On Finding an irrational number excluded from a countable union of intervals. I recently was quite amazed in class by the fact that taking $A = [0,1] \cap \mathbb{Q} $ we can find a cover of $A$ that is missing some irrationals on the interval $[0,1]$.
The way to do this is to call $A = \{ a_k | k \in \mathbb{N} \}$ (we can do this because the rationals are a countable set), take $\epsilon > 0$ and take $I_k =[ a_k - \epsilon/ 2^k, a_k + \epsilon/ 2^k ]$ for every $k \in \mathbb{N}$.
Clearly $A \subset \bigcup_{k \in N} I_k$ but the sum of the lengths of all $I_k$ is 
$$\sum_{k = 0}^{\infty} \frac{2 \epsilon}{2^k} = 2 \epsilon \frac{1}{1 - 1/2} =  4 \epsilon$$
and so for $\epsilon < 1/4$ there must be some irrational points in $[0,1]$ left uncovered.
Now I was trying to see if I can find a sequence of the $a_k$ s.t. I could explicitly prove that a given irrational is excluded from the covering.
As an example I was interested in excluding $\pi / 4$ from the covering, my reasoning to show this goes as follows:
I know that there exists a subsequence of the $a_{k_i}$  s.t. $a_{k_i} \rightarrow \pi / 4 $ for $i$ going to infinity.
Now I choose the ordering of the  $a_k$ s.t. the $a_{k_i}$ subsequence are the last terms. Now $\forall \epsilon_1$ I can find an $k_\epsilon$ s.t. $|a_{k_i} - \pi / 4 | < \epsilon_1 \quad \forall{k > k_\epsilon}$. Now it would remain to show that  for $k_\epsilon + 1$ the length of $I_{{k_i}}$ is less than $\epsilon_1$ to show that  $\pi / 4$ is excluded from the covering. Is my reasoning correct? how could I prove this last step?
 A: You cannot have the $a_{k_i}$ as "the last terms". If so, the sequence $(a_k)$ would converge to $\pi/4$ so that $(a_k)$ has no chance to cover all rationals.
A: It would be simpler just to truncate $I_k$ if $\pi/4 \in I_k$. I.e., if $\pi/4 \in I_k$ and $\pi/4 > a_k$, replace $I_k$ by $I_k \cap (0, \pi/4)$; if $\pi/4 \in I_k$ and $\pi/4 < a_k$, replace $I_k$ by $I_k \cap (\pi/4, 1)$.
[Your approach won't work, because you can't enumerate $[0, 1] \cap \Bbb{Q}$ by a convergent sequence - a convergent sequence lies inside a small neighbourhood of its limit for all but finitely many terms.]
A: You just need to assure that each rational in order is far enough from $\frac \pi 4$.  Start with some order of the rationals in $[0,1]$.  Now make a new one by taking the earliest term in the old order that 1)hasn't been used yet and 2)isn't too close to $\frac\pi 4$.  This gives an explicit construction based on $\epsilon$ and the initial sequence.  
Added:  If you sequence starts off $\frac 34, \frac 78, \frac {22}7, \frac 12, \frac 14$ and $\epsilon=\frac 15$ then $\frac 34$ is too close, so the first element of the new sequence is $\frac 78$.  $\frac 34$ is too close for the second term as well because $\epsilon/2^2=\frac 1{20} \gt \frac {\pi-3}4 \approx \frac 1{28}$, so the second element is $\frac 12$.  Then $\frac 34$ is OK and we use it.  $\frac {22}7$ has to be held for a while, so we use later entries, but eventually we will get around to it.
