# Thread: Closed, non-empty and bounded sets.

1. ## Closed, non-empty and bounded sets.

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.

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

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

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

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

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

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

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

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

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