Timeline for Finitely additive measure on Cartesian square of countable set
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 17, 2019 at 3:15 | review | Close votes | |||
| Apr 17, 2019 at 13:27 | |||||
| Mar 21, 2019 at 13:10 | review | Close votes | |||
| Mar 21, 2019 at 17:02 | |||||
| Mar 18, 2019 at 23:28 | answer | added | Andreas Lietz | timeline score: 2 | |
| Mar 18, 2019 at 13:28 | comment | added | Michael Greinecker | It is not $2^{\omega\times\omega}$. Indeed, $\{(m,n)\mid m\leq n\}\notin\mathcal{U}$. For if $A\times B\subseteq \{(m,n)\mid m\leq n\}$, then $A$ must be finite and therefore $\mu^2(A\times B)=0$. A similar argument applies to the complement $\{(m,n)\mid m> n\}$. | |
| Mar 18, 2019 at 11:35 | history | asked | ar.grig | CC BY-SA 4.0 |