# Showing this set $A$ is closed, bounded and not compact?

Let $$l^1(\mathbb{N}) = \left\{ (x_n)_n \mid \sum_{n = 0}^{\infty} | x_n | \ \text{converges} \right\},$$ the space of all sequences whose associated series converge absolutely. On this space we consider the metric $$d_1((x_n)_n, (y_n)_n) = \sum_{n = 0}^{\infty} |x_n - y_n|.$$ Now consider the subset $$A = \left\{x \in l^1(\mathbb{N}) \mid d_1(x,0) = 1 \right\}.$$ I need to prove that $A$ is closed, bounded and not compact.

I'd appreciate it if you point out any mistakes I made.

Attempt:

Closed: Let $(x_n)_n$ be a sequence in $A$, such that $(x_n) \rightarrow (y_n)$ for the $d_1$ metric, where $(y_n)_n \in l^1(\mathbb{N})$. I want to prove that $(y_n)_n \in A$, i.e. that $d_1(y_n,0) = 1$. Let $\epsilon > 0$. Since $(x_n) \rightarrow (y_n)$, there exists an $n_0 \in \mathbb{N}$ such that $$\sum_{k=n_0 +1}^{\infty} |x_k - y_k| < \epsilon/2.$$ So by the triangle inequality $$\sum_{k=n_0 +1}^{\infty} |y_k| \leq \sum_{k=n_0 + 1}^{\infty} |y_k - x_k | + \sum_{k=n_0 +1}^{\infty} |x_k| < \epsilon/2 + \sum_{k=n_0 +1}^{\infty} |x_k|$$ It follows that $$\sum_{k=0}^{\infty} |y_k| = \sum_{k=0}^{n_0} |y_k| + \sum_{k=n_0+1}^{\infty} |y_k| < \epsilon/2 + 1$$ since $\sum_{k}^{\infty} |x_k| = 1.$ Since $\epsilon > 0$ was arbitrary, this shows that $\sum_k |y_k| = 1$ and so $(y_n)_n \in A.$

Bounded: Let $M=2$. Let $x,y \in A$. Then $d_1(x,y) \leq d_1(x,0) + d_1(0,y) = 1 + 1 = 2 = M.$ Hence $A$ is bounded.

Not compact: I want to find a sequence in $A$ that has no convergent subsequence, or an open cover of $A$ that has no finite subcover. But I'm lacking inspiration here at the moment. Any help is appreciated.

Note that $d_1(x,0) = \sum_{n=0}^\infty |x_n|$ is the $\ell^1$ norm $\|x\|_1$ of $x$, so $A$ is the unit sphere (the boundary of the unit ball). In general, the unit sphere and unit ball of a normed vector space are compact if and only if the space is finite-dimensional. Since $\ell^1(\mathbb N)$ is infinite dimensional, $A$ is not compact.
To see this, recall that sequential compactness is equivalent to compactness in metric spaces. So $\ell^1(\mathbb N)$ if and only if every sequence has a convergent subsequence. Let $\{e^{(n)} : n\in\mathbb N\}$ be the canonical basis for $\ell^1(\mathbb N)$. Then $\|e^{(n)}\|_1=1$ for all $n$, so $e^{(n)}\in A$, but $\{e^{(n)}\}$ clearly has no convergent subsequence.
Hint for non-compactness: Let $x_n=(0,0,\ldots,0,1,0,\ldots)$ be the sequence of only zeroes except at component number $n$, where there is a $1$.