Timeline for Existence of scales with special properties
Current License: CC BY-SA 3.0
9 events
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| Jan 23, 2013 at 18:19 | comment | added | Chris Lambie-Hanson |
Also, re."having a small cofinally interleaved sequence implies that it holds", are you talking about a sequence cofinally interleaved with the entire scale or with an initial segment of the scale? In either case, I don't see how such a sequence contradicts the failure of my property. It seems quite possible that there is a small cofinally interleaved family and $\kappa$-many $\alpha$ such that $f_\alpha <_i f_\beta$. For example, a member of this cofinally interleaved family could be $<^*$-above $\kappa$-many of the relevant $f_\alpha$s.
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| Jan 23, 2013 at 18:16 | comment | added | Chris Lambie-Hanson | A scale is always linearly ordered by $<$ mod $I$, though, so it certainly won't produce an Aronszajn tree. And while the ultrafilter in the trichotomy theorem does extend the dual filter to the ideal, it is still the case that being cofinally interleaved modulo the ultrafilter is a weaker statement than being cofinally interleaved modulo the ideal. | |
| Jan 23, 2013 at 17:12 | comment | added | Eran | - The $<$ in my comments is a typo, I meant $<$ mod $I$. - Please read the Trichotomy theorem carefully - the Ultrafilter agrees with the give Ideal. | |
| Jan 23, 2013 at 17:06 | comment | added | Chris Lambie-Hanson | Even in the Trichotomy theorem, the small cofinally interleaved family of functions is only cofinally interleaved modulo an ultrafilter, not necessarily the bounded ideal. Also, the scale ordered by $<$ is not necessarily a tree - it is quite possible that the $<$-predecessors of a given $f_\alpha$ are not linearly ordered. Even if it were a tree, my condition would not imply that it had levels of size $<\kappa$. In fact, the tree would have to have height $<\kappa$. | |
| Jan 23, 2013 at 12:12 | comment | added | Eran | Re. "having a small cofinally interleaved sequence implies that it holds" - if you have such a (short - usually $2^{cf(κ)}$) interleaved sequence $S$, then you cannot have a $\kappa$-size set of $\lt_i f_\beta$ for any $\beta$ as you will have to have at least $\kappa$ elements in the short sequence $S$ - impossible. | |
| Jan 23, 2013 at 11:56 | comment | added | Eran | - Filters vs. ideals - a similar theorem (Trichotomy) exists for ideal increasing sequences of functions. - Re. e.u.b - you are right. In case of a scale each least upper bound of a sequence modulo an ideal, is also an exact upper bound (removing 3. in your question is more interesting). In that case let me suggest the following: order the scale on a tree, using the $\lt$ order. Thus assuming your condition, each level has size $\lt \kappa$. The fact that the scale is $\kappa^+$-long entails the $\kappa^+$-Aronszajn property, which we know is independent. | |
| Jan 23, 2013 at 4:00 | comment | added | Chris Lambie-Hanson | I'm not entirely sure what you're saying here. The entire scale certainly does have an exact upper bound, namely the function $g$ with $g(i)=\kappa_i$. On the other hand, I don't see how an initial segment $\langle f_\alpha \mid \alpha < \beta \rangle$ for $\beta < \kappa^+$ of the scale having an e.u.b. (and it will for stationarily many $\beta$) implies that my condition fails or that having a small cofinally interleaved sequence implies that it holds. Also, the Dichotomy theorem is about functions increasing modulo an ultrafilter, not modulo the bounded ideal. Please elaborate. | |
| Jan 22, 2013 at 23:26 | comment | added | Asaf Karagila♦ | Isn't it the trichotomy theorem? | |
| Jan 22, 2013 at 23:20 | history | answered | Eran | CC BY-SA 3.0 |