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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.

prosyanoy
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  • Tricky example. Let $C$ be the usual middle-thirds Cantor set. Then the translate $\pi+C$ is perfect, and (conjecturally) consists only of irrational numbers. – GEdgar Nov 29 '23 at 13:22

2 Answers2

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If you know that fact that the set of irrational numbers $\mathbb I$ is homeomorphic to the Baire space, then the example has an easy description. [Disclaimer: Having been a practicing research topologist for 40 years now, I only just learned this fact in the last few years, from various stackexchange posts].

The Baire space is $\mathbb N^{\mathbb N}$, using the discrete topology on $\mathbb N$ and the product topology on $\mathbb N^{\mathbb N}$. One can use continued fractions to construct a homeomorphism $\mathbb I \to \mathbb N^{\mathbb N}$. Restricting the base of the product $\mathbb N^{\mathbb N}$ to the subset $\{1,2\} \subset \mathbb N$, one obtains an embedding $\{1,2\}^{\mathbb N} \hookrightarrow \mathbb N^{\mathbb N}$. And from the standard construction of the Cantor middle thirds set $C$ one obtains a homeomorphism $C \to \{1,2\}^{\mathbb N}$.

So, putting this altogether, the image of the composed maps $$C \to \{1,2\}^{\mathbb N} \hookrightarrow \mathbb N^{\mathbb N} \to \mathbb I $$ gives a perfect subset which is compact and therefore closed. It is furthermore nowhere dense, because the embedding $\{1,2\}^{\mathbb N} \hookrightarrow \mathbb N^{\mathbb N}$ is nowhere dense: for any given element $f \in \{1,2\}^{\mathbb N}$, define a sequence $g_i \in \mathbb N^{\mathbb N}$ where $g_i(i)=3$ and $g_i(j)=f_i(j)$ for $j \ne i$. The sequence $g_i$ clearly converges to $f$, but it is contained in $\mathbb N^{\mathbb N} - \{1,2\}^{\mathbb N}$.

Lee Mosher
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    I don't like the notation $\mathbb{I}$ since its used for the interval $[0, 1]$. I saw notation $\mathbb{P}$ before though... – Jakobian Nov 29 '23 at 13:08
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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 $E$.

Jim Belk
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