I'm doing mathematical analysis and I'm stuck with the following question:
Give an example of a nowhere dense closed set of irrational numbers with no isolated points (that is, a perfect set of irrational numbers)
I read this: Is this a perfect set? so apparently we can find such a set through specific enumeration of rational numbers, taking their vicinities union and taking its complement, but I struggle to find such enumeration that would not create isolated points.
Best Answer
Let $E_0$ be the closed interval $[\pi,2\pi]$, and let $q_1,q_2,\ldots$ be an enumeration of $\mathbb{Q}\cap I$.
Choose an irrational $\epsilon>0$ so that $[q_1-\epsilon,q_1+\epsilon]\subseteq \mathrm{int}(E)$, and let $E_1$ be the complement of $(q_1-\epsilon,q_1+\epsilon)$ in $E_0$. Note that $E_1$ is a disjoint union of two closed intervals, each of which has irrational endpoints.
If $q_2\notin E_1$, let $E_2=E_1$. If $q_2\in E_1$, the since the intervals of $E_1$ have irrational endpoints, $q_2$ must lie in the interior of $E_1$. In particular, we can choose a new irrational $\epsilon>0$ so that $[q_2-\epsilon,q_2+\epsilon]\subseteq \mathrm{int}(E_1)$. Let $E_2$ be the complement of $(q_2-\epsilon,q_2+\epsilon)$ in $E_1$, and note that $E_2$ is a union of finitely many closed intervals with irrational endpoints.
Continuing in this fashion, we obtain a nested sequence of closed sets $E_0\supseteq E_1\supseteq E_2 \supseteq \cdots$ such that each $E_n$ is a finite union of closed intervals with irrational endpoints, and $E_n$ does not contain $q_n$. The intersection $E=\bigcap_{n=0}^\infty E_n$ is the desired set. It is nonempty by compactness, nowhere dense since its complement contains an open neighborhood of the rationals, and perfect since each interval from each $E_n$ contains infinitely many points of $E$.