Closed ball in $\mathbb{R}^{\infty}$. How do I argue that the closed ball of radius 1 isn't compact in $\mathbb{R}^{\infty}=\{ (x_1,x_2,\cdots) : x_i \in \mathbb{R} \ not \ all\ x_i \neq 0 \}$?
This set isn't metric space, so I can't use the fact that a compact set in a metric space is closed and bounded.
Perhaps it isn't closed in $\mathbb{R}^{\infty}$.
Any hints?
Thanks in advance.
 A: First I’m going to fix the definitions:
$$\Bbb R^\infty=\{\langle x_k:k\in\Bbb Z^+\rangle:\exists m\in\Bbb Z^+\forall k\ge m(x_k=0)\}\;,$$
the set of all real sequences that are eventually $0$. For each $n\in\Bbb Z^+$ we define
$$R_n=\{\langle x_k:k\in\Bbb Z^+\rangle:x_k=0\text{ for all }k>n\}\;.$$
There is a natural bijection $\pi_n:R_n\to\Bbb R^n:\langle x_k:k\in\Bbb Z^+\rangle\mapsto\langle x_1,\ldots,x_n\rangle$. A set $U\subseteq\Bbb R^\infty$ is open iff $\pi_n[U\cap R_n]$ is open in the usual topology of $\Bbb R^n$. Finally, $$D=\left\{\langle x_k:k\in\Bbb Z^+\rangle\in\Bbb R^\infty:\sum_{k\ge 1}x_k^2\le 1\right\}\;,$$ and the problem is to prove that $D$ is not compact.

With that out of the way, here’s a large HINT:
For $n\in\Bbb Z^+$ let $x^{(n)}=\langle x_k^{(n)}:k\in\Bbb Z^+\rangle$, where
$$x_k^{(n)}=\begin{cases}
1,&\text{if }k=n\\
0,&\text{otherwise}\;;
\end{cases}$$
clearly each $x^{(n)}\in\Bbb R^\infty$. For each $n\in\Bbb Z^+$ let
$$U_n=\left\{\langle x_k:k\in\Bbb Z^+\rangle:\frac12<x_n<\frac32\text{ and }|x_k|<\frac12\text{ if }k\ne n\right\}\;.$$


*

*Show that each $U_n$ is open in $\Bbb R^\infty$.


Now let $E=\{x^{(n)}:n\in\Bbb Z^+\}$. 


*

*Show that $E$ is closed in $\Bbb R^\infty$. Conclude that if $D$ were compact, $E$ would also be compact.  

*Show that $\{U_n:n\in\Bbb Z^+\}$ is an open cover of $E$ with no finite subcover, thereby showing that $E$ is not compact.

