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| bio | website | |
|---|---|---|
| location | ||
| age | 31 | |
| visits | member for | 9 months |
| seen | Jan 1 at 11:12 | |
| stats | profile views | 23 |
I'm a grad student in mathematics (very near graduation) and have been employed as an analyst since graduating college in 2003.
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Oct 22 |
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Combinatorial Identity Starting with $n+r-1$ & maybe picking someone NOT on one team & using alg identity is overly complicated. Simpler: start with $n+r$, $n$ can be umpire, $r$ cannot. RHS: pick ump, then from $n+r-1$ left, pick $2r$ & divide them into 2 teams of $r$. LHS: pick the ump, then pick $r$ from $n+r-1$ left to be on a team. From the $n$ not on team one, pick $r$ to be on team 2. Whoops, our ump might be on team two. Forget him for a bit (divide by $n$). Pick someone from the $n-r$ left. If he can be ump, good. Otherwise swap him with the orig ump. All combos from RHS can be gotten this way. |
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Sep 26 |
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Discussion: Differing definitions for the rank of a set @AsafKaragila: I guess I don't see what you mean by there are new sets using the second definition. When I construct $V_\omega$, $\omega$ is not yet a set so I wouldn't be asking what its rank is. When I construct $V_{\omega+1}$ I now have the set $\omega$ and it's a subset of $V_\omega$ so it's rank is $\omega$. So I don't ever have a set whose rank has not been defined yet. It looks like a "labeling" difference to me. I don't see a hierarchical or structural difference. |
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Sep 3 |
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Discussion: Differing definitions for the rank of a set @AsafKaragila: lol, that's the opposite of what I was asking. I know and understand that. I was asking how does that relate to def 2? How does def 2 seem to introduce new info at limit points? |
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Sep 2 |
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Discussion: Differing definitions for the rank of a set @AndréNicolas: Ah the shortcomings of the written language. So much non-verbal communication is lost. Actually under the "Ordinals" and "Usefulness in proofs" sections above, I consider it a strength of the first definition that $\text{rank}(\alpha)=\alpha+1$ and every rank is a successor ordinal. Those conditions are universal. I think a hybrid definition would sacrifice the strengths of both definitions. |
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Sep 2 |
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Discussion: Differing definitions for the rank of a set @AndréNicolas: Before there is the empty set, there is nothing. The empty set is something - it's that nothingness put into a set. You can think of is as zero is the rank of nothingness: $\text{rank}(\ \ \ )=0$ |
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Sep 2 |
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Discussion: Differing definitions for the rank of a set @AsafKaragila: The part about sets existing once they are elements instead of subsets (hence subclasses) makes sense. Could you expound upon how the second definition implies (or seems to imply) new information is added at limit stages? (Not the philosophy - not why you want limits to be accumulations instead of introducing new material - that I already understand, but how the definition relates to this philosophy.) |
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Aug 26 |
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What is the name of $V_\alpha$? Well "the $\alpha$th level ..." is more creative than anything I was able to come up with just now. |
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Aug 25 |
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What does $H(\kappa)$ mean? Ah. I've been stuck in large-cardinal land for the past several days. Good catch. |
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Aug 25 |
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What does $H(\kappa)$ mean? @Asaf: Why did you change the (large-cardinals) tag to (cardinals)? Since $\kappa$ is a strongly inaccessible cardinal, which ZFC can't prove exists, doesn't that meet the criteria for (large-cardinals)? |
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Aug 25 |
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What does $H(\kappa)$ mean? @BrianM.Scott: Thanks. It was the revision history that showed me my view-source was out of date (even though I opened view-source after SE dynamically refreshed the page). |
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Aug 25 |
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What does $H(\kappa)$ mean? @BrianM.Scott: Much easier to read. :-) I was struggling to see what change you made to the TeX, but then I figured out that somehow with SE's dynamic updating, the page had refreshed but view-source was still giving me the old source until I did a manual refresh. |
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Aug 25 |
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What does $H(\kappa)$ mean? @BrianM.Scott, I'm new to TeX. Is there a way to add more space between the two rows of your piecewise definition for ${\bigcup}^n(x)$? |
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Aug 25 |
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Predicting the next vector given a known sequence @Ang Zhi Ping: I wasn't able to add this comment to his answer, but the Richardson extrapolation that 'example' showed is one of the "more advanced" Numerical Analysis extrapolation methods I mentioned. |
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Aug 21 |
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How does the supremum of the cardinalities of a set of ordinals relate to the supremum of the ordinals? Not necessary, but equivalent and useful. Thanks for all your help! Your posts are very clear and insightful. |
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Aug 21 |
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How does the supremum of the cardinalities of a set of ordinals relate to the supremum of the ordinals? @Arthur: What is the significance of continually saying "increasing sequence that is cofinal". Is it not sufficient to say "set that is cofinal"? The set has no sense or order between elements, so it's meaningless to say they are increasing, but doesn't "cofinal" capture everything that "increasing sequence" does - that for every element in parent set, there is an element in the subset that is greater? |
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Aug 21 |
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How does the supremum of the cardinalities of a set of ordinals relate to the supremum of the ordinals? @Asef: Ah, yes. Your large expression makes sense now. :-) I'm trying to wrap my head around your ordinal sum, but I need to take a break and rest my brain. After lunch I'll review this and read what Arthur wrote about cofinality and decide if I need to rewrite my proof with cases or not. Thanks for the help. |
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Aug 21 |
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How does the supremum of the cardinalities of a set of ordinals relate to the supremum of the ordinals? Well, both I guess; I suppose I've not fully grasped the implications of a sum of cardinals in the RHS. As I wrote, I've nearly proven everything just comparing the LHS and RHS for both regular and singular cardinals, not comparing either to $\kappa$. I see that sandwiching the RHS between the LHS and $\kappa$ gives equality (for regular cardinals), but I don't see why it is. I've almost been able to show the LHS$\ge$RHS relationship, but not quite. Doing so for both regular and singular cardinals would finish my proof, but if I need to rewrite using cases I will. |
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Aug 21 |
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How does the supremum of the cardinalities of a set of ordinals relate to the supremum of the ordinals? Everything is clear (including for regular $\kappa$, LHS=$\kappa$), except for the part you call trivial. Proofs gloss over and I'm missing something. Rather than using 2 cases, I can show for both cases that LHS$\le$RHS. Trying to show that LHS$\ge$RHS, I take the smallest $\beta$ from the LHS and use it in place of $\delta$ in the RHS. Showing $\beta\in\{RHS\}$ shows $\beta\ge\inf\{RHS\}$. Showing $\sum_{\xi\lt\beta}\kappa_\xi\le\kappa$ is easy. Trying to show $\kappa\le\sum_{\xi\lt\beta}\kappa_\xi$ is what led to my question above. |
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Aug 21 |
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How does the supremum of the cardinalities of a set of ordinals relate to the supremum of the ordinals? So it does depend on whether $\kappa$ is a limit cardinal or not. It seems to me the proof was applying the concept indiscriminately to all infinite cardinals. Either I'm still missunderstanding the proof, or it was hastily done (for classroom display). I'm going to take some time and digest the rest of what you wrote on cofinality, but since you addressed the actual question, I'll mark it as answered. You two know each other IRL? Cool. :-) |
