Timeline for Are there insane families in $L$?
Current License: CC BY-SA 3.0
11 events
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| May 20, 2014 at 19:50 | comment | added | Asaf Karagila♦ | Joel, I was talking with someone about something related to this question. They remarked that a tower is usually well-ordered by $\supsetneq^*$. Do you have any suggestions for a name for an increasing sequence? Maybe "inverse tower" or "co-tower" or "pyramid"? | |
| Sep 10, 2013 at 11:24 | comment | added | Joel David Hamkins | I agree with Garrett. The issue is that $P(\omega)/\text{Fin}$ is not a complete Boolean algebra. One can take $A_\omega'$ in Garrett's argument as $A_\omega-B$ in my construction. | |
| Sep 10, 2013 at 5:34 | comment | added | Garrett Ervin | @AsafKaragila: I agree: $A_{\omega}'$ will almost contain each of the $B_n$. But the difference $X=A_{\omega} \setminus A_{\omega}'$ will then be almost disjoint from each $B_n$, and also from the $B_{\beta}$ for $\beta \geq \omega$, since $A_{\omega}$ is so, and $X \subseteq A_{\omega}$. | |
| Sep 10, 2013 at 4:46 | comment | added | Asaf Karagila♦ | @Garret, it is somewhat related to my previous question. However, I don't see why your argument works. $A'_\omega$ won't be almost disjoint from any $B_n$. | |
| Sep 10, 2013 at 4:14 | comment | added | Garrett Ervin | This question reminds me of your previous question about the countable completeness of $\mathcal{P}(\omega)$/fin. If an insane family exists, then $A_{\omega}$ (for example) is a least upper bound for the countable chain $\langle A_n: n < \omega \rangle$. For if $A_{\omega}' \subsetneq_* A_{\omega}$ was a strictly smaller upper bound, the difference set $X=A_{\omega} \setminus A_{\omega}'$ would be almost disjoint from all the $B_{\beta}$, contradicting maximality. But countable chains never have least upper bounds in $\mathcal{P}(\omega)$/fin, so it must be that there is no insane family. | |
| Sep 9, 2013 at 21:53 | comment | added | Joel David Hamkins | In the case of $P(\kappa)$, if you use an ideal $I$ for which $P(\kappa)/I$ is a complete Boolean algebra, then you can realize any maximal antichain as the difference antichain of a tower in the Boolean algebra. But you've got to work modulo $I$ instead of modulo finite. | |
| Sep 9, 2013 at 21:34 | vote | accept | Asaf Karagila♦ | ||
| Sep 9, 2013 at 21:34 | comment | added | Asaf Karagila♦ | Hmmm, yeah. Well, I guess this means that tomorrow there's a lot of work cut out for me. Thank you very much! | |
| Sep 9, 2013 at 20:37 | comment | added | Joel David Hamkins | I find the issue related to the connection between maximal antichains and difference antichains in a complete Boolean algebra. See mathoverflow.net/a/139743/1946 and a few other places. | |
| Sep 9, 2013 at 20:33 | comment | added | Asaf Karagila♦ | Oh. I suspected as much. Drats. This means that I am going to have so much more work... Oh well. Thanks. I suppose this will be the same argument if we require $\leq$ and $>$ instead... | |
| Sep 9, 2013 at 20:23 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |