If $[a_n, b_n] \cap [a_m, b_m] \neq \emptyset$ then $\bigcap_{1}^{\infty} [a_n,b_n] \neq \emptyset$ Let $[a_n,b_n]$, $n=1,2,3,\ldots$, be closed intervals with $[a_n,b_n] \bigcap [a_m,b_m] \neq \emptyset$ for all $n$, $m$. Prove $\bigcap_{1}^{\infty} [a_n,b_n] \neq \emptyset$.
I can show by induction that $\bigcap_{1}^N [a_n,b_n] \neq \emptyset$. But I am not sure about the infinity bit, maybe, I am missing something obvius.
Any hint guys?
Thanks
 A: The intersection $\bigcap_{n=1}^N [a_n,b_n]$ is itself a closed interval $[c_N,d_N]$.  If you can prove that, you can probably show that $c_1\le c_2 \le c_3 \le \cdots$ and $d_1 \ge d_2 \ge d_3 \ge \cdots$.  Then think about $\sup\{c_1,c_2,c_3,\ldots\}$ and $\inf\{d_1,d_2,d_3,\ldots\}$.
A: Define $A_n = \max_{i=1}^n a_i, B_n = \min_{i=1}^n b_i$. Show that
$$
\bigcap_{n=1}^{\infty} [a_n, b_n] = \bigcap_{n=1}^{\infty} [A_n,B_n].
$$
Since $[A_n,B_n] \subseteq [a_i,b_i]$ for $i=1, \dots, n$, one side of the inclusion is trivial. For the other side, suppose $x$ is on the LHS. Then it is in every of the $[a_n,b_n]$'s, and thus $a_n \le x$ for every $n$. But that means $A_n \le x$ for every $n$, and a similar argument shows that $x \le B_n$ for every $n$, thus giving you the reverse inclusion.
The $[A_n,B_n]$'s have a nice property : they are nested intervals, i.e. $[A_n,B_n] \supseteq [A_{n+1},B_{n+1}]$. This argument up there means that without loss of generality, we can suppose that the intervals are nested (well, we almost can, but I'll deal with it).
Now by construction $A_n$ is an increasing sequence, i.e. $A_n \le A_{n+1}$. Also, $B_{n+1} \le B_n$ by construction. We also have $A_n \le B_n$. To see this, suppose $A_n > B_n$. This means there exists $i,j$ such that $a_i > b_j$, but then $[a_i, b_i] \cap [a_j, b_j] = \varnothing$, which is excluded. Therefore we have $A_n \le A_{n+1} \le B_{n+1} \le B_n$. Therefore $A_n$ is increasing and bounded above, and $B_n$ is decreasing and bounded below, so both sequences $A_n$ and $B_n$ are convergent, call their limits $\alpha$ and $\beta$ respectively. Clearly we cannot have $\alpha > \beta$, because then there would exists $n$ such that $B_n < A_n$, which is excluded. This means $\alpha \le \beta$. Therefore, 
$$
\bigcap_{n=1}^{\infty} [A_n, B_n] = [\alpha, \beta] \neq \varnothing. 
$$
(This is because $x \ge \alpha$ if and only if $x \ge A_n$ for every $n$, and similarly for $B_n$.)
Hope that helps!
A: You have to have $a_n \leq b_m$ for all $m$ and $n$, otherwise $[a_n,b_n]$ and $[a_m,b_m]$ are disjoint. So $\sup_n a_n \leq \inf_m b_m$. Letting $S = \sup_n a_n$ and $I = \inf_m b_m$, any $x$ with $S \leq x \leq I$ will be in $\cap_n [a_n,b_n]$. 
