# Tag Info

2

Because $a=\lim_{n\to\infty}S_n$, and because multiplication is continuous: For all $n\in\Bbb{N}$ we have $$xS_n=S_{n+1}-1_A,$$ as you already noted. We may rewrite this to get $S_{n+1}-xS_n=1_A$, and hence $$\lim_{n\to\infty}(S_{n+1}-xS_n)=\lim_{n\to\infty}1_A=1_A.$$ Because addition and multiplication are continuous we have ...

1

To determine whether the map is injective, we first need to clarify the domain. If we let the domain be $\mathscr{O}(\sigma(a)) = \bigcup_{\sigma(a) \subset U} \mathscr{O}(U)$, where the $U$ are open sets, then the map $\Phi \colon f \mapsto \tilde{f}(a)$ is trivially not injective, we can let $U = D_1 \cup D_2$ consist of two disjoint open disks, with ...

Top 50 recent answers are included