bio  website  math.huji.ac.il/… 

location  Jerusalem  
age  28  
visits  member for  2 years, 5 months 
seen  5 hours ago  
stats  profile views  216 
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 HarringtonShelah 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 