Timeline for Existence of a strong antichain
Current License: CC BY-SA 4.0
13 events
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| Oct 22, 2020 at 13:57 | history | edited | Attila Joó | CC BY-SA 4.0 |
added 40 characters in body
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| Oct 22, 2020 at 13:54 | comment | added | Attila Joó | I think I figured out what is the reason of the misunderstanding. An antichain of a poset is a set of pairwise incomparable elements in my question. In the context of forcing this term is used differently: set of pairwise incompatible elements. Since you wrote 'stronger condition' I guess you used the forcing related interpretation. | |
| Oct 22, 2020 at 13:28 | comment | added | Attila Joó | Why do you think that there is no $ a'\in A$ which is comparable with both $p$ and $q$? For example $q$ itself can be in $A$ and hence is suitable choice for $a'$. | |
| Oct 22, 2020 at 10:51 | comment | added | Johannes Schürz | Then $q \subset p$, but there cannot exist an $a' \in A$ such that both $p$ and $q$ are comparable with $a'$. So either I misunderstand your definition of strong antichain, or non-trivial antichains can NEVER exist in $\mathcal{P} (\omega)\,/\, \text{fin}$. | |
| Oct 22, 2020 at 10:50 | comment | added | Johannes Schürz | No, I never said that my poset (actually this is also your poset, which you consider in your question) has a largest or smallest element (I just forgot to exclude them). Assume that $A \subset \mathcal{P} (\omega)\,/\, \text{fin}$ is a non-trivial strong antichain. In particular there exists $a \in A$ such that $\omega \setminus a$ is infinite. Now let $b \subset a$ be infinite such that also $a \setminus b$ is infinite. Define $q:=b$ and $p:= b \cup (\omega \setminus a)$. | |
| Oct 21, 2020 at 21:38 | comment | added | Attila Joó | If I understood correctly in your first comment you consider a poset which has a largest and smallest element and claim that it has no strong antichain. In my reaction pointed out that it is wrong since in this case we always have. The existence of a strong non-trivial antichain in your poset is consistent with ZFC, my question is if it is actually provable in ZFC or independent of it. | |
| Oct 20, 2020 at 20:17 | comment | added | Johannes Schürz | Sure, $a$ must be co-infinite in my comment. But I don't see why there should be a non-trivial strong antichain. My comment above seems to prove the opposite, right? | |
| Oct 20, 2020 at 19:29 | comment | added | Attila Joó | If the poset has a largest (or smallest) element $a$, then $A:=\{a \}$ is a strong antichain since $a$ is comparable with everybody. (The deletion of the largest and smallest element is to exclude these trivial strong antichains.) | |
| Oct 19, 2020 at 14:00 | comment | added | Johannes Schürz | I don't quite get your definition of strong antichain: The way I understand your definition, it cannot exist in $\mathcal{P}(\omega) \, /\, \text{fin}$: Let $A \subset P$ be a strong antichain and pick an $a \in A$. Define $q$ to be a stronger condition such that $a \setminus q$ is infinite. Set $p:=\omega \setminus (a \setminus q)$. Then (if your notation is to force upwards) $p \leq q$, but there cannot exist $b \in A$ which is both comparable with $p$ and $q$. | |
| S Oct 15, 2020 at 21:31 | history | suggested | CommunityBot | CC BY-SA 4.0 |
Corrected typo, added name of paper
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| Oct 15, 2020 at 19:45 | review | Suggested edits | |||
| S Oct 15, 2020 at 21:31 | |||||
| Oct 15, 2020 at 18:11 | review | First posts | |||
| Oct 15, 2020 at 18:25 | |||||
| Oct 15, 2020 at 18:10 | history | asked | Attila Joó | CC BY-SA 4.0 |