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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?

Thanks

Rashmirathi
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3 Answers3

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

Michael Hardy
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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!

Patrick Da Silva
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    The maximum and the minimum need not exist. – Did Nov 27 '11 at 07:36
  • 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
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    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
  • Sorry for sounding nasty, but the shortness of your comment also felt aggressive, so I just childishly reacted. No offense taken. But it's true that I inverted some of the inequalities in there... I'll just correct that, the idea is still there. Thanks Didier – Patrick Da Silva Nov 28 '11 at 00:24
  • @Didier : There you go, corrected. Since you're saying you weren't the downvoter, I hope downvoter comes back and explains himself now, perhaps he was wondering about the same thing as you did. – Patrick Da Silva Nov 28 '11 at 00:26
  • @PatrickDaSilva I have read your answer and I think that it is an excellent answer. My answer too got downvoted (though the downvoter has now retracted his downvote I think); I hope it is not one person who just downvoted several answers at once... –  Nov 28 '11 at 00:37
  • @Benjamin : Thanks. I upvoted your answer too ; had it been the first one it would've been a good hint-type of answer. – Patrick Da Silva Nov 28 '11 at 00:55
  • @PatrickDaSilva I don't understand one thing, can we always take the maximum of all the $a_i's$? Don't we require that there only be a finite number of them? E.g. to prove that the finite intersection of open sets is open, it is essential that we use finiteness to take the minimum of the radii of a finite number of open balls. –  Nov 28 '11 at 01:38
  • I am using finiteness to define my maximum... don't you see that my $A_n$'s are defined as the max of the $n$ first $a_i$'s? It might happen though that the $n$ first $a_i$'s are all $-\infty$, but that doesn't cause any trouble and you can read the proof as if $A_n = -\infty$ was a possibility. – Patrick Da Silva Nov 28 '11 at 02:17
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    Ahhhh you are defining the maximum for a finite number of $a_n's$ at each step, facepalm* –  Nov 28 '11 at 02:43
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    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-Álvarez Nov 28 '11 at 06:28
  • @Mariano : Indeed it is. Thanks for noticing, I appreciate it. – Patrick Da Silva Nov 28 '11 at 06:59
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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 $\cap_n [a_n,b_n]$.

Zarrax
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