Timeline for The sum of two well-ordered subsets is well-ordered
Current License: CC BY-SA 4.0
16 events
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| Jan 20, 2021 at 22:37 | comment | added | SSequence | I guess another alternative definition for $first(r)$ could go like: "Find the smallest value $\alpha<p$ such that $a_\alpha+y=r$ (where $y \in B$). Then $first(r)=a_\alpha$." | |
| Jan 20, 2021 at 21:44 | history | edited | SSequence | CC BY-SA 4.0 |
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| Jan 20, 2021 at 21:44 | comment | added | SSequence | Sorry for too much bumping, but I think I see the mistake made (and hence reason for downvote). The definition for first, second was too loose/incorrect. | |
| Jan 20, 2021 at 21:39 | history | edited | SSequence | CC BY-SA 4.0 |
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| Jan 20, 2021 at 9:40 | history | edited | SSequence | CC BY-SA 4.0 |
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| Jan 20, 2021 at 9:31 | history | edited | SSequence | CC BY-SA 4.0 |
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| Jan 20, 2021 at 4:47 | comment | added | SSequence | Is there a specific reason for the downvote? What is the specific issue (or a major mistake) in this answer? | |
| Jan 19, 2021 at 17:49 | comment | added | SSequence | It seems that instead of writing $\alpha_1=min\{\alpha \in Ord:first(f(x))=a_{\alpha} \wedge x \in \mathbb{N} \wedge x>n_0 \}$ it would probably be better to write something like: $\alpha_1=min\{\alpha \in Ord: \exists x \in \mathbb{N} (first(f(x))=a_{\alpha} \wedge x>n_0) \}$. | |
| Jan 18, 2021 at 20:02 | history | edited | SSequence | CC BY-SA 4.0 |
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| Jan 18, 2021 at 19:40 | history | edited | SSequence | CC BY-SA 4.0 |
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| Jan 18, 2021 at 19:33 | history | edited | SSequence | CC BY-SA 4.0 |
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| Jan 18, 2021 at 19:23 | history | edited | SSequence | CC BY-SA 4.0 |
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| Jan 18, 2021 at 19:22 | history | undeleted | SSequence | ||
| Jan 18, 2021 at 18:24 | history | deleted | SSequence | via Vote | |
| Jan 18, 2021 at 18:05 | history | edited | SSequence | CC BY-SA 4.0 |
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| Jan 18, 2021 at 17:59 | history | answered | SSequence | CC BY-SA 4.0 |