User chris lambie-hanson - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T15:54:03Z http://mathoverflow.net/feeds/user/26002 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/119506/existence-of-scales-with-special-properties Existence of scales with special properties Chris Lambie-Hanson 2013-01-21T19:39:50Z 2013-01-30T19:28:06Z <p>Let $\kappa$ be a singular cardinal, and let $\langle \kappa_i \mid i&lt;\mathrm{cf}(\kappa) \rangle$ be an increasing sequence of regular cardinals cofinal in $\kappa$. Recall that a scale on $\Pi_{i&lt;\mathrm{cf}(\kappa)} \kappa_i$ is a sequence $\langle f_\alpha \mid \alpha &lt; \kappa^+ \rangle$ such that:</p> <ol> <li>For every $\alpha &lt; \kappa^+$, $f_\alpha \in \Pi_{i&lt;\mathrm{cf}(\kappa)} \kappa_i$.</li> <li>For every $\alpha &lt; \beta &lt; \kappa^+$, there is $i &lt; \mathrm{cf}(\kappa)$ such that <code>$f_\alpha &lt;_i f_\beta$</code>, i.e. for every $j\geq i$, $f_\alpha(j) &lt; f_\beta(j)$.</li> <li>For every $g\in \Pi_{i&lt;\mathrm{cf}(\kappa)} \kappa_i$, there is $\alpha &lt; \kappa^+$ and $i &lt; \mathrm{cf}(\kappa)$ such that <code>$g &lt;_i f_\alpha$</code>.</li> </ol> <p>Question: Is it consistent that there is a scale on $\Pi_{i&lt;\mathrm{cf}(\kappa)} \kappa_i$ such that, for every $\beta &lt; \kappa^+$ and every $i&lt;\mathrm{cf}(\kappa)$, <code>$\left|{\{\alpha &lt; \beta \mid f_\alpha &lt;_i f_\beta\}}\right| &lt; \kappa$</code> ?</p> <p>My intuition is that the answer should be no, but I haven't been able to find a proof.</p> http://mathoverflow.net/questions/119506/existence-of-scales-with-special-properties/120353#120353 Answer by Chris Lambie-Hanson for Existence of scales with special properties Chris Lambie-Hanson 2013-01-30T19:28:06Z 2013-01-30T19:28:06Z <p>I have a negative answer assuming some mild cardinal arithmetic assumptions. Namely, if $(\kappa_i)^i &lt; \kappa$ for every $i&lt;\mathrm{cf}(\kappa)$, then there can be no scale with the desired property. This is true, for example, whenever $\mathrm{cf}(\kappa) = \omega$ or $\kappa$ is strong limit. We also make the harmless assumption that $\mathrm{cf}(\kappa) &lt; \kappa_0$.</p> <p>Assume for sake of contradiction that $\langle f_\alpha \mid \alpha &lt; \kappa^+ \rangle$ is such a scale. For $j&lt;\mathrm{cf}(\kappa)$, define $g_j \in \Pi_{i&lt;\mathrm{cf}(\kappa)}\kappa_i$ as follows: Using the fact that $(\kappa_j)^j &lt; \kappa$, fix $B_j \subseteq \kappa^+$ and $f \in \Pi_{i\leq j}\kappa_i$ such that $\left|{B_j}\right|=\kappa_j$ and, for every $\alpha \in B_j$ and $i\leq j$, $f_\alpha(i)=f(i)$. For $i\leq j$, let $g_j(i)=f(i)+1$. For $i>j$, let <code>$g_j(i)=\sup(\{f_\alpha(i)+1 \mid \alpha \in B_j \})$</code>. Now define $g \in \Pi_{i&lt;\mathrm{cf}(\kappa)}\kappa_i$ by letting <code>$g(i)=\sup(\{g_j(i) \mid j&lt;\mathrm{cf}(\kappa) \})$</code>. Finally, find $\beta &lt; \kappa^+$ and $i&lt;\mathrm{cf}(\kappa)$ such that <code>$g &lt;_i f_\beta$</code>. Letting $B = \bigcup_{j&lt;\mathrm{cf}(\kappa)}B_j$, we have that $\left|{B}\right| = \kappa$ and <code>$f_\alpha &lt;_i f_\beta$</code> for every $\alpha \in B$. Contradiction.</p> http://mathoverflow.net/questions/112733/relation-between-neg-square-kappa-and-the-tree-property-at-kappa/113777#113777 Answer by Chris Lambie-Hanson for Relation between $\neg \square(\kappa)$ and the tree property at $\kappa$. Chris Lambie-Hanson 2012-11-18T22:27:56Z 2012-11-18T22:27:56Z <p>The answers to the second and third questions are no and yes, respectively. I don't know the answer to the first question.