I would like to show that a dense set in $[0,1]$ is not equal to $[0,1]$ without measure theory

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.

• 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. – GEdgar Jun 9 '14 at 21:30
• @GEdgar exactly, by explicit I meant an element with certain properties or a construction such that one can deduce that it is not in $A$. – Thorben Jun 9 '14 at 21:52
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 $j\ge n\ge 1$ $$\left|\frac{n}{j}-\frac{1}{\sqrt{2}} \right|\ge \frac{1}{4j^2} > \frac{1}{4^{n+j}} \tag3$$