Uncountable closed set of irrational numbers

Question

Could you construct an actual example of a uncountable set of irrational numbers that is closed (in the topological sense)?

I can find countable examples that are closed, like $\{ \sqrt{2} + \sqrt{2}/n \}_{n=1}^\infty \cup \{ \sqrt2 \}$ , but how does one construct an uncountable example?

At least one uncountable example must exist, since otherwise the rational numbers form a Berstein set and are non-measurable.

Answer 1

If enumerate the rationals $\mathbb{Q}=\{q_1,q_2,\ldots\}$ and define the intervals $I_n=(q_n-2^{-n},q_n+2^{-n})$, we have that the set $S=\bigcup_{n=1}^\infty I_n$ contains every rational number and $$m(S)\leq\sum_{n=1}^\infty m(I_n)=\sum_{n=1}^\infty 2^{-\left(n-1\right)}=2$$ Therefore, the complement of $S$ in $\mathbb{R}$ is of infinite measure (hence uncountable) and contains only irrationals.

Answer 2

List the rationals as $r_0,r_1,\dots,r_n,\dots$.

Around the $n$-th rational, make an open interval of width $\epsilon/2^n$. Take the union of these. The complement does the job.

Answer 3

The irrational numbers are homeomorphic to $\Bbb N^{\Bbb N}$. Let $C=\{0,1\}^{\Bbb N}$, viewed as a subset of $\Bbb N^{\Bbb N}$. By the Tikhonov product theorem $C$ is compact, and $\Bbb N^{\Bbb N}$ is Hausdorff, so $C$ is closed in $\Bbb N^{\Bbb N}$. (In fact $C$ is homeomorphic to the middle-thirds Cantor set.) Finally, $C$ is clearly uncountable. Now if $h:\Bbb R\setminus\Bbb Q\to\Bbb N^{\Bbb N}$ is any homeomorphism, $h^{-1}[C]$ is an uncountable closed subset of $\Bbb R\setminus\Bbb Q$, the irrationals. Being compact, it’s also closed in $\Bbb R$.

Answer 4

Here's one example that requires neither measure theory nor an enumeration of the rational numbers. The set of positive irrationals is in bijection with the set of infinite sequences of positive integers through their continued fraction representation (which is unique for irrationals). Take the subset for which all of the terms of the continued fraction representation are $1$ or $2$. The complement in the set of positive reals is open, since for every irrational that has some continued fraction coefficient${}>2$ one easily finds a neighborhood where the coefficients up to and including this one are unchanged, and in a sufficiently small neighborhood of every rational number $\alpha$, some continued fraction coefficient (beyond the finite representation of $\alpha$ itself) is necessarily very big.

I may add that if the set ${\Bbb N_{>0}}^\Bbb N$ of infinite sequences of positive integers is equipped with the product topology for the discrete topology of $\Bbb N_{>0}$ (so that subsets of sequences in which a finite initial subsequence is fixed form a basis of open sets), then the continued fraction representation is a homeomorphism of the positive irrationals (with their topology restricted from $\Bbb R$) to ${\Bbb N_{>0}}^\Bbb N$. Therefore this is just a concrete version of the answer by Brian M. Scott (but with positive irrationals only, and with a more elementary argument to show the set is closed).

Answer 5

I think with the Axiom of Choice one can construct such a set. For each irrational number $x$ in the interval $[0,1]$, associate a sequence of irrational numbers $\{x_n\}$ that converges to $x$. Do you think this is sufficiently constructive?

EDIT: @Tom Oldfield pointed out that terms from different sequences could converge to rational numbers, and this sounds rather difficult to prevent, so the set could well be not closed.

Answer 6

Here is an alternative construction that is a bit more general. It works for any comeager subset $X \subset \mathbb{R}$. Let $(U_i : i \in \mathbb{N})$ be a sequence of dense open sets such that $X \subset \bigcap_{i\in \mathbb{N}} U_i$. (If $X$ is the set of irrational numbers then we can take $U_i = \mathbb{R} \setminus \{q_i\}$ where $q_i$ is the $i^\text{th}$ rational number in some fixed enumeration.) We can define a family of non-degenerate closed intervals $I_s$, one for each finite binary sequence $s$, with the following properties:

• If the sequence $s$ extends the sequence $t$ then $I_s \subseteq I_t$

• If neither sequence $s$ or $t$ extends the other then $I_s \cap I_t = \emptyset$

• If the sequence $s$ has length $n$ then the interval $I_s$ has length $\le 2^{-n}$ and is contained in the set $U_n$

(It is easy to define $I_s$ by induction on the length of $s$. The exact choice of definition is not important.) We can use this "Cantor scheme" to define an embedding $f$ of the Cantor space $2^\mathbb{N}$ into $X$ by mapping an infinite binary sequence $x$ to the unique point $f(x)$ in the intersection of the decreasing sequence of closed intervals $(I_{x \restriction n} : n \in \mathbb{N})$.

Answer 7

Explicit example: translation of a Cantor-like set.

Consider the Cantor set $$C := \Big\{ \sum \limits_{n=1}^{+\infty} \frac{\varepsilon_n}{4^n}\ \mid\ (\varepsilon_n)_n \in \{0,1\}^{\mathbb{N}}\Big\}.$$

It is uncountable and closed. Consider now the number $$x := \sum \limits_{n=1}^{+\infty} \frac{2}{4^{n^2}}.$$

The closed set which will answer the question is $$K := x + C = \{x+c,\ c\in C\}$$

Indeed, let us take an element $c$ of $C$, and distinguish two cases:

• $c$ is rational, in which case $c+x$ is irrational since $x$ is irrational

• $c$ is irrational and can be written uniquely as $c = \sum \limits_{n=1}^{+\infty} \frac{\varepsilon_n}{4^n}$ with $\varepsilon_n \in \{0,1\}$ for all $n$. Then the base $4$ representation of $c+x$ is $\sum \limits_{n=1}^{+\infty} \frac{\varepsilon_n + 2\cdot 1_{\sqrt{n} \in \mathbb{N}}}{4^n}$. Thus the coefficients at non perfect-square-positions are $0$ or $1$, while the coefficients at perfect-square-positions are $2$ or $3$. Hence, the base $4$ representation cannot be periodic, so $c+x$ is not rational.