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  • Consider the family $A_n = (0, \frac{1}{2n}]$. We have $A_{n+1} \subset A_n$ for every $n \in \mathbb{N}$, and the length of $A_n$ approaches zero as $n$ approaches infinity, is $\bigcap_{n=1}^{\infty} A_n$ empty or not? Please give a proof !

  • Every $A_n$ above is of positive length, thus I guess it is impossible for the intersection of all nonempty interval $A_n$ to be empty, is my reasoning right ?

P.S. Since the case here doesn't satisfy the conditions required for the nested interval theorem, so I cannot use the theorem to judge whether $\bigcap_{n=1}^{\infty} A_n$ is empty or not here.

Andrés E. Caicedo
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1 Answers1

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You can just directly check if it is empty or not. For an element to be in $\cap^\infty_{n=1} A_n$ it needs to be in each $A_n$.

Let $x \in \mathbb R$. If $x \le 0$ then $x \not \in A_n$ for any $n$, so $x \not\in\cap^\infty_{n=1} A_n.$ If $x > 0$, then there is some $N$ so that $\tfrac 1 {2N} < x$. But then $x \not \in A_N$ and so $x \not\in\cap^\infty_{n=1} A_n.$ Thus the set is empty.

User8128
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