Timeline for Is the axiom $\Diamond\Box\varphi\to\Box\Diamond\varphi$ in c.c.c. forcing potentialism equivalent to the productivity of c.c.c. forcing?
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| Feb 9, 2018 at 21:28 | history | edited | Not Mike | CC BY-SA 3.0 |
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| Feb 9, 2018 at 0:47 | history | edited | Not Mike | CC BY-SA 3.0 |
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| Feb 7, 2018 at 13:44 | history | edited | Not Mike | CC BY-SA 3.0 |
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| Feb 7, 2018 at 2:26 | history | edited | Not Mike | CC BY-SA 3.0 |
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| Feb 7, 2018 at 2:22 | comment | added | Not Mike | @JoelDavidHamkins thanks for pointing that out. Updated. | |
| Feb 7, 2018 at 2:18 | history | edited | Not Mike | CC BY-SA 3.0 |
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| Feb 7, 2018 at 1:46 | history | edited | Not Mike | CC BY-SA 3.0 |
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| Feb 7, 2018 at 1:40 | history | edited | Not Mike | CC BY-SA 3.0 |
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| Feb 6, 2018 at 22:09 | history | edited | Not Mike | CC BY-SA 3.0 |
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| Feb 6, 2018 at 22:00 | history | edited | Not Mike | CC BY-SA 3.0 |
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| Feb 6, 2018 at 20:57 | history | edited | Not Mike | CC BY-SA 3.0 |
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| Feb 6, 2018 at 20:40 | comment | added | Joel David Hamkins | Probably you know this, but for everyone else, it may help to point out that to show every c.c.c. partial order is c.c.c.necessarily c.c.c., as in the question, it suffices to consider only c.c.c. squares $\mathbb{P}^2$, where $\mathbb{P}$ is c.c.c., since for any two c.c.c. partial orders, you can take the lottery sum (side by side forcing), and the square of that includes the product. | |
| Feb 6, 2018 at 20:39 | comment | added | Joel David Hamkins | I see, that is $\Box(\mathbb{P}$ is ccc), or $\mathbb{P}$ is c.c.c. absolutely c.c.c., aka c.c.c.-necessarily c.c.c. | |
| Feb 6, 2018 at 20:36 | comment | added | Not Mike | @JoelDavidHamkins I've always used it to mean that the product of $\mathbb{P}$ with any other c.c.c. partial order, is again c.c.c. | |
| Feb 6, 2018 at 20:33 | comment | added | Joel David Hamkins | For me, the phrase "$\mathbb{P}$ is productively c.c.c." means merely that $\mathbb{P}\times\mathbb{P}$ is c.c.c., but clearly you are using this term to mean something else. What does it mean for you? Do you mean arbitrary finite-support products of $\mathbb{P}$ are c.c.c.? | |
| Feb 6, 2018 at 20:22 | history | edited | Not Mike | CC BY-SA 3.0 |
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| Feb 6, 2018 at 20:17 | comment | added | Not Mike | @JoelDavidHamkins added a partial result on the productivity of the c.c.c. under $(.2)_{ccc}$ | |
| Feb 6, 2018 at 20:16 | history | edited | Not Mike | CC BY-SA 3.0 |
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| Feb 6, 2018 at 20:02 | history | edited | Not Mike | CC BY-SA 3.0 |
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| Feb 6, 2018 at 14:45 | comment | added | Not Mike | @JoelDavidHamkins The partial order of finite homogeneous sets/cliques is c.c.c. and being powerfully c.c.c. just means the corresponding property holds for the clique po-set. I think this terminology was initially used by Todorčević. However I'm not certain | |
| Feb 6, 2018 at 14:42 | comment | added | Joel David Hamkins | Thank you for the answer -- I am totally excited by the idea that $(.2)_{ccc}$ implies $\neg$CH. Meanwhile, I'm also sorry to be slow, but what does it mean to say that a graph is c.c.c.? | |
| Feb 6, 2018 at 14:09 | history | edited | Not Mike | CC BY-SA 3.0 |
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| Feb 6, 2018 at 14:01 | history | answered | Not Mike | CC BY-SA 3.0 |