Consider the following set $$ I_{n,j}=[\frac{n}{j}-\frac{1}{4^{n+j}},\frac{n}{j}+\frac{1}{4^{n+j}}]$$ for some integers $n$ and $j$. Now let $$A:=\bigcup_{n\geq 1}\bigcup_{j\geq1}I_{n,j}$$

The goal of some exercise I was working on was to show that $A$ is dense in $[0,1]$ and in next step to show that $[0,1]\setminus A\neq \emptyset$.

The first part follows directly as rationals are dense in $\mathbb R$ and I managed to show the second part by using that the Lebesgue-measure of $A$ is strictly smaller then 1.

And so the question appeared if it is possible to show $[0,1]\setminus A\neq \emptyset$ without measure theory and in particular if it is possible to find an explicit element in $[0,1]$ which is not in $A$.

I would appreciate any help.

Thanks in advance!

  • 1
    $\begingroup$ You mean you want to show something like $(\sqrt{5} - 1)/2$ is not in your set? Base it on the continued fraction for that number. $\endgroup$ – GEdgar Jun 9 '14 at 21:30
  • $\begingroup$ @GEdgar exactly, by explicit I meant an element with certain properties or a construction such that one can deduce that it is not in $A$. $\endgroup$ – Thorben Jun 9 '14 at 21:52
  • $\begingroup$ Well, the golden section is known to have approximation by rationals as bad as possible. That's why I suggested using it. $\endgroup$ – GEdgar Jun 10 '14 at 1:30
  • $\begingroup$ @GEdgar Would you like to show in an answer how this works in the upper context? It's not urgent, just out of interest so just in case you can find time... $\endgroup$ – Thorben Jun 10 '14 at 19:47

For example, $1/\sqrt{2} \notin A$. Indeed, for every rational $n/j$ we have $$ \left|\frac{n}{j}-\frac{1}{\sqrt{2}} \right| \, \left|\frac{n}{j}+\frac{1}{\sqrt{2}} \right| = \frac{|2n^2-j^2|}{2j^2}\ge \frac{1}{2j^2} \tag1$$ There is no point to consider $n>j$. So, $n/j\le 1$, hence $\left|\frac{n}{j}+\frac{1}{\sqrt{2}} \right|\le 2$. It follows that $$ \left|\frac{n}{j}-\frac{1}{\sqrt{2}} \right|\ge \frac{1}{4j^2} \tag2$$ It's easy to prove (by induction, say) that $j^2<4^j$ for all $j\ge 1$. Hence, for $ j\ge n\ge 1$ $$ \left|\frac{n}{j}-\frac{1}{\sqrt{2}} \right|\ge \frac{1}{4j^2} > \frac{1}{4^{n+j}} \tag3$$


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