# Give an example of set with irrational numbers which has measure … [closed]

Task: Give an example of closed set $$A \subset [0,1]$$, which only consists of irrational numbers and has measure not less than $$0.9$$ .

I just have started taking course of Lebesgue measure. I found this task in my book, and can't get it out of my mind. I would really appreciate if you could help me solve this. Thanks in advance.

• If I understand your question correctly you can simply take the set of irrationals $I\mathbb{R}$ and then $A=I\mathbb{R}\cap[0,1]$ has measure $1>0.9$, because countable subsets like $\mathbb{Q}$ have measure $0$. – freakish May 12 at 20:51
• Of possible historical interest is that in 1884 (a time in which Cantor sets had only appeared in 3 or 4 papers), Ludwig Scheeffer (1859-1885) proved on pp. 291-293 of Zur Theorie der Stetigen Funktionen einer Reellen Veränderlichen that for each (generalized) Cantor set $C,$ there exists a dense set (hence, an infinite set) of real numbers $r$ such that the $r$-translate of $C$ contains no rational numbers. (continued) – Dave L. Renfro May 12 at 21:51
• For more such results, see this 11 May 2000 sci.math post. – Dave L. Renfro May 12 at 21:52
• @freakish: your set $A$ is not closed. – TonyK May 13 at 9:58

## 2 Answers

Pick your favourite enumeration $$(q_n)$$ of the rationals in $$[0,1]$$, and define the set $$S=\cup (q_n-\epsilon^n,q_n+\epsilon^n)$$ Then $$S$$ is open, so the complement of $$S$$ in $$[0,1]$$ is closed and contains only irrational numbers; and you can make the measure of $$S$$ as small as you like by choosing small enough $$\epsilon$$.

• This is not an example. – ajotatxe May 12 at 20:17
• Sure it is. ${}{}$ – copper.hat May 12 at 20:17
• TonyK has proved the existence of such a set, but he hasn't defined any set. I insist: this is not an example. – ajotatxe May 12 at 20:18
• That does not mean that the set is not defined. – copper.hat May 12 at 20:23
• @ajotatxe: I'm suggesting that I don't know if $1/e$ or $1/\pi$ are in that set or not --- No one presently knows whether either of these numbers belongs to the set $E$ consisting of all real numbers whose decimal expansions contain at most finitely many $2$'s, and $E$ is about as explicitly definable (from decimal representations of reals) as possible without being trivial. – Dave L. Renfro May 12 at 20:47

We are going to start with $$[0,1]$$ and then remove an open ball around each rational point in the range $$[0,1]$$ so that the the remaining set is closed and has no rational points.

If the sum of all of the diameters is less than $$0.1$$ then our set will have measure at least $$0.9$$.

We know that the set of rational numbers in that interval is countable, so let the rational numbers be $$q_1,q_2,\dots$$

We just need to select diameters $$d_i$$ such that $$\sum\limits_{i=1}^\infty d_i < 0.1$$

We can take $$d_i = \frac{1}{10\cdot 2^i}$$

• As the sum is not finite I would suggest writing $\infty$ at the top of the sum or take the limit of the sum. Besides that the solution seems fine – LegNaiB May 12 at 20:26
• Oh yes, thank you ! – sorryifslow May 12 at 20:27