1

$ E(n) $ is a subset of $\mathbb{R}$. If $E(n)$ is open when $n$ is even and closed when $n$ is odd, and $E(n+1) \subseteq E(n)$ then $\cap_{n=1}^{\infty}E(n)=?$

Tried: It's essentially the same as asking $\lim_{n\to\infty} E(n)=?$

Let $E(1) = [0,1],E(2)=(0,1),E(3)=[\frac{1}{100},\frac{99}{100}]...$

First, I have no idea what $E(n)$ would be open or closed as $n \to \infty$. Second, I don't think it should be empty set, because this is a countably infinite union of dense sets...

MatheMagic
  • 247
  • 1
    You can’t say anything in general without more information. Are these sets nested? Are they arbitrary open and closed sets, or are they specifically open and closed intervals? – Brian M. Scott Oct 18 '14 at 22:22
  • Let $E(n) = \varnothing$. Then $\bigcap E(n) = \varnothing$. Let $E(n) = (-\infty,\infty)$. Then $\bigcap E(n) = (-\infty,\infty)$. If the openness or closedness of the $E(n)$ is all you have to go on, nothing can be said about the infinite intersection. – Bruce Zheng Oct 18 '14 at 22:27
  • @BrianM.Scott You're right, I forgot the very important condition that they are nested: $E(n+1) \subseteq E(n)$. Now it's corrected. – MatheMagic Oct 19 '14 at 00:10
  • @BruceZheng Sorry I should have mentioned that they are nested sets. – MatheMagic Oct 19 '14 at 00:13

1 Answers1

1

HINT: Note that since $E(n+1)\subseteq E(n)$ for each $n$,

$$\bigcap_{n\ge 1}E(n)=\bigcap_{n\ge 1}E(2n-1)\;.$$

Brian M. Scott
  • 616,228
  • So $\cap_{n \geq 1} E(n)$ is actually $\emptyset$ right? – MatheMagic Oct 20 '14 at 23:47
  • @MatheMagic: Not necessarily. You might have $E(2n)=\left(-1-\frac1{2n},1+\frac1{2n}\right)$ and $E(2n-1)=\left[-1-\frac1{2n-1},1+\frac1{2n-1}\right]$ for $n\in\Bbb Z^+$, in which case the intersection is $[-1,1]$. You can say something about it, however: what can you always say about an intersection of closed sets? – Brian M. Scott Oct 20 '14 at 23:54
  • I see...the only thing we are certain is that it must be closed, since any arbitrary intersection of closed sets is always closed. – MatheMagic Oct 21 '14 at 00:14
  • @MatheMagic: That’s right. – Brian M. Scott Oct 21 '14 at 00:15