Is the set of all binary sequences compact in $l^{\infty}$? Here, the metric space is defined as the set of all bounded sequences, with the distance function defined as the supremum of the absolute value of the difference between corresponding elements.
Intuitively, it seems that this set is not compact, although my intuition for compactness isn't very strong. I've been trying to show that the set doesn't contain all its limit points and is therefore not closed, and hence not compact, but I'm not sure how to proceed. I'm able to show that no element of this set is a limit point, but I'm not sure how to handle limit points outside of the set.
I'd appreciate any advice on how to proceed, or more broadly, intuition for compactness and advice on how to think about compactness in non-traditional metric spaces.
 A: Consider the collection of open sets $\mathcal{B}:=\left\{B_{l^\infty}(x,1): x \in \{0,1\}^{\mathbb N} \right\}$. Notice that for $x, y \in \{0,1\}^{\mathbb N}$ we have $\left\| x - y \right\|<1$ if and only if $x=y$. So $\mathcal B$ covers $\{0,1\}^{\mathbb N}$ but there is no way to refine $\mathcal B$ to a finite subcover.
Note: $\{0,1\}^{\mathbb N}$ denotes the set of all binary sequences.
A: Let $(x^{(n)})_{n \geq 1} \subset l^\infty$ defined by
$$
x^{(n)}_i = \chi_{[n,+\infty)}(i) = \cases{0 \text{ if } i < n \\ 1 \text{ otherwise}}
$$
Now, if the space of binary sequences were compact, this sequence should have a convergent subsequence $(x^{(n_k)})_{k \geq 1} \to x$. In particular, we should have that 
$$
\chi_{[n_k,+ \infty)}(i) = x^{(n_k)}_i \xrightarrow{k \to +\infty} x_i
$$
and so necessarily, $x \equiv 0$. However, the original subsequence does not converge to zero, since
$$
d(x^{(n_k)},0) = \sup_{i \geq 1}x^{(n_k)}_i = 1 \quad  (\forall n \in \mathbb{N}).
$$
A: Let $e_n$ be the binary sequence with $1$ in the $n$th slot and zeros everywhere else. If $m\ne n,$ then $\|e_m-e_n\|_{l^\infty} = 1.$ Thus $(e_n)$ has no convergent subsequences. It follows that the set of binary sequences is not compact in $l^\infty.$
