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# Questions tagged [ultrafilters]

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$\scr{B}$ is the base of a nonprincipal ultrafilter $\scr{U}$ on $\omega$ if 1. $\forall U,V\in\mathscr{B}~\exists T\in\mathscr{B}:~T\subset U\cap V$, 2. $\forall X\in\mathscr{U}~\exists U\in\mathscr{... 1answer 254 views ### Number of ultrafilters in an extender Assume GCH. Suppose$j: V\to M$is an elementary embedding such that$\mathrm{crit}(j)=\kappa$,${}^\kappa M \subset M$,$M\supset V_{\kappa+2}$. We can assume$j$is defined from some extender of ... 0answers 135 views ### Asymptotically discrete ultrafilters Definition 1. A ultrafilter$\mathcal U$on$\omega$is called discrete (resp. nowhere dense) if for any injective map$f:\omega\to \mathbb R$there is a set$U\in\mathcal U$whose image$f(U)$is a ... 1answer 187 views ### Relation between ultrafilters${\scr U}$and${\scr U} \otimes {\scr U}$[closed] If${\scr U}$and${\scr V}$are ultrafilters on non-empty sets$A$and$B$respectively, then the tensor product${\scr U}\otimes{\scr V}$is the following ultrafilter on$A\times B$: $$\big\{X\... 0answers 200 views ### Ultrapower of a field is purely transcendental Let F be a field, I a set, and U an ultrafilter on I. Is the ultrapower \prod_U F a purely transcendental field extension of F? According to Chapter VII, Exercise 3.6 from Barnes, Mack "... 1answer 75 views ### Dense subfilter of selective ultrafilter Given selective ultrafilter \mathcal{U} on \omega and dense filter \mathcal{F_1}=\{A\subset\omega~|~\rho(A)=1\}, where \rho(A)=\lim_{n\to\infty}\frac{|A\cap n|}{n} if the limit exists. Let \... 1answer 245 views ### Models of \mathsf{ZFC} with neither P- nor Q-points A P-point is an ultrafilter \scr Uon \omega such that for every function f:\omega\to\omega there is x\in {\scr U} such that the restriction f|_x is either constant, or finite-to-one. A Q... 1answer 97 views ### Dense filter and selective ultrafilter We say that \rho(A)=\lim_{n\to\infty}\frac{|A\cap n|}{n} is the density of subset A\subset\omega if the limit exists. Let us define the filter \mathcal{F_1}=\{A\subset\omega~|~\rho(A)=1\}. ... 0answers 197 views ### Metrically Ramsey ultrafilters On Thuesday I was in Kyiv and discussed with Igor Protasov the system of MathOverflow and its power in answering mathematical problems. After this discussion Igor Protasov suggested to ask on MO the ... 1answer 164 views ### The property of the dense subfilter of a selective ultrafilter Let us define the density of subset A\subset\omega :$$\rho(A)=\lim_{n\to\infty}\frac{|A\cap n|}{n}$$if the limit exists. Let \mathcal{F_1}=\{A\subset\omega~|~\rho(A)=1\}. \mathcal{F_1} is the ... 1answer 101 views ### Some kind of idempotence of dense filter In discussion of following questions question1, question2, question3 became clear (see definitions in question3 ) that for the Frechet filter \mathcal{N} we have \mathcal{N}\nsim\mathcal{N}\otimes\... 1answer 102 views ### Maximal elements in the Rudin-Keisler ordering Let \text{NPU}(\omega) be the set of non-principal [ultafilters] on \omega. The Rudin-Keisler preorder on \text{NPU}(\omega) is defined by$${\cal U} \leq_{RK} {\cal V} :\Leftrightarrow (\... 1answer 143 views ### Non-tensor-representable ultrafilters on$\omega$If${\cal U}$and${\cal V}$are ultrafilters on non-empty sets$A$and$B$respectively, then the tensor product${\cal U}\otimes{\cal V}$is the following ultrafilter on$A\times B$:$$\big\{X\... 1answer 125 views ### Minimal cardinality of a filter base of a non-principal uniform ultrafilters Let$\kappa$be an infinite cardinal. An ultrafilter${\cal U}$on$\kappa$is said to be uniform if$|R|=\kappa$for all$R\in{\cal U}$. If${\cal U}$is a non-principal ultrafilter on$\kappa$, ... 2answers 2k views ### Ultrafilters as a double dual Given a set$X$, let$\beta X$denote the set of ultrafilters. The following theorems are known:$X$canonically embeds into$\beta X$(by taking principal ultrafilters); If$X\$ is finite, then there ...

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