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For each $n \in \mathbb{N}$ a closed interval $[x_n, y_n]$ is given. Assume that $[x_m, y_m]\, \cap \, [x_n, y_n] \neq \emptyset$ for all $m,n \in \mathbb{N}$. Show that $\bigcap_{n=1}^{\infty} [x_n, y_n] \neq \emptyset$.

What I've done is, assumed for contradiction that $\bigcap_{n=1}^{\infty} [x_n, y_n] = \emptyset$ which implies the existence of some disjoint intervals, i.e: that there exists $m,n \in \mathbb{N}$ such that $[x_m, y_m] \, \cap \, [x_n, y_n] = \emptyset$ which contradicts our hypothesis.

Now, I think this is wrong because it (i) makes the problem far too easy and (ii) infinities are fishy, I'm not convinced that the infinite intersection of intervals being empty implies that there exists at least two disjoint intervals.

Question: Is my "proof" completely off the right train of thought? How do I fix it?

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    $\begingroup$ How can you immediately conclude that there must be a pair of disjoint intervals? $\endgroup$ – carmichael561 Dec 14 '16 at 2:05
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It is not obviously true that $\bigcap_{n=1}^{\infty} [x_n, y_n] = \emptyset$ implies there exist $m$ and $n$ such that $[x_m, y_m] \, \cap \, [x_n, y_n] = \emptyset$. Notice, for instance, that the intersection of the three sets $\{0,1\}$, $\{0,2\}$, and $\{1,2\}$ is empty, even though the intersection of any two of them is nonempty. Or notice that $\bigcap_{n=1}^\infty(0,1/n)$ is empty, even though these intervals not only have nonempty pairwise intersections but actually are nested.

Here's how I would suggest to approach it. First, show that the intersection of any finite collection of your intervals is nonempty (for this, you will need to use the fact that they are intervals, rather than just random sets). Second, show that this implies the infinite intersection is nonempty (for this you will need to use that the sets are not just intervals but closed and bounded intervals; for instance, you could use compactness).

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  • $\begingroup$ Right, thanks. So my "proof" works for showing that the intersection of a finite collection of my intervals is non-empty, right? I'm not sure how to show that this implies infinite intersection is non-empty. We haven't covered anything close to compactness yet. $\endgroup$ – Zain Patel Dec 14 '16 at 2:10
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    $\begingroup$ No, it doesn't: see the example of $\{0,1\}$, $\{0,2\}$, and $\{1,2\}$. To show the intersection of a finite collection is nonempty, I would suggest figuring out what the intersection actually looks like. For instance, if you have three intervals $[a,b]$, $[c,d]$, and $[e,f]$, can you give a simple description of the set $[a,b]\cap[c,d]\cap[e,f]$? (If you have trouble with this, try it with just two intervals first.) Once you have this description, you can use it to handle the infinite case as well. $\endgroup$ – Eric Wofsey Dec 14 '16 at 2:12
  • $\begingroup$ I must be being particularly thick today... anywho, assume wlog $x_m \leq x_n$ then since the intersection is non-empty we require $x_n \leq y_m$ and so we have $[x_m, y_m] \, \cap [x_n, y_n] = [x_n, y_m] $ if $y_m < y_n$ and $[x_n, y_n]$ otherwise (i.e: the first interval is contained in the second). So in general, we can write $[x_m, y_m] \, \cap \, [x_n, y_n] = [\max\{x_m, x_n\}, \min\{y_m, y_n\} $ as per mathbeing's comment. $\endgroup$ – Zain Patel Dec 14 '16 at 2:39
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Define $C_1:=[x_1,y_1]$ and, recursively, $$ C_n:=C_{n-1}\cap[x_n,y_n]. $$

This gives a sequence of non-empty compact sets satisfying $C_1\supseteq C_2\supseteq C_3\cdots$, so $$ \bigcap_{n=1}^{\infty}C_n=\bigcap_{n=1}^\infty[x_n,y_n]\neq\phi $$ by Cantor's Intersection Theorem.

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  • $\begingroup$ The proof of the Cantor's Intersection Theorem (which is done in three lines in the Wikipedia link I posted) is actually the argument you tried to start. $\endgroup$ – user378947 Dec 14 '16 at 2:20
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    $\begingroup$ It is not obvious that each $C_n$ is nonempty. $\endgroup$ – Eric Wofsey Dec 14 '16 at 2:22
  • $\begingroup$ @EricWofsey We are dealing with a finite collection of intervals with non-empty pairwise intersection, so why do you say that? We actually have $C_n=[\max_{1\leq j\leq n}x_j,\min_{1\leq j\leq n}y_j]$... $\endgroup$ – user378947 Dec 14 '16 at 2:34
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    $\begingroup$ I didn't say it's not true; just that it's not obvious (and appears to be something that OP doesn't know how to prove yet). In any case, it takes a little thought to explain why $\max_{1\leq j\leq n}x_j\leq \min_{1\leq j\leq n}y_j$. $\endgroup$ – Eric Wofsey Dec 14 '16 at 2:37
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    $\begingroup$ Thanks to the both of you, I've repped you both. I've managed to work it out after staring at a bunch of poorly drawn intervals with Eric's hint and your technique. Thanks again. $\endgroup$ – Zain Patel Dec 14 '16 at 3:01

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