Prove that the following set is uncountable Good evening to everybody. Today I was trying to find a solution to the following exercise:
Let $$  A_\epsilon= \bigcup_{n=1}^{\infty} ( q_n-\frac{\epsilon}{2^n} , q_n + \frac{\epsilon}{2^n} ) $$ where $ q_n$ are the rational numbers of $ [0,1]$ . Let $$A=\bigcap_{j=1}^{\infty} A_{1/j} $$ The question was to prove that : i)
$   \lambda(A_{\epsilon}) \leq 2\epsilon $
ii)  For all $\epsilon < \frac{1}{2}$ , it holds that $[0,1]\backslash A_\epsilon $ is non-empty and $A $ is a subset of $ [0,1]$
iii) it holds $ \lambda(A)=0$
iv) $ \mathbb{Q}\cap[0,1] \subset A $ and that the set $ A $ is uncountable.
Ok, I have already proved quite easily the first 3 parts and the first relation of part iv) but I am stuck on the proof of the  uncountability of this set. From what I have already thought, we can identify the rational number $q_1$ with the sequence $ 1, 1 , ... $ (meaning that $q_1$ belongs to the first set of the union in $A_1$ , to the first set of the union in $A_\frac{1}{2}$ etc..) and the rational number $q_2$ with the sequence $ 2, 2 , ... $ (meaning that $q_2$ belongs to the second set of the union in $A_1$ , to the second set of the union in $A_\frac{1}{2}$ etc..),so we can exclude all the rationals in the set $ A $ and, if we show that the set $ A $ contains also some irrational , let say it $ X $ , then we can pick the first integer $N_1 $ to be the natural number ( here $N_1\geq 0 $ )  such that $ X $ belongs to the first set of the union of $A_1 , A_\frac{1}{2} , A_\frac{1}{3},..., A_\frac{1}{N_1-1}$ but NOT in $A_\frac{1}{N_1}$ , then the integer $N_2$ to be the natural number (here we need also $N_2\geq0$) such that $ X $ belongs to the second set of the union of $A_1 , A_\frac{1}{2} , A_\frac{1}{3},..., A_\frac{1}{N_2-1}$ but NOT in $A_\frac{1}{N_2}$  , etc..., and thus identify each non rational number of $ A $ by the sequence $ N_1 , N_2 , ...$. Then assuming that $A$ is countable , say $ \phi_1 , \phi_2 , ...$ we can use the diagonal argument and take the element $ ( \phi_1(1)+1, \phi_2(2)+1 , \phi_3(3)+1 , ... ) $ . . This element is in $ A $ but it is not any of the sequences $ \phi_1 , \phi_2 , ... $ So I only need to prove that $ A $ does not contain ONLY the rationals $q_1 , q_2 ... $ Any ideas would be really helpful.
 A: You almost got part iv), especially when you mentioned "diagonal argument".
The crux of diagonal argument to prove a set is uncountable is assuming the set is countable and then picking a diagonal that

*

*avoids all elements in the set one by one and

*identifies a particular element in the set,

which is a contradiction.
In order to satisfy requirement 1, for $i$-th diagonal element, we can select an interval in some $A_{1/j}$ that does not include $u_i$, where we assume elements in $A$  are listed as $u_1,u_2,\cdots$. Here we do not care whether $u_i$ is a rational number or not (which is a concern that might have distracted you).
In order to satisfy requirement $2$, each interval that we will select should be contained in the previous interval we have selected.
I hope the hints above are enough for you to make progress. Once you got a proof, you may add it to this answer.
A: Let $A_{n,j} = ( q_n-\frac{1}{2^nj} , q_n + \frac{1}{2^nj} )$. So
$  A_{1/j}= \bigcup_{n=1}^{\infty} ( q_n-\frac{1}{2^nj} , q_n + \frac{1}{2^nj} )=\bigcup_{n=1}^{\infty}A_{n,j}$.
Note that if $k>j$, $A_{1/k} \subseteq A_{1/j}$.
Let $A=\bigcap_{j=1}^{\infty} A_{1/j}$
Note that $\overline{A_{n,j}}= [ q_n-\frac{1}{2^nj} , q_n + \frac{1}{2^nj} ]$.
Suppose that $A$ is countable. So $A= \{a_1, a_2,...\}$. Now:

*

*for $i=1$, choose $n_1$ such that $\overline{A_{n_1,3}}\subset (0,1)$ and $|q_{n_1}-a_1|> \frac{1}{3}$ (this is always possible). It follows immediately that $a_1 \notin \overline{A_{n_1,3}}  $.

*for each $i \geqslant 1$, choose $n_{i+1}$ such that $n_{i+1}> n_i$, $\overline{A_{n_{i+1},3^{i+1}}}\subset A_{n_i,3^i}$ and $|q_{n_{i+1}}-a_{i+1}|> \frac{1}{3^{i+1}}$ (this is always possible). It follows immediately that $a_{i+1} \notin \overline{A_{n_{i+1},3^{i+1}}}  $.

It follows immediately that $\left \{\overline{A_{n_i,3^i}} \right\}_i $ is decreasing sequence of compact set such that any finite intersection is non-empty.
So, $\bigcap_{i=1}^\infty \overline{A_{n_i,3^i}}\neq \emptyset$. So there is $b$ such that
$$ b \in \bigcap_{i=1}^\infty \overline{A_{n_i,3^i}} = \bigcap_{i=2}^\infty \overline{A_{n_i,3^i}}= \bigcap_{i=1}^\infty \overline{A_{n_{i+1},3^{i+1}}} \subseteq \bigcap_{i=1}^\infty A_{n_i,3^i} \subseteq \bigcap_{i=1}^\infty A_{1/3^i}=\bigcap_{j=1}^{\infty} A_{1/j} = A$$
But, for all $i$, $b \neq a_i$. Contradiction. So, $A$ is not countable.
