Timeline for Bi-annihilator of a subspace of the dual of an infinite-dimensional vector space
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Aug 16, 2020 at 15:09 | comment | added | lefuneste | Thanks again for your clear explanations, LSpice. I have upvoted your answer and your comments. | |
| Aug 16, 2020 at 15:01 | comment | added | LSpice | The topology @user131781 mentions is the weak* topology, weakest making all eval. fncls. continuous. Since $\lim_{F \subseteq \mathcal B} \bigl(\sum_{b \in F} e_b\bigr)(v)$ equals $\bigl(\sum_{b \in \mathcal B} e_b\bigr)(v)$ for all $v \in V$, this topology must declare that $\lim_{F \subseteq \mathcal B} \sum_{b \in F} e_b$ equals $\sum_{b \in \mathcal B} e_b$ (the limit taken over the finite subsets $F$ of $\mathcal B$—the so called "unordered sum"). | |
| Aug 16, 2020 at 14:54 | comment | added | lefuneste | OK, I agree, what you write makes sense. However I don't know about weak topology and I thank you for your definition which doesn't make reference to topology. | |
| Aug 16, 2020 at 14:47 | comment | added | LSpice | Sure it does: $\bigl(\sum_{b \in \mathcal B} e_b\bigr)(v) = \sum_{b \in \mathcal B} e_b(v)$ is a finitely supported sum for each $v \in V$. I think that, as @user131781 points out, this is a convergent sum in a weak topology; but anyway it can be defined without direct reference to topology. | |
| Aug 16, 2020 at 14:36 | comment | added | lefuneste | Your symbol $\sum_{b \in \mathcal B} e_b$ does not make sense because you cannot sum infinitely many non-zero vectors in a vector space. | |
| Aug 16, 2020 at 14:12 | history | answered | LSpice | CC BY-SA 4.0 |