</p> <p>For the second question, let $\lambda$ be regular and let $\kappa > \lambda$ be weakly compact. Then forcing with $\mathrm{Coll}(\lambda, &lt;\kappa$) yields a model in which $\kappa = \lambda^+$, $\square(\kappa)$ fails, and, since $\lambda^{&lt;\lambda}=\lambda$, there is a special $\kappa$-Aronszajn tree, so the tree property fails.</p> <p>For the third question, the usual construction of a special Aronszajn tree from a weak square sequence using minimal walks (see, for example, section 5.1 of Cummings' "Notes on Singular Cardinal Combinatorics") still yields a $\kappa$-Aronszajn tree when applied to a $\square(\kappa)$-sequence when $\kappa$ is regular, so $\square(\kappa)$ implies the failure of the tree property. </p> http://mathoverflow.net/questions/105655/square-and-stationary-reflection Square and stationary reflection Chris Lambie-Hanson 2012-08-27T20:10:33Z 2012-08-27T20:10:33Z <p>It is easily shown that, for any uncountable infinite cardinal $\kappa$, $\square_\kappa$ implies that for any stationary $S\subseteq \kappa^+$, there exists a stationary $T\subseteq S$ such that $T$ does not reflect at (i.e. is not stationary in) any $\alpha&lt;\kappa$ of uncountable cofinality. The standard proof does not go through, however, when $\square_\kappa$ is replaced by the weaker notion of $\square(\kappa^+)$. Is $\square(\kappa^+)$ compatible with stationary reflection? More precisely, if $\kappa$ is an uncountable infinite cardinal, is $\square(\kappa^+)$ consistent with the statement "every stationary $S\subseteq \kappa^+$ consisting of ordinals of cofinality $&lt;\kappa$ reflects at some $\alpha&lt;\kappa^+$"?</p> http://mathoverflow.net/questions/119506/existence-of-scales-with-special-properties/119609#119609 Comment by Chris Lambie-Hanson Chris Lambie-Hanson 2013-01-23T18:19:11Z 2013-01-23T18:19:11Z Also, re.&quot;having a small cofinally interleaved sequence implies that it holds&quot;, 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 <code>$f&#95;\alpha &lt;&#95;i f&#95;\beta$</code>. For example, a member of this cofinally interleaved family could be <code>$&lt;^&#42;$</code>-above $\kappa$-many of the relevant $f_\alpha$s. http://mathoverflow.net/questions/119506/existence-of-scales-with-special-properties/119609#119609 Comment by Chris Lambie-Hanson Chris Lambie-Hanson 2013-01-23T18:16:24Z 2013-01-23T18:16:24Z A scale is always linearly ordered by $&lt;$ 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. http://mathoverflow.net/questions/119506/existence-of-scales-with-special-properties/119609#119609 Comment by Chris Lambie-Hanson Chris Lambie-Hanson 2013-01-23T17:06:22Z 2013-01-23T17:06:22Z 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 $&lt;$ is not necessarily a tree - it is quite possible that the $&lt;$-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 $&lt;\kappa$. In fact, the tree would have to have height $&lt;\kappa$. http://mathoverflow.net/questions/119506/existence-of-scales-with-special-properties/119609#119609 Comment by Chris Lambie-Hanson Chris Lambie-Hanson 2013-01-23T04:00:01Z 2013-01-23T04:00:01Z 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 $\langle f_\alpha \mid \alpha &lt; \beta \rangle$ for $\beta &lt; \kappa^+$ 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.