Timeline for When are two forcing posets "the same"?
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| 23 hours ago | vote | accept | new account | ||
| Aug 3, 2023 at 6:01 | comment | added | new account | I guess this is what you mean: $E$ has a maximal antichain $\{B_b:b\in B,b\neq 0\}$ where $B_b\simeq B\upharpoonright b$; if $e\in A$ then it has nonempty intersection with some $B_b$, say $e\land B_b\simeq B\upharpoonright b'$; there must be some $e'\in A$ such that $e'\land B_{b'}\neq 0$; if $e=e'$ then no $c\in B$ satisfy $B\upharpoonright c\simeq E\upharpoonright e$; if $e\neq e'$ and $B\upharpoonright c\simeq E\upharpoonright e$ and $B\upharpoonright c'\simeq E\upharpoonright e'$ then $b'\leq c$ and $c'\land b'\neq 0$. We would get a contradiction if $c\neq c'$, but why is that so? | |
| Aug 2, 2023 at 11:27 | comment | added | Joseph Van Name | Joel finally upgraded his site from http to https. | |
| Aug 2, 2023 at 11:03 | history | edited | Calliope Ryan-Smith | CC BY-SA 4.0 |
Included Andreas's finishing of my partial solution
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| Aug 2, 2023 at 8:00 | comment | added | Calliope Ryan-Smith | @AndreasLietz Nice! I'll amend the answer to include your contribution. | |
| Aug 1, 2023 at 14:33 | comment | added | Andreas Lietz | @CalliopeRyan-Smith If Im not mistaken, your strategy to show (a) to be strictly stronger than (ii) should work if you take $B$ a rigid atomless cBa and set $E=B$. | |
| Jul 27, 2023 at 8:07 | comment | added | Calliope Ryan-Smith | I've swapped back the construction and implemented Joel's fix instead, since my original "fix" still had the problem of having a summand of $B$ not be embed-into-able, so you can have a $B$-generic with no $C$-generic inside. | |
| Jul 27, 2023 at 8:05 | history | edited | Calliope Ryan-Smith | CC BY-SA 4.0 |
Using $\leq$ means the forcings are not atomic
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| Jul 26, 2023 at 15:12 | comment | added | Calliope Ryan-Smith | $\bigoplus$ to mean lottery sum, yes. I suppose even in my new fix, we need to stick a bit on the end- Redefine $C=P_0\oplus\bigoplus_{n<\omega}P_{2n+1}$. | |
| Jul 26, 2023 at 14:52 | comment | added | new account | I suppose $\bigoplus$ means lottery sum (instead of finite support product)? And I don't quite get how $B$ completely embeds into $C$; your original $\prod_{m<n}$ makes more sense to me, since then we have a sequence of complete embeddings $P_0\rightarrow P_1\rightarrow\cdots$; the even and odd lottery sums completely embed into each other. | |
| Jul 26, 2023 at 13:55 | comment | added | Calliope Ryan-Smith | @JoelDavidHamkins Yes that's a much cleaner fix, d'oh! | |
| Jul 26, 2023 at 13:51 | comment | added | Joel David Hamkins | I guess one should add an extra lottery component of adding a Cohen real to $Q$, for the same reason as in your EDIT. And couldn't you have done that in your edit instead of using the infinite product? | |
| Jul 26, 2023 at 13:30 | history | edited | Calliope Ryan-Smith | CC BY-SA 4.0 |
Correcting a mistake
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| Jul 26, 2023 at 13:16 | comment | added | Joel David Hamkins | I like your (iii) but not (i) example—I had been looking at more complicated examples. A similar perhaps simpler example would be to let $P$ be the lottery sum of the collapses of $\omega_k$ for even $k$ and $Q$ the same for odd $k$. | |
| Jul 26, 2023 at 13:07 | comment | added | Calliope Ryan-Smith | @newaccount You're right, thanks for pointing it out. I believe that (iii) is independent of (i), and have updated my answer to reflect this. | |
| Jul 26, 2023 at 13:07 | history | edited | Calliope Ryan-Smith | CC BY-SA 4.0 |
Error in the answer, corrected.
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| Jul 26, 2023 at 10:26 | comment | added | new account | Why is (iii) stronger than (i) and (ii)? In (i) there is the extra requirement that $V[G]=V[H]$, which isn't clearly implied by (iii). | |
| Jul 26, 2023 at 9:13 | history | answered | Calliope Ryan-Smith | CC BY-SA 4.0 |