Open set containing rationals but complement non-denumerable

Question

I am taking Real Analysis classes and I got a homework that asks me:

Give an example of an open set $\mathcal{A}$ such that $\mathcal{A}\supset\mathbb{Q}$ but $\mathbb{R}-\mathcal{A}$ is non-denumerable.

My attempt: First let $\mathcal{A} = \bigcup(r_n-1/2^n,r_n+1/2^n)$ where $r_n$ is the $n$-th rational, this is a union of open sets so $\mathbb{R}-\mathcal{A}$ is closed. I have reasons to believe that such set is also non-denumerable (as seen here: Uncountable closed set of irrational numbers but I have no experience in measure theory, is there other way to prove it's non-denumerability? Is that an answer at all?

Please excuse my bad english, thank you.

score 9 · Answer 1 · answered Dec 25 '15 at 05:51

9

The sum of the measures of these sets is $\sum\limits_{n=1}^\infty \dfrac 1 {2^{n-1}} = 2$. Consequently the measure of its complement is $\infty$. That the measure of the complement is positive is enough to entail that the complement is uncountable.

answered Dec 25 '15 at 05:51

Michael Hardy

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I understood OP to mean that they'd like an answer that doesn't involve measure theory at all. – Travis Willse Dec 25 '15 at 05:53
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Absolutely right, but OP asked for a non-measure-theoretic proof, the which I for one can’t think of. – Lubin Dec 25 '15 at 05:54
Thank you Michael Hardy,your answer is quite helpful and i appreciate your support but in this specific question i'll accept Noah's answer becuase i'm not allowed to use measure theory. – Alan246 Dec 25 '15 at 10:02

score 6 · Accepted Answer · edited Dec 25 '15 at 08:15

For a non measure-theoretic proof you can construct a Cantor set containing no rationals and then take its complement. Here's a sketch of how to make it:

Start with an interval $[a,b]$ with $a,b$ irrational. Let $\{q_n\}$ be an enumeration of the rationals in this interval. Now remove from $[a,b]$ an interval $(c,d)$ with irrational endpoints containing $q_1$. Now remove open intervals with irrational endpoints from the two remaining closed intervals so that $q_2$ is no longer in the set, and so on.

The result is a closed set avoiding every rational number which has cardinality equal to that of the continuum (equivalently, to the set of all infinite binary sequences). The proof that Cantor sets have cardinality continuum is very similar to the proof that the standard Cantor ternary set consists of all numbers with only 0 and 2 in their ternary expansion (Or since its a perfect set the Baire category theorem shows that it's uncountable).

If the Baire Category Theorem is available, one can prove that there is a translation of the Cantor set which contains no rationals. This takes away the need for the "new" construction. — Aloizio Macedo, Dec 25 '15 at 06:28

Open set containing rationals but complement non-denumerable

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