How can I prove that this set is finite? Let $\{I_k\}_{k=1}^n$ be a finite collection of open intervals such that $\mathbb{Q} \cap [0,1] \subset \bigcup_{k=1}^n I_k$. How can I prove that the set $A=\{x\in \mathbb{I} \cap [0,1] : x \notin \bigcup_{k=1}^n I_k \}$ is finite?
($\mathbb{I}$ is the set of irrational numbers).
 A: The key idea is that there are only finitely many of these open intervals $I_k$.
Since they are intervals, we can write $I_k = (a_k, b_k)$ with $a_k < a_{k+1}$ and $b_k < b_{k+1}$ w.l.o.g. and also, w.l.o.g. each $b_k \in [0,1]$ for $k\ne n$ and $a_k \in [0,1]$ for $k\ne 1$
Now since the intervals cover all of $\mathbb{Q} \cap [0,1]$ we know we must have $b_k \ge a_{k+1}$ for each $k$. (what would happen if it was not?) But since there are only finitely many intervals this condition immediately implies that there can only be finitely many points in $A$ (can you see why?) 
Notice that finitely many is crucial here, as we can fix $\epsilon$ arbitrarily small, enumerate the rationals $q_1, q_2, \cdots$ and take intervals $I_k = (q_k - \frac{\epsilon}{2^k}, q_k + \frac{\epsilon}{2^k})$ and the total measure of the union is at most $\epsilon$, and hence there are tons of irrationals in $A$. 
A: Since $\bigcup_{k=1}^{n}I_{k}$ contains all rational points of $[0,1]$, then $[0,1]\setminus \bigcup_{k=1}^{n}I_{k}$ is a set consisting of the irrational endpoints of those $I_{k}$ that are not covered by any other $I_{m}$. Finiteness comes from the fact that there are only finitely many $I_{k}$, and thus only finitely many of their endpoints. 
