Re: Construction of a non-empty perfect set of real numbers without rationals

Question

We consider the set $[e, \pi]$.

Let $\{x_1,x_2,x_3,...\}$ be the enumeration of the rationals in $[e, \pi]$. $r_n= \min\{|x_n-e|,|\pi-x_n|\} $

We can enclose $x_n$ by $I_n=(a_n,b_n)=(x_n-\frac{r_n}{2^{100n}},x_n+\frac{r_n}{2^{100n}})\subsetneq [e, \pi] $.

[Note: The irrationality of the endpoints will be maintained for every $n \in \mathbb{N}$, since $r_n$ is irrational].

$A=\displaystyle[e, \pi] \setminus\{\bigcup_{n=1}^\infty I_n\}$ is closed.

Suppose, $N'_\delta(c)\cap A=\emptyset$ for some $\delta>0$, i.e $c $ must lie on the "edge" of two consecutive intervals. .

At least for two intervals $I_s$ and $I_t$, it happens. For this to occur, the intervals must lie on two "opposite sides" of $c$ since $c \in A$ and $c \notin \bigcup I_n$.

We must have $\displaystyle x_s-\frac{r_s}{2^{100s}}=x_t+\frac{r_t}{2^{100t}}$. But it is not possible in any case since $x_s-x_t$ is rational and $\displaystyle\frac{r_s}{2^{100s}}+\frac{r_t}{2^{100t}}$ is irrational.

Answer 1

Make it more abstract and the construction is easier.

All you need is a family of compact subsets (intervals if you like) $A_s$ where $s$ is running over all finite sequences of $0$'s and $1$'s and say $A_\varepsilon = [0,1]$, say (where $\varepsilon$ is the empty sequence) and such that

• $A_{s0} \cap A_{s1} = \emptyset$ and $A_{s0} \subseteq A_s$ and $A_{s1} \subseteq A_s$ for all sequences $s$.
• $\operatorname{diam}(A_s) \le \frac{1}{2^{|s|}}$ where $|s|$ is the length of the sequence $s$.
• $f(s) \notin A_{s}$ for some bijection $f$ from the finite $0$-$1$ sequences to $\Bbb Q$ (which exists as both sets are countable).

(The usual Cantor set construction has $A_{0}=[0,\frac13]$, $A_{00}=[0,\frac19]$, $A_{01}=[\frac{2}{9},\frac13]$ etc.) It's not hard to see how to fulfill the first 2 conditions recursively (take two disjoint closed subintervals of $A_s$) and the third condition takes almost no effort: just shrink a bit two avoid two additional points when splitting $A_s$ into $A_{s0}$ and $A_{s1}$ in the recursivion step.

Then if $x$ is an infinite sequence of $0$'s and $1$'s (so $s \in 2^\omega$) and $x\mid n$ is the finite sequence of the first $n$ of them, $$A:= \bigcup_{x \in 2^\omega} \bigcap_{n \in \omega} A_{s\mid n}$$ is homeomorphic to the Cantor set (and thus perfect) and the final condition ensures that no member of $\Bbb Q$ lies in $A$.

This borrows from standard techniques from descriptive set theory (see Kechris' book "Classical Descriptive Set Theory" for more explanations and in depth results in this spirit).

Answer 2

You can do this with decimal expansions. Let $E$ be the set of real numbers whose decimal expansions have the form

$$.1 Z_1 a_1 1 Z_2 a_2 1 Z_3 a_3 \dots.$$

Here $Z_n$ is a block of $n$ zeros and $a_n\in \{0,1\}.$

Each number in $E$ is irrational because its decimal expansion has infinitely many $1$'s and arbitrarily large blocks of $0$'s. And $E$ is closed because if a sequence $x_n$ in $E$ converges to $x,$ then the expansions of the $x_n$ converge in each slot to that of $x.$ This implies $x$ has an expansion of the required form, hence $x\in E.$

Let $x = .1 Z_1 a_1 1 Z_2 a_2 \dots \in E.$ For $n=1,2,\dots,$ define $x_n\in E$ to be the number whose expansion is the same as that for $x,$ except that $a_n$ is replaced by $1-a_n.$ Then the $x_n$ are distinct and converge to $x.$ Thus $E$ is perfect as desired.