# Constructing a non-empty perfect set of real numbers that does not contain rationals.

Duplicate: Perfect set without rationals

My approach: We consider the set $$[e, \pi]$$. I am trying to "cover" the rationals by enclosing each one of them by open intervals with irrational endpoints, then remove them.

Let $$\{x_1,x_2,x_3,...\}$$ be the enumeration of the rationals in $$[e, \pi]$$.

Let $$x_n$$ be any rational in the interval and let the minimum of the distances of $$x_n$$ from the endpoints be $$r_n$$ . We can enclose $$x_n$$ by $$I_n=(a_n,b_n)=\displaystyle(x_n-\frac{r_n}{2^{100n}},x_n+\frac{r_n}{2^{100n}})\subsetneq [e, \pi]$$.

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

Now, $$\sum_{n=1}^\infty |I_n| < 1.33639×10^{-30}$$ as well as $$A=\displaystyle[e, \pi] \setminus\{\bigcup_{n=1}^\infty I_n\}$$ is closed.

Now, how do I show that any point $$c \in A$$ is a limit point of $$A$$? The goal is to find out an irrational that lies arbitrarily close to $$c$$ and is not "trapped" by any $$I_n$$.

Is my approach even valid?

Let $$c$$ be an element in $$A \subset [e,\pi]$$, which is an irrational number. Assume for contradiction that $$c$$ is not a limit point of $$A \subset [e,\pi]$$. Then there is a small real number $$\delta > 0$$ such that $$(c-\delta,c+\delta) \subset [e,\pi]$$ and contains no point of $$A \subset [e,\pi]$$ other than $$c$$. By the construction of $$A$$, this means that there exists 2 connected subsets, $$(c-\delta,c)$$ and $$(c,c+\delta)$$, of rational numbers $$\mathbb{Q}$$, which is a contradiction to the fact that rational numbers $$\mathbb{Q}$$ is totally-disconnected.