A subset of $\ell^2$

Question

I want to show that $A=\{(x^{(n)})\in \ell^2:|x_m^{(n)}|\leq \frac{1}{m}\}$ is a complete, convex subset of $\ell^2$ with empty interior.

It is not difficult to show that $A$ is convex. Also I could show that $A$ is closed in $\ell^2$. Since $\ell^2$ is a Banach space, therefore, $A$ is complete. But how to show that $ A^{0}=\emptyset $?

Answer 1

One can show that for all $x = (x_n) \in A$, for all $\epsilon >0$, there is $y\notin A$ and $d(x, y) <\epsilon$. Let $m \in \mathbb N$ so that $\frac{3}{m}<\epsilon$. Define $y = (y_n)$, where

$$y_n = \begin{cases} x_n &\text{if } n\neq m \\ 2/m &\text{if }m=n.\end{cases}$$

Then $y\in \ell^2$ and $y\notin A$. Also $d(x, y) \le |2/m - x_m| \le 3/m <\epsilon$ as $|x_m|\le 1/m$.

The above implies in particular that $A^0 = \emptyset$.

Indeed, one can show that $A$ is compact and is homeomorphic to $[0,1]^\mathbb{N}$.