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I don't know how to find the intersection in these two problems.

The first problem asks for an example of a decreasing nest of open finite intervals

($a_1$, $b_1$) $\supseteq$ ($a_2$, $b_a$) $\supseteq$...

such that $\bigcap_{k=1}^{\infty}$($a_k$, $b_k$)=$\emptyset$.

The second problems asks for an example of a decreasing nest of open intervals

($a_1$, $b_1$) $\supseteq$ ($a_2$, $b_a$) $\supseteq$...

such that $b_k$-$a_k$$\rightarrow$$0$ yet $\bigcap_{k=1}^{\infty}$($a_k$, $b_k$)$\neq$$\emptyset$.

For the first problem, the book chooses the interval ($0$, $\frac 1k$) where the decreasing nest is

($0$, $1$) $\supseteq$ ($0$, $\frac12$) $\supseteq$ ($0$, $\frac13$)$\supseteq$...

And for the second problem, the book states that the correct interval is (-$\frac1n$, $\frac1n$) where the decreasing nest is

($-1$, $1$)$\supseteq$(-$\frac12$, $\frac12$)$\supseteq$(-$\frac13$, $\frac13$)$\supseteq$...

I understand the basic definition of an intersection to be the set of all the elements that belong to both A and B.

In the first problem, the book claims that the intersection is $\emptyset$. I'm not sure how this answer was achieved since the number $0$ occurs in all the intervals. Wouldn't the intersection be $0$?

I have the same dilemma for problem two. The books states that the intersection of the intervals is $0$. I'm not sure how that answer was reached either.

Can someone give me a better definition of the intersection of intervals and help me understand why these intersections were achieved?

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    $\begingroup$ "I'm not sure how this answer was achieved since the number 0 occurs in all the intervals." No, it doesn't. 0 doesn't appear in any of the intervals. the intervals consist of numbers between 0 and 1/k. They do not include 0 or 1/k. $\endgroup$ – fleablood May 4 '17 at 0:23
  • $\begingroup$ It's not intersection you are having a problem with. It is intervals. (a,b) is all the numbers x so that a< x < b. a and b are not in the intervals. So the first example. There is no element in all the intervals. The second 0 is in all the intervals. $\endgroup$ – fleablood May 4 '17 at 0:26
  • $\begingroup$ $\bigcap_{k=1}^{\infty}(0,\frac1k)=\emptyset$, where $(0,\frac1k)=\{x:0<x<\frac1k\}$. Indeed if $y\le0$ then $y$ does not belong to any $(0,\frac1k),$ and hence does not belong to the intersection. If $y>0$ then take $k>\frac1y$, then $y>\frac1k$ hence $y$ does not belong to $(0,\frac1k)$ (for this particular $k$), hence $y$ does not belong to the intersection. Note also that $\bigcap_{k=1}^{\infty}[0,\frac1k]=\bigcap_{k=1}^{\infty}[0,\frac1k)=\{0\}$ (singleton set containing the only element $0$), and the latter is different from $0$.(There is a convention that $0$ is the same as $\emptyset$) $\endgroup$ – Mirko May 4 '17 at 1:25
  • $\begingroup$ I'm starting to understand. But I have one more question: after looking at a few more examples, it seems that the intersection must be an integer. In the first problem, $\frac14$ appears to occur in every interval. Yet the intersection is empty. Does the intersection have to be an integer? $\endgroup$ – ErinA May 4 '17 at 1:26
  • $\begingroup$ No the intersection isn't an interger; it's a set of numbers. 1/4 is not in (0,1/8). The ... means that the sets never end. (0,1/n), (0, 1/(n+1)) and so on ad infinitum. What are the next trillion intervals? $\endgroup$ – William Elliot May 4 '17 at 2:48

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