bio  website  math.huji.ac.il/… 

location  Jerusalem  
age  28  
visits  member for  2 years, 9 months 
seen  29 mins ago  
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2d

answered  Graph on the set of all functions $f:\mathbb{N}\to\mathbb{N}$ 
2d

awarded  Enlightened 
2d

awarded  Nice Answer 
2d

awarded  Yearling 
2d

awarded  Editor 
2d

revised 
Graph $G$ with $\omega(G) = 2$ but $\chi(G) \geq \aleph_0$
added 293 characters in body 
2d

answered  Graph $G$ with $\omega(G) = 2$ but $\chi(G) \geq \aleph_0$ 
May 9 
answered  monochromatic cyclefree colouring of the complete graph on R? 
Apr 24 
comment 
name for an intermediate notion between huge and 2huge
In Magidor and Shelah's paper, "The tree property at successors of singular cardinals," a cardinal $\kappa$ is called $\tau$huge if there is an elementary embedding $j:V \rightarrow M$ with critical point $\kappa$ such that $\kappa < \tau < j(\kappa) < j(\tau)$ and $M^{j(\tau)} \subseteq M$. 
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 $<\kappa$. 