# Compact set of infinitely many irrational numbers [closed]

What is a set of infinitely many irrational numbers that is also compact?

• $\{1/n+\sqrt 2: n\in\Bbb N\}\cup\{\sqrt2\}$. You could also construct an uncountable Cantor set with no rationals. Commented Dec 12, 2016 at 17:37
• @DavidMitra I don't believe that set is compact - you need to include $\sqrt{2}$. (Consider the open set $U_n=({1\over n}-{1\over 2n^2}, {1\over n}+{1\over 2n^2}).$) Commented Dec 12, 2016 at 17:38
• @NoahSchweber Oh yes, thanks... Commented Dec 12, 2016 at 17:39
• To the OP: what have you tried, and where did you get stuck? Commented Dec 12, 2016 at 17:40

Here's a nice fact: suppose $$C$$ is a compact set with empty interior (like the Cantor set). Then there is a translate of $$C$$ - that is, a set of the form $$C+r=\{c+r: c\in C\}$$ - such that $$C$$ contains no rationals.

The proof of this is via the Baire category theorem. For $$q\in\mathbb{Q}$$, let $$A_q=\{r: q\not\in C+r\}.$$ Since $$C$$ is closed, $$A_q$$ is open, and since $$C$$ has empty interior $$A_q$$ is dense. Finally, since there are only countably many rationals, this means $$B=\bigcap_{q\in\mathbb{Q}} A_q$$ is dense by Baires category theorem, hence not empty. Let $$b\in B$$; then $$C+b$$ contains no rationals (do you see why?).

Of course, there are examples not involving the Baire category theorem, but this is a very useful fact to know.

• Two answers for the price of one! (I’m working through my backlog, so I saw the other question first.) I’m lazy: when I see this question my first thought is always that $2^\omega\subseteq\omega^\omega$. Commented Dec 12, 2016 at 21:58

Try $$K=\{\pi\}\cup\Big\{\frac{n}{n+1}\pi\,\big|\, n=1,2,\ldots\Big\}$$