Timeline for Counting copies of a BA within a BA: arbitrarily many vs infinitely many
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Nov 9, 2019 at 0:00 | answer | added | YCor | timeline score: 2 | |
| Nov 8, 2019 at 22:52 | history | edited | YCor | CC BY-SA 4.0 |
made topological reformulation explicit
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| Nov 8, 2019 at 22:42 | history | edited | YCor | CC BY-SA 4.0 |
fixed notation, added gn tag because of Stone duality interpretation
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| Sep 29, 2011 at 10:51 | comment | added | Asher M. Kach | Yes, I am using BA for Boolean algebra and $\mathcal{A} \cong b_i$ for $\mathcal{A} \cong \mathcal{B} \upharpoonright b_i$. | |
| Sep 29, 2011 at 5:26 | comment | added | Stefan Geschke | Yes, Joel, I meant $\mathcal B\restriction b_i$. | |
| Sep 29, 2011 at 1:50 | comment | added | Joel David Hamkins | Stefan, I think you should refer instead to ${\cal B}\upharpoonright b_i$. | |
| Sep 28, 2011 at 21:13 | comment | added | Stefan Geschke | Do I assume correctly that BA stand for Boolean algebra? And when you write $\mathcal A\cong b_i$, do you mean $\mathcal A\cong\mathcal A\restriction b_i$ where $\mathcal A\restriction b_i$ is the BA of all elements of $\mathcal A$ that are $\leq b_i$? | |
| Sep 28, 2011 at 19:50 | history | asked | Asher M. Kach | CC BY-SA 3.0 |