Suppose that for each $\displaystyle n \in \mathbb{N} $ we have a non-empty closed and bounded set $\displaystyle A_n \subset \mathbb{C} $ and

$\displaystyle A_1 \supseteq A_2 \supseteq ... \supseteq A_n \supseteq A_{n+1} \supseteq ... $

Prove that $\displaystyle \bigcap_{n=1}^{\infty} A_n $ is non empty. [Hint: use Bolzano-Weierstrass]

Solution:

Let $\displaystyle x_n \in A_n $. Then there exists a bounded sequence $\displaystyle \{ x_n \} $ with a convergent subsequence $\displaystyle \{x_{k_n} \} $ whose limit is $\displaystyle x \in A_n $. I need to show that $\displaystyle x \in A_m \forall m \in \mathbb{N} $. I can see that $\displaystyle \forall m \le n, x \in A_m $ since $\displaystyle A_1 \supseteq A_2 \supseteq ... \supseteq A_n \ni x $. But how do I show that if $\displaystyle x \in A_n $ then $\displaystyle x \in A_{n+1} $? Any help would be appreciated.