328 reputation
24
bio website math.cmu.edu/~clambieh
location Pittsburgh
age 27
visits member for 2 years
seen 25 mins ago

Nov
6
awarded  Yearling
Nov
5
answered A Special Pair of Models for ZFC (New Version)
Oct
28
accepted Square and stationary reflection
Oct
28
comment Square and stationary reflection
Yes, a Harrington-Shelah style forcing construction works for every regular, uncountable $\kappa$. In fact, it can be shown that, in the generic extension, we have a $\square(\kappa^+)$ sequence whose clubs avoid a stationary subset of $S^{\kappa^+}_\kappa$. I'm not sure about the situation for singular $\kappa$, though.
May
26
awarded  Enthusiast
Jan
31
awarded  Scholar
Jan
31
accepted Existence of scales with special properties
Jan
30
answered Existence of scales with special properties
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.
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.
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 $<\kappa$.
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.
Jan
21
asked Existence of scales with special properties
Nov
18
awarded  Teacher
Nov
18
answered Relation between $\neg \square(\kappa)$ and the tree property at $\kappa$.
Nov
4
awarded  Supporter
Aug
27
awarded  Student
Aug
27
asked Square and stationary reflection