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Jan 23, 2014 at 20:09 vote accept Joseph Van Name
Oct 31, 2013 at 22:06 comment added Joseph Van Name Here the partition relation $\kappa\rightarrow[\alpha]^{2}_{\lambda,<\mu}$ means that if $f:[\kappa]^{2}\rightarrow\lambda$, then there is some subset $K\subseteq\kappa$ with $|K|=\alpha$ where $|f[[K]^{2}]|<\mu$. I am unsure if the condition that these conditions are necessary though.
Oct 31, 2013 at 22:06 comment added Joseph Van Name More generally, assume that if $\kappa,\mu,\lambda$ are cardinals with $\kappa$ regular, $\mu$ measurable or $\mu=\aleph_{0}$, and $\kappa\rightarrow[\kappa]_{\lambda,<\mu}^{2}$. Furthermore, assume that $\mathcal{U}$ is a $\mu$-complete ultrafilter on a set $I$ generated by a set of $\lambda$ many elements and $T_{i}$ is a tree on $\kappa$ of height $\kappa$ for $i\in I$. Then the only $\kappa$ branches on the trimmed ultraproduct of trees $\prod_{i\in I}T_{i}/(\mathcal{U})$ are the ones induced by the ultraproducts of the $\kappa$-branches of the $T_{i}'s$.
Oct 31, 2013 at 21:43 comment added Joseph Van Name Contrasting the case of trees of height $\omega_{1}$, if $\kappa$ is a weakly compact cardinal, $|I|<\kappa$, $\mathcal{U}$ is an ultrafilter on $I$ and $T_{i}$ is a tree of height $\kappa$ for $i\in I$, then the $\kappa$-branches of the trimmed ultraproduct of trees $\prod_{i\in I}T_{i}/(\mathcal{U})$ are simply the ultraproducts of the $\kappa$-branches of the trees $T_{i}$ (i.e. taking an ultraproduct does not add any more branches then we have to). The proof of this fact is a straightforward application of the partition relation for weakly compact cardinals.
Oct 31, 2013 at 21:17 answer added Ali Enayat timeline score: 7
Oct 31, 2013 at 1:32 answer added Andreas Blass timeline score: 10
Oct 31, 2013 at 0:37 answer added Goldstern timeline score: 16
Oct 30, 2013 at 19:55 comment added saf of interest: matwbn.icm.edu.pl/ksiazki/fm/fm118/fm118113.pdf
Oct 30, 2013 at 19:37 history asked Joseph Van Name CC BY-SA 3.0