Timeline for Maximum size of a union of incomparable chains
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Apr 23, 2019 at 22:17 | vote | accept | CommunityBot | ||
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Apr 22, 2016 at 6:23 | answer | added | bof | timeline score: 2 | |
| Apr 21, 2016 at 12:52 | answer | added | Fedor Petrov | timeline score: 4 | |
| Sep 24, 2015 at 3:24 | answer | added | Noah Schweber | timeline score: 1 | |
| Aug 25, 2015 at 2:15 | answer | added | Josh Brown Kramer | timeline score: -1 | |
| Aug 21, 2015 at 11:37 | comment | added | user78375 | Yes. I want a collection $(s_i)_{i=1}^n$ of subsets of $A$ such that each set $s_i$ is a chain, while no member of $s_i$ is comparable to any member of $s_j$ when $i\neq j$, and the cardinality of $\cup_{i=1}^n s_i$ is as large as possible. | |
| Aug 20, 2015 at 22:30 | comment | added | Aaron Meyerowitz | So do you intend that the subsets $s_i \subseteq A \subset \mathbb{N}^{<\mathbb{N}}$ you are considering are required to be chains? | |
| Aug 20, 2015 at 18:12 | comment | added | user78375 | To clarify, a collection of sets being incomparable means that if we pick any members of two different sets from that collection, those members are incomparable. | |
| Aug 20, 2015 at 18:07 | comment | added | user78375 | "Comparable" is not defined for sets. Only incomparable. The sets you give fail to be incomparable, since $(5,5)$, a member of the second set, is comparable to a member of the first set, $(5,5,100)$, since $(5,5)$ is an initial segment of $(5,5,100)$. | |
| Aug 20, 2015 at 18:05 | comment | added | Boris Bukh | Still confused: An example of a subset of $\mathbb{N}^{<\mathbb{N}}$ is $\{ (1,5,3), (5,5,100)\}$. Another example is $\{(5,5)\}$. Are these comparable according to your definition? | |
| Aug 20, 2015 at 18:00 | comment | added | user78375 | I mean subsets. The definition of incomparable states that the members of $s_i$ are incomparable to the members of $s_j$ when $i$ and $j$ are distinct. | |
| Aug 20, 2015 at 17:55 | comment | added | Boris Bukh | I am confused --- do you really mean "... subsets of $\mathbb{N}^{<\mathbb{N}}$", or rather "... elements of $\mathbb{N}^{<\mathbb{N}}$". If it is former, then how do you compare two subsets of $\mathbb{N}^{<\mathbb{N}}$? | |
| Aug 20, 2015 at 17:47 | review | First posts | |||
| Aug 20, 2015 at 18:07 | |||||
| Aug 20, 2015 at 17:42 | history | asked | user78375 | CC BY-SA 3.0 |