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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?

justanotheruser
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1 Answers1

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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.