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!
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$ $$ \left|\frac{n}{j}-\frac{1}{\sqrt{2}} \right|\ge \frac{1}{4j^2} > \frac{1}{4^{n+j}} \tag3$$