Uncountable closed set of irrational numbers 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.

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.
