Show that the closed ball $B[x,1]$ in $c_0$ is not compact.

Question

Consider $c_0 = \{(a_n)_{n \in \mathbb{N}}\subset \mathbb{R}:a_n \to 0\}$. Show that the closed ball $B[x,1]=\{y \in c_0 : d_{\infty}(x,y)\leq 1\}$ is not compact in $c_0$. Were $d_{\infty}(x,y)=sup_{n \in \mathbb{N}}|x_n - y_n|$.

I considered an open cover $\left(c_0 - B[x,\frac{1}{n}]\right)_{n \in \mathbb{N}}$. However I can only cover $c_0 - \{x\}$. I've been trying to fix it, but I wasn't able, I thought of including $\{x\}$ in the open cover but is closed. And $\{x\}^c$ doesn't work either. Am I on the correct track? Any other possible strategy?

Answer 1

Hint: consider an open cover consisting of open balls of radius $1/3$ around every point of $B[x,1]$. Find infinitely many points whose pairwise distances are all $> 2/3$.

Answer 2

Hint: Recall that for metric spaces, compactness is equivalent to sequential compactness. Consider the sequence of sequences $(x_n)_m = 1$ if $n=m$ and $0$ if $n\not=m$. Every element has distance $1$ from the $0$ sequence. Can it have a convergent subsequence?