Given an interval around each rational number: How to find a real number which is in none of these intervals? There are countable infinite many rational numbers, so it is possible to "count" them: There are many ways to do this; just pick one.
If an order is chosen, there is an $n$th rational number. Each of these numbers is now embedded into an interval $I_n$ which has the width $2^{-n}$. The rational number is exactly in the middle of this interval. Such an interval is created for each rational number.
Is it possible to name a real number which is in none of these intervals (given a ordering for the rational numbers)?
Thank you
 A: As mentioned in the comments by @lulu, the total length of the intervals is $$\sum_{n=0}^{\infty} {2^{-n}} = 2$$
So even if those intervals did not overlap at all, there is no way they can cover the whole real line, and there will be uncountable many real numbers to choose from that are not covered by those intervals.
It might seem counter-intuitive that you can have these intervals around the rationals and still not cover all the reals even though the rationals are dense in the reals, so let me write a bit more about that so that it becomes less weird.
The ordering you choose will for most rationals $p/q$ give an index that is at least $q$. So the interval around $p/q$ will generally be smaller than $2^{-q}$ in length.
(Note that this is not rigorous, and that you can construct orderings where this is not true for infinitely many rationals in the ordering, but let's go with it for now just to get a better feel for what is happening.)
Would you always expect to be able to approximate a real that closely by a rational? For example, $355/113$ is a very good approximation to $\pi$, as it is correct to about 6 decimals. That falls far short however of the $2^{-113}$ accuracy that would be required for it to fall within that intervals in this question. That would require 34 decimals of accuracy.
It would be asking a lot for all reals to have an approximation $p/q$ that is accurate to $2^{-q}$. There are in fact numbers that are hard to approximate. These are numbers where there is no approximation $p/q$ that is more than $c/q^2$ away for some constant $c$. So even putting intervals of length $c/q^2$ around rationals $p/q$ is not sufficient to cover all the reals.
