let $(l^{1}(\mathbb Z), || .||_{1})$ be a norm space where $l^{1}(\mathbb Z):=\{x=(x_{n})_{n \in \mathbb Z}\subset\mathbb C : ||x||_{1}<\infty \}$

a.) Show that $\overline{B_{1}(0)}$ is closed and bounded.

b.) Show that $\overline{B_{1}(0)}$ is not compact.

My thoughts: In terms of $a.)$ proving that it is bounded is easiest. Consider $\overline{B_{1}(0)}:=\{x\in l^{1}(\mathbb Z):||x-0||_{1}\leq1\}$, this means $\exists c > 1 $: $\forall x\in \overline{B_{1}(0)}$: $||x||_{1}<1<c$, so it is bounded.

I'm struggling in terms of closedness, how do I prove that $l^{1}(\mathbb Z)/ (\overline{B_{1}(0)})$ is open?

Next, how can I prove that $\overline{B_{1}(0)}$ is not compact? Any hints appreciated.

  • $\begingroup$ Suppose x is a sequence that is NOT in $l^1(Z)$. What does that tell you about $x_1$? $\endgroup$ – user247327 May 21 '18 at 11:05
  • $\begingroup$ That $||x||_{1}=\infty$ $\endgroup$ – SABOY May 21 '18 at 11:07
  • $\begingroup$ You state this as if it is for an arbitrary norm on $\ell_1(\Bbb Z)$, but you use the particular notation $\|\cdot\|_1$, which normally applies to a specific norm: the supremum norm. Do you really mean an arbitrary norm, or do you want the supremum norm here? $\endgroup$ – Paul Sinclair May 21 '18 at 20:03

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