Subset of metric space is bounded and closed but not compact.

Question

We consider the set $l^1({\mathbb{N}})$ of the sequences whose associated series are absolute convergent: $$ l^{1} (\mathbb{N}) = \left\{ (x_n)_{n \in \mathbb{N}} \in \mathbb{R}^{\mathbb{N}} \mid \sum_{n = 0}^{\infty} | x_n | \ \text{converges} \right\}. $$

with the metric $$ d_{1} ((x_n)_n, (y_n)_n) = \sum_{n = 0}^{\infty} | x_n - y_n | . $$

We now consider the subset $A = \left\{x \in l^1({\mathbb{N}}) \mid d_1(x,0) = 1\right\}$.

I would like to proof that A is bounded and closed, but not compact. I can see why it is bounded because the sum of the $|x|$ values converges to 1 so x can never be bigger than |1|, but I don't know how to write it rigorously and proof why it's closed but not compact.

Thanks in advance

Answer 1

Hint:

The norm $\|\cdot\|_1$ is a continuous function and $A$ is a preimage of the closed set $\{1\}$ with respect to it.

The sequence of canonical vectors $(e_n)_n$ is in $A$, but has no convergent subsequence because $d_1(e_m, e_n) = 2$ for all $m \ne n$.

Answer 2

It is closed because it is the set $f^{-1}\bigl(\{1\}\bigr)$, where $f$ is the continuous function $x\mapsto d_1(x,0)$.

And it is not compact because if you define\begin{align}x(1)&=(1,0,0,0,\ldots)\\x(2)&=(0,1,0,0,\ldots)\\x(3)&=(0,0,1,0,\ldots)\\&\vdots\end{align}then, if $m\neq n$, $d_1\bigl(x(m),x(n)\bigr)=2$. Therefore, the sequence $\bigl(x(n)\bigr)_{n\in\mathbb N}$ (which is a sequence of elements of $A$) has no convergent subsequence. It follows from this that your set is not compact.