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We know that $[0,1]\cap \mathbb{Q}$ is a dense subset of $[0,1]$ and has measure zero, but what about $([0,1]\cap \mathbb{Q})\times([0,1]\cap \mathbb{Q})$? Is it also a dense subset of $[0,1]\times[0,1]$ and has measure zero too?

Besides, what about its complement? Is it dense in $[0,1]\times[0,1]$ and has measure zero?

  • 1
    Futhermore, the measure of $([0,1] \cap \mathbb Q) \times [0,1]$ is zero.2012-08-22

3 Answers 3

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I'll post a slick proof here.

Let $d(r)$ denote the denominator of the rational number $r$. In other words, if $r=p/q$ holds for positive integer $q$ and integer $p$, where $p,q$ are co-prime, we have $d(r)=q$. Then $d(r)\ge1$ for all $r\in\Bbb Q$.

Suppose that $D(m)=\left\{\,x\,\big|\,0\le x\le1\land d(x)=m\,\right\}$, we have $D(1)=\{0,1\}$, and $0<|D(m)| for $m>1$, therefore $\sum_{x\in D(m)}\frac1{d(x)^4}=\sum_{x\in D(m)}m^{-4} Thus $\sum_{x\in [0\mathinner{..}1]}\frac1{d(x)^4}=2+\sum_{m>1}\sum_{x\in D(m)}\frac1{d(x)^4}<2+\sum_{m>1}m^{-3}=C$

For all $0\le x,y\le1$ and $x,y\in\Bbb Q$, we draw a circle whose center is $(x,y)$ and radius is $(d(x))^{-2}(d(y))^{-2}\epsilon$. The total area is \begin{align} \sum_{\substack{0\le x,y\le1\\x,y\in\Bbb Q}}\frac{\pi\epsilon^2}{d(x)^4d(y)^4} &=\pi\epsilon^2\sum_{\substack{0\le x,y\le1\\x,y\in\Bbb Q}}\frac1{d(x)^4d(y)^4}\\ &=\pi\epsilon^2\sum_{\substack{0\le x\le1\\x\in\Bbb Q}}\frac1{d(x)^2}\sum_{\substack{0\le y\le1\\y\in\Bbb Q}}\frac1{d(y)^2}\\ &\le\pi\epsilon^2C^2 \end{align} Therefore the measure is not greater than $\pi\epsilon^2C^2$. Let $\epsilon\to0$, we get the answer.

Note The summation-interchanging works well because the terms are all nonnegative.

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Any countable set is of zero Lebesgue measure, and the product of (finitely many) countable sets is countable. Also, a product of finitely many dense sets is dense. Thus, the answer is yes.

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To give a somewhat comprehensive answer:

  • the set in question is countable (as a product of countable sets), so it is of measure zero (because any countable set is zero with respect to any continuous measure, such as Lebesgue measure).
  • it is also dense, because it is a product of dense sets.
  • it has measure zero, so its complement has full measure.
  • its complement has full measure with respect to Lebesgue measure, so it's dense in $[0,1]^2$
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    @user38448: Notice that every open set in $[0,1]^2$ has positive measure, so any set of full measure must be dense.2012-08-23