Closed, non-empty and bounded sets.

**slevvio** · October 21st 2009, 02:18 PM

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

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

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

Solution:

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

**Jose27** · October 21st 2009, 03:41 PM

Let $(x_n)$ be a sequence such that $x_n\in A_n$ for all $n$ then since $A_n \subset A_1$ for all $n$ and $A_1,A_n$ are closed and bounded there is a subsequence $x_{n_k}$ with $n_k<n_{k+1}$ such that $x_{n_k} \rightarrow x$ . And by construction, and since the sequence is nested, $x \in A_n$ for all $n$ .

**slevvio** · October 21st 2009, 03:51 PM

sorry could you explain nested is that a precise mathematical definition?

**Jose27** · October 21st 2009, 03:53 PM

The fact that $...\subset A_{n+1} \subset A_n \subset ... \subset A_1$ is expressed saying that it's a sequence of nested sets.

**slevvio** · October 21st 2009, 03:56 PM

ok i will have a look a this once i have had some sleep. thanks for the help.

**slevvio** · October 22nd 2009, 02:42 AM

I have thought about this, does the way I have written this make sense and is it a valid proof?

Let $\{x_n\}_{n=1}^{\infty}$ be a sequence such that $x_n \in A_n \forall n \in \mathbb{N}$ (i.e. $x_1 \in A_1, x_2 \in A_2$ ... )

$\exists k_1 < k_2 < ...$ such that $\{ x_{k_n} \}_{n=1}^{\infty}$ converges to limit $x \in A_1$ since $A_1$ closed.

Since the sets are nested the sequence $\{x_n \}_{n=m}^{\infty}$ is in $A_m$ with subsequence $\{x_{k_n} \}_{n=m}^{\infty}$ which converges to x as $\{x_{k_n} \}_{n=m}^{\infty}$ is a subsequence of a convergent sequence (the first m terms have simply gone).

Hence $x \in A_m \forall m \in \mathbb{N}$ and so $x \in \bigcap_{n=1}^{\infty} A_n$ .

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