Constructing a sequence of nonempty, bounded, closed sets in $\mathscr l^2$ such that $A_{n+1} \subset A_n$, but $\cap_{n=1}^\infty A_n = \emptyset$. 
Given the space of square summable sequences $\mathscr l^2= \{ (x_n)_{n \in \mathbb{N}}: x_n \in \mathbb R, \, \sum_{n=1}^\infty |x_n|^2 < \infty\}$. Construct a sequence of nonempty, bounded, closed sets in $\mathscr l^2$ such that $A_{n+1} \subset A_n$, but $\cap_{n=1}^\infty A_n = \emptyset$. The hint I was given is that diam$(A_n) \geq 1$ for all $n.$

Would somebody be able to give me some direction on how to construct such a sequence?
Does bounded in this case mean that there exists an $r\in \Bbb R$ and $(m_n)\in\mathscr l^2$ such that $\sqrt{\sum_{n=1}^\infty |m_n - x_n|^2}<r$ for all $(x_n)\in A_n?$
Also, am I correct in assuming that closed implies that for some sequence of elements in $A_n$, say $(y_{m,k})$, if $y_{m,k}\rightarrow x_k$, then $(x_k)\in A_n?$
 A: Denote $\mathcal{H}=l^{2}$, which is a Hilbert space under the usual
inner product. For each $n\in\mathbb{N}$, let $e_{n}\in\mathcal{H}$
be defined by $e_{n}(k)=\begin{cases}
1, & \mbox{ if }k=n\\
0, & \mbox{ if }k\neq n
\end{cases}.$ It is well-known that $\{e_{n}\mid n\in\mathbb{N}\}$ is an orthonormal
base for $\mathcal{H}$. Now, for each $n\in\mathbb{N}$, define $A_{n}=\{e_{m}\mid m\geq n\}$.
Clearly $A_{1}\supseteq A_{2}\supseteq\ldots$. We go to prove that:


*

*For each $n\in\mathbb{N}$, $A_{n}$ is non-empty, closed and bounded
subset of $\mathcal{H}$,

*$\cap_{n}A_{n}=\emptyset$.
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*

*Let $n\in\mathbb{N}$. Clearly $A_{n}$ is non-empty and $||x||\leq1$
for any $x\in A_{n}$. Therefore, $A_{n}$ is non-empty and bounded.
Observe that for any $x,y\in A_{n}$ with $x\neq y$, $||x-y||=\sqrt{2}$.
Let $(x_{m})$ be a sequence in $A_{n}$ such that $x_{m}\rightarrow x$
for some $x\in\mathcal{H}$. Let $\varepsilon=0.1$, then there exists
$N$ such that $||x_{m}-x||<\varepsilon$ whenever $m\geq N$. In
particular, for any $m_{1},m_{2}\geq N$, we have 
\begin{eqnarray*}
||x_{m_{1}}-x_{m_{2}}|| & \leq & ||x_{m_{1}}-x||+||x-x_{m_{2}}||\\
 & < & 2\varepsilon\\
 & < & \sqrt{2}.
\end{eqnarray*}
Therefore, $x_{m_{1}}=x_{m_{2}}$. That is $x_{m}=x_{N}$ for all
$m\geq N$. Hence $x=\lim_{m\rightarrow\infty}x_{m}=x_{N}\in A_{n}$.
This shows that $A_{n}$ is closed.

*It is trivial. For, suppose the contrary that there exists $x\in\cap_{n}A_{n}$.
Since $x\in A_{1}$, so $x=e_{k}$ for some $k$. By the very definition
of $A_{k+1}$, $e_{k}\notin A_{k+1}$. This contradicts to $x\in A_{k+1}$.
A: Start with the sequence of standard basis vectors $(e_i)_{i \in \Bbb{N}}$, where each $e_i$ takes the value $1$ in the $i$th position, and $0$ elsewhere. Then, consider the nested sequence of sets
$$A_n = \{e_i : i \ge n\}.$$
