Let $\displaystyle E=\bigcap_{n=1}^{\infty}E_{n}$. $E$ is clearly nonempty, does not contain any rational number, and also it is compact, being an intersection of compact sets.

Now let us see that $E$ does not have any isolated points. Let $x\in E$, and $\epsilon>0$ be given. If $x\neq a_{j}$ for any $j\in\{0,1,2,\ldots\}$, choose a rational number $q_{k}$ such that $x. Then $q_{k}\in(a_{k},b_{k})$ and since $x\in E$ we must have $x, which means $a_{k}\in(x,x+\epsilon)$. Since we know that $a_{k}\in E$, this shows that $x$ is a limit point. Otherwise, if $x=a_{j}$ for some $j$, then choose a $q_{k}$ such that $x-\epsilon. Similarly, $q_{k}\in(a_{k},b_{k})$ and it follows that $b_{k}\in(x-\epsilon,x)$. We have shown that any point of $E$ is a limit point, hence $E$ is perfect.

Title nonempty perfect subset of $\mathbb{R}$ that contains no rational number, a NonemptyPerfectSubsetOfmathbbRThatContainsNoRationalNumberA 2013-03-22 15:26:10 2013-03-22 15:26:10 Gorkem (3644) Gorkem (3644) 18 Gorkem (3644) Example msc 54A99