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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?


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Do you know the Cantor intersection theorem? This follows from that theorem. – Srivatsan Nov 27 '11 at 0:54
Nop, never seen that before. – Rashmirathi Nov 27 '11 at 1:01
Show that $\bigcap_1^N[a_n,b_n]$ is not only nonempty but in fact a closed interval $[a'_N,b'_N]$. Consider the supremum of the $a'_N$s. – Henning Makholm Nov 27 '11 at 1:22
up vote 4 down vote accepted

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 $\inf\{d_1,d_2,d_3,\ldots\}$.

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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!

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The maximum and the minimum need not exist. – Did Nov 27 '11 at 7:36
Uh, you definitively need to get some sleep and read that proof once again. And then take off your downvote. Even if the maximum is worth $-\infty$ and the minimum $+\infty$, the proof goes the same and everything I said is true. Those max and mins do not need to be real numbers. You should wait a little before downvoting just like that... – Patrick Da Silva Nov 27 '11 at 18:22
You are right that $A_n$ and $B_n$ do exist. My mistake, sorry about that. Due to the exceptionally nasty tone of your last comment, I cannot signal that you mixed up persistently increasing with decreasing and $\le$ with $\ge$, nor the confusion between increasing and nondecreasing and between decreasing and nonincreasing, can I? Nice proof, anyway (and, since this seems to obsess you, I mention that I did not downvote it). – Did Nov 27 '11 at 19:00
Ahhhh you are defining the maximum for a finite number of $a_n's$ at each step, facepalm* – user38268 Nov 28 '11 at 2:43
Dear Patrick, please do try not to react childishly like this: even if one does not like a comment —for whatever reason— it is much better to keep calm. – Mariano Suárez-Alvarez Nov 28 '11 at 6:28

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 $\cap_n [a_n,b_n]$.

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This is a homework problem. – user38268 Nov 27 '11 at 5:55
It was already answered when I wrote this. – Zarrax Nov 27 '11 at 7:03
Thanks for the clarification. I accept your reason. – user38268 Nov 27 '11 at 8:46

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