Chris Lambie-Hanson
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Registered User
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Jan 31 |
awarded | ● Scholar |
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Jan 31 |
accepted | Existence of scales with special properties |
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Jan 30 |
answered | Existence of scales with special properties |
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Jan 23 |
comment |
Existence of scales with special properties 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 |
comment |
Existence of scales with special properties 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. |
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Jan 23 |
comment |
Existence of scales with special properties 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 $<\kappa$. |
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Jan 23 |
comment |
Existence of scales with special properties 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. |
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Jan 21 |
asked | Existence of scales with special properties |
