Give an infinite collection of intervals such that $A_{n+1} \subset A_n$ and $\cap^{\infty}_{n=1} A_n = \varnothing$

Question

Give an example of an infinite collection $A_n $ of intervals such that $A_{n+1} \subset A_n$ and $\bigcap^{\infty}_{n=1} A_n = \varnothing$.

I have come to the following collection of intervals, but dont know if it is correct nor how to prove it. $A_n = \left(0, \frac{1}{n^2}\right)$. How do I prove my example works?

EDIT: The intervals are on the real line!

Let $x\in \mathop{\cap}_{n =1}^{+\infty}A_n$. Then $0<x$ and $\forall n\in \Bbb N\left(x<\dfrac 1{n^2}\right)$. Conclude. — Git Gud, Nov 03 '13 at 15:31
@GitGud Sorry, the intervals are on the real line. My mistake! — user1090614, Nov 03 '13 at 15:33
Suppose $x\in\bigcap^{\infty}{n=1}$ and then we can prove it by using the squeeze theorem. The lim$\frac{1}{n^2}=0$ and lim$0=0$ Which forces the lim $x=0$. But $0\not \in\bigcap^{\infty}{n=1}$. Thus we conclude... — user60887, Nov 03 '13 at 15:36

score 1 · Answer 1 · answered Nov 03 '13 at 17:53

1

if $x \leq 0$, $x$ belongs to none of the sets. If $x>0$, take $n \in \mathbb{N}^+$ with $x > 1/n^2$. Then $x$ does not belong to $\mathcal{A}_n$, so $x$ does not belong to the intersection.

answered Nov 03 '13 at 17:53

Stefan Smith

8,192

Don Larynx · Answer 2 · 2013-11-03T15:40:27.690

HINT: The set containing sequences of $\frac{1}{n}$ approaching 1 as its limit (BUT IT DOESN'T CONTAIN $1$) from the LHS and the RHS. The answer to the sequence contained in your metric space is below.

$$S = \{1 - \frac{1}{n}\} \bigcup \{1 + \frac{1}{n}\} : n \in \mathbb{N}$$

Generally, just unite two open sets, and don't include their limit point (This could make it closed and have a nonempty intersection).

If you know about separability, just make sure the intersection contains none of the sets closures!

score 0 · Answer 3 · answered Nov 03 '13 at 15:42

0

a(n)=[n,+∞)

so An+1⊂An

n=1⋂∞ An=∅

answered Nov 03 '13 at 15:42

Khosrotash

24,922

Give an infinite collection of intervals such that $A_{n+1} \subset A_n$ and $\cap^{\infty}_{n=1} A_n = \varnothing$

3 Answers3