Is the unit ball in this sequence space compact?

I have a set $X=\{\text{complex sequences } \{x_n\}: \sup\limits_{n}\sqrt{n}\left|x_n\right|\leq 1\}$ equipped with a metric $d(\{x_n\},\{y_n\})=\sup\limits_{n}|x_n-y_n|+\sup\limits_{n}\sqrt{n}|x_n-y_n|$. I want to show that the closed unit ball centered around $\{0\}$ is either compact or not compact, i.e. $N_1=\{\{x_n\}\in X: d(\{x_n\}, \{0\})\leq 1\}$. Closed unit balls are typically not compact (the only case of this is when $X$ is finite dimensional).

I'd like to find a sequence in $N_1$, that is a sequence of sequences, that does not have a convergent subsequence. Alternatively, I can show that it isn't complete or totally bounded. The abundance of possibilities, as well as my unfamiliarity with sequence spaces, is confounding me. Can anyone help?

• Try to mimic the approach for $\ell^p$ spaces: a sequence of unit vectors obtained from the standard basis $e_n$. Some scalar multiples will be needed. – user147263 Feb 1 '15 at 19:49
• For each $m$, find a sequence $(x_n)$ of unit norm where all of the terms, except the $m$th, are equal to $1$. Once you have it, show that $d(x_k,x_n)\ge 1$ whenever $k\ne n$. – user147263 Feb 1 '15 at 19:55

I recommend using function notation for sequences in this context; after all, a sequence is just a function whose domain is $\mathbb{N}$. This avoids the awkward language of sequences of sequences, and associated notational mishaps.
So, the distance between two functions $f:\mathbb{N} \to \mathbb{C}$ and $g:\mathbb{N}\to \mathbb{C}$ is $$d(f,g) = \sup_{n\in\mathbb{N}} |f(n)-g(n)|+ \sup_{n\in\mathbb{N}} \sqrt{n} |f(n)-g(n)|$$ You can find, for each $m\in\mathbb{N}$, a function $f_m$ of unit distance from $0$ such that $f_m(n)=0$ whenever $n\ne m$. Indeed, it is $$f_m(n) = \begin{cases} \dfrac{1}{1+\sqrt{m}},\quad &n=m \\ 0,\quad & n\ne m \end{cases}$$
The idea is that $f_1,f_2,\dots$ belong to the closed unit ball but are uniformly separated. Indeed, $$d(f_m,f_k) \ge 1\quad \text{whenever } m\ne k \tag{1}$$ (I leave it for you to check this.)
Property $(1)$ makes it impossible for the sequence $(f_m)$ to have a convergent subsequence.