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.
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
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.