Timeline for Definition of $S_1(A,B)$
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 23, 2017 at 6:58 | vote | accept | Alexander Osipov | ||
| S Jul 23, 2017 at 0:47 | history | suggested | Linus Rastegar | CC BY-SA 3.0 |
Fixed grammar and formatting
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| Jul 23, 2017 at 0:20 | review | Suggested edits | |||
| S Jul 23, 2017 at 0:47 | |||||
| Jul 22, 2017 at 22:05 | answer | added | Boaz Tsaban | timeline score: 2 | |
| Jul 22, 2017 at 19:39 | review | First posts | |||
| Jul 22, 2017 at 19:41 | |||||
| Jul 22, 2017 at 19:21 | comment | added | LSpice | If I understood @GeraldEdgar's point properly, he was pointing out that you were asking whether you could take the elements $x_n$ to be empty, which seems "ill typed" (a priori they are just elements, not sets). Also, you have only defined $x_n$ in your statement; what is $x_{n_k}$? | |
| Jul 22, 2017 at 18:46 | comment | added | Alexander Osipov | Yes, in this interpretation we get that the answer is positive.... Since $ B_{n_k}$ can be empty .... | |
| Jul 22, 2017 at 18:35 | comment | added | Renan Mezabarba | If $A$ and $B$ are families of subsets of a set $S$ and $B$ is "closed upward", then I think $S_1(A,B)$ is equivalent to the following: for each sequence $(A_n:n\in\omega)$ of elements of $A$ there is a sequence $(B_n:n\in\omega)$ such that $B_n\subset A_n$, $|B_n|\leq 1$ and $\bigcup_{n\in\omega}B_n\in B$. | |
| Jul 22, 2017 at 18:22 | comment | added | Alexander Osipov | Thank you. It seems that it is so. It is interesting that in the definition $S_{fin}(A,B)$ we can choose empty subsets $X_n\subset A_n$. | |
| Jul 22, 2017 at 17:53 | comment | added | Gerald Edgar | You say $x_n$ is an element of $A_n$. I would think that in most cases $\emptyset$ is not an element of $A_n$. | |
| Jul 22, 2017 at 17:44 | history | asked | Alexander Osipov | CC BY-SA 3.0 |