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Let the square $S$ be the set of points $(x,y) \in [0,1]^2$
Let $R \subset S = S \cap \mathbb{Q}^2$, that is, the "rational pairs" in the square.

To each of these points $r_i \in $ R, we can associate a small square $s_i$ of area $\epsilon / 2^i$, centered at $r_i$. Now the collection $\{s_i\}$ must cover $S$ because if any region of $S$ is uncovered, then that region contains a rational pair that is uncovered which is a contradiction.

So since we covered $S$ with a buch of small squares $s_i$, then $\text{area}(S) \le \sum \text{area}(s_i) = \epsilon$
Since $\epsilon$ was arbitrary, the area of a square is $0$!

So what went wrong here?

leo
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Mark
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    You proved that the set of points in a square with rational coordinates has measure zero. This is a different, true, statement. – Potato Mar 08 '13 at 03:46
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    There are no "regions" left uncovered, but what about points? What prevents an uncountable set of points of nontrivial measure from escaping the little squares? – Shaun Ault Mar 08 '13 at 03:48
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    So it seems that there must be points left uncovered by the small squares whose area is greater than 1 - $\epsilon$. – Mark Mar 08 '13 at 03:52
  • I gave an answer to another quesetion that might help you think through things. –  Mar 08 '13 at 16:53

2 Answers2

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The same argument can be made in one dimension. It is true that the set of points left includes no rectangular region (in 2D) or interval (in 1D) but it is still 'most' (assuming $\epsilon \ll 1$) of the square. The logic is fine. The error is thinking that the complement containing no rectangle means the complement has area zero.

Ross Millikan
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$\mathbb R^{2}$ is a separable metric space, since it contains a countable dense subset, $\mathbb Q^{2}$. This is exactly the phenomena you're seeing in this example. Since $\mathbb Q^{2}$ is countable, it has measure zero, and thus the complement, $\mathbb R^{2} - \mathbb Q^{2}$ has full measure, but contains no rectangles, since it doesn't even contain any open sets, since its complement, $\mathbb Q^{2}$ is dense. Now if you restrict everything to $[0,1]^{2}$ you get exactly what you've already shown.

Dylan Yott
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