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      Questions tagged [ultrafilters]

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      97 views

      Non-existence of countable base of arbitrary ultrafilter

      $\scr{B}$ 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{...
      5
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      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 ...
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      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 ...
      5
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      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\...
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      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 "...
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      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 $\...
      4
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      1answer
      245 views

      Models of $\mathsf{ZFC}$ with neither $P$- nor $Q$-points

      A $P$-point is an ultrafilter $\scr U$on $\omega$ 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 $Q$...
      3
      votes
      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 $\mathcal{F_1}=\{A\subset\omega~|~\rho(A)=1\}$. ...
      6
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      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 ...
      4
      votes
      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 ...
      1
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      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\...
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      1answer
      102 views

      Maximal elements in the Rudin-Keisler ordering

      Let $\text{NPU}(\omega)$ be the set of non-principal [ultafilters][1] on $\omega$. The Rudin-Keisler preorder on $\text{NPU}(\omega)$ is defined by $${\cal U} \leq_{RK} {\cal V} :\Leftrightarrow (\...
      7
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      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\...
      3
      votes
      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 $\kappa$, ...
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      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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