nonempty perfect subset of $ℝ$ that contains no rational number, a

We will construct a nonempty perfect set contained in $ℝ$ thatcontains no rational number.

We will begin with a closed interval, and then, imitating theconstruction of Cantor set, we will inductively delete eachrational number in it together with an open interval. We will doit in such a way that the end points of the open intervals willnever be deleted afterwards.

Let $E_0=[b_0,a_0]$ for some irrational numbers $a_0$ and $b_0$,with . Let $\{q_1,q_2,q_3,\mathrm{\dots }\}$ be an enumeration of therational numbers in $[b_0,a_0]$. For each $q_i$, we will define anopen interval $(a_i,b_i)$ and delete it.

Let $a_1$ and $b_1$ be two irrational numbers such that. Define $E_1=E_0\backslash \backslash (a_1,b_1)$.

Having defined $E_1,E_2,\mathrm{\dots },E_n$, $a_1,a_2,\mathrm{\dots },a_n$ and $b_1,b_2,\mathrm{\dots },b_n$, let’s define $a_{n+1}$ and $b_{n+1}$:

If $q_{n+1}\in {\displaystyle \bigcup _{k=1}^n}(a_k,b_k)$ then there existsan $i\le n$ 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 numberssuch that , and which satisfy:

and

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

Let $E={\displaystyle \bigcap _{n=1}^{\mathrm{\infty }}}{E}_n$. $E$ is clearlynonempty, does not contain any rational number, and also it iscompact, being an intersection of compact sets.

Now let us see that $E$ does not have any isolated points. Let$x\in E$, and $ϵ>0$ be given. If $x\ne a_j$ for any $j\in \{0,1,2,\mathrm{\dots }\}$, choose a rational number$q_k$ such that . Then $q_k\in (a_k,b_k)$ andsince $x\in E$ we must have , which means $a_k\in (x,x+ϵ)$. 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 . Similarly, $q_k\in (a_k,b_k)$ and it follows that $b_k\in (x-ϵ,x)$. We have shown thatany point of $E$ is a limit point, hence $E$ is perfect.