TITLE: I would like to show that a dense set in $[0,1]$ is not equal to $[0,1]$ without measure theory
QUESTION [6 upvotes]: 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!

REPLY [3 votes]: 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$
$$
 \left|\frac{n}{j}-\frac{1}{\sqrt{2}} \right|\ge \frac{1}{4j^2} > \frac{1}{4^{n+j}}
\tag3$$