Timeline for Descending almost-contained subsets of $\omega$
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 11, 2018 at 15:27 | comment | added | Joel Adler | @MonroeEskew Thanks! That's what I thought. I was a bit confused as we use ordinals, so the index set is assumed to be well-ordered. | |
| Dec 11, 2018 at 12:28 | comment | added | Monroe Eskew | @JoelAdler in the latter case, we are saying that the index set is well-ordered. | |
| Dec 11, 2018 at 11:43 | comment | added | Joel Adler | @MonroeEskew Sorry for asking the following stupid question: What is the difference between a sequence $(A_\alpha)$ and a well-ordered sequence $(A_\alpha)$? | |
| Dec 11, 2018 at 10:58 | comment | added | Monroe Eskew | @JoelAdler what? | |
| Dec 11, 2018 at 6:03 | comment | added | Joel Adler | Sorry for asking the following stupid question: What is the difference between a sequence $(A_\alpha)$ and a sequence $(A_\alpha)$? | |
| Dec 10, 2018 at 18:19 | history | edited | Nik Weaver | CC BY-SA 4.0 |
deleted 135 characters in body
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| Dec 10, 2018 at 18:18 | comment | added | Nik Weaver | YCor is right, you want to require $A_\beta\setminus A_\alpha$ to be finite for all $\beta > \alpha$. | |
| Dec 10, 2018 at 17:56 | comment | added | YCor | So you just consider the poset $\omega_1$ as a disjoint union of plenty of copies of the poset $\omega$. | |
| Dec 10, 2018 at 17:53 | comment | added | Forever Mozart | @YCor For my particular purposes I'm not sure I need any particular relation between non-consecutive sets. But you suggest an interesting variation. | |
| Dec 10, 2018 at 17:51 | comment | added | YCor | It's not what I mean. I mean that one should ask some relation (other than just being distinct) between $A_\alpha$ and $A_\beta$ for all $\alpha<\beta$, not only between $A_\alpha$ and $A_{\alpha+n}$ for $n<\omega$. | |
| Dec 10, 2018 at 17:49 | comment | added | Forever Mozart | @Nik Thank you, your assumption was correct too. I would also be interested in a more constructive (inductive) proof if it exists, because I am trying to apply this to construct a particular example. | |
| Dec 10, 2018 at 17:44 | history | edited | Nik Weaver | CC BY-SA 4.0 |
added 134 characters in body
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| Dec 10, 2018 at 17:44 | comment | added | Asaf Karagila♦ | One should add, CH has nothing to do with it. Although it is consistent that a maximal one does have length $\omega_1$ which is strictly smaller than the continuum. | |
| Dec 10, 2018 at 17:42 | history | answered | Nik Weaver | CC BY-SA 4.0 |