Questions tagged [ultrafilters]

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A nonprincipal ultrafilter that is not a $p$-point

On pg 76 of Jech's Set Theory, he proves the existence of a nonprincipal ultrafilter on $\omega$ that is not a $p$-point. Given a partition $\{A_n\}$ of $\omega$ into $\aleph_0$ infinite pieces, ...
andpe's user avatar
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7 votes
1 answer
294 views

Pseudo-intersections, splitting families, and ultrafilters

Suppose $U$ is a non-principal ultrafilter on $\omega$, and let us define $\tau(U)$ to be the minimum cardinality of a family $\mathcal{X}\subseteq U$ such that $\mathcal{X}$ does not have an infinite ...
Todd Eisworth's user avatar
10 votes
1 answer
539 views

Is the product of commuting ultrafilters an ultrafilter?

If $U$ is a filter on $X$ and $W$ is a filter on $Y$, their product is the filter $U\times W$ on $X\times Y$ generated by rectangles $A\times B$ where $A\in U$ and $B\in W$. In certain circumstances ...
Gabe Goldberg's user avatar
6 votes
1 answer
259 views

A question on simple $P_{\aleph_2}$-points

This question is motivated by discussion surrounding this MO question. An ultrafilter $U$ on $\omega$ is a simple $P_{\aleph_2}$-point if it is generated by a sequence $\langle X_\alpha:\alpha<\...
Todd Eisworth's user avatar
3 votes
0 answers
122 views

The existence of $T$-ultrafilters in ZFC

Looking at this MO-problem, my collegue Igor Protasov suggested to ask on Mathoverflow his old question on $T$-ultrafilters hoping that somebody on MO can solve it. First I recall the necessary ...
Taras Banakh's user avatar
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7 votes
1 answer
422 views

Ultrafilter on the ordinal $\omega^\omega$

For any ultrafilter $\mathcal{U}$ on $\omega$ and any finite $k$ we can construct tensor power $\mathcal{U}^{\otimes k}$ which is ultrafilter on $\omega^k$. Does there exist some natural extension of ...
ar.grig's user avatar
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3 votes
1 answer
235 views

The intersection of all normal ultrafilters on a measurable cardinal

Suppose $\kappa$ is a measurable cardinal. Let $W$ be the intersection of all normal ultrafilters on $\kappa$. I am interested in a precise characterization of the filter $W$. One sure way to ...
D. Hershko's user avatar
5 votes
1 answer
222 views

Club filter basis in $\omega_1$

My question is about existing of basis of club filter club($\omega_1$) with cardinality $c$. Does it exist?
ar.grig's user avatar
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3 votes
1 answer
239 views

Ultrapower of amenable group

Let $\Gamma$ be an amenable group. Consider its ultrapower $^*\Gamma$. It is known that $^*\Gamma$ need not be amenable. In fact, there is a stronger notion of uniform amenability for $\Gamma$ (...
BharatRam's user avatar
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Trying to construct the ultrafilter 2-monad on $\mathbf{Cat}$

By which I mean, following Bôrger's paper Coproducts and Ultrafilters, the terminal monad among those that preserve finite coproducts, if such a thing can be constructed. So far, what I have is, ...
twocubes's user avatar
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What is the "right" notion of exponentiation in $\beta \mathbb N$?

The Stone–Čech compactification $\beta \mathbb N$ of the positive integers has extensive applications in combinatorial number theory. A feature of $\beta \mathbb N$ that makes these applications ...
Jakub Konieczny's user avatar
2 votes
1 answer
167 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{...
user avatar
5 votes
1 answer
340 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 ...
Otto's user avatar
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7 votes
0 answers
172 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 ...
Taras Banakh's user avatar
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5 votes
1 answer
209 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\...
Dominic van der Zypen's user avatar
5 votes
0 answers
322 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 "...
Lisa S.'s user avatar
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2 votes
1 answer
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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 $\...
ar.grig's user avatar
  • 1,133
4 votes
1 answer
380 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 $Q$...
Dominic van der Zypen's user avatar
3 votes
1 answer
153 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 $\mathcal{F_1}=\{A\subset\omega~|~\rho(A)=1\}$. ...
ar.grig's user avatar
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8 votes
0 answers
224 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 ...
Taras Banakh's user avatar
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4 votes
1 answer
190 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 ...
ar.grig's user avatar
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117 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\...
ar.grig's user avatar
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1 answer
190 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 (\...
Dominic van der Zypen's user avatar
7 votes
1 answer
192 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\...
Dominic van der Zypen's user avatar
4 votes
1 answer
244 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 $\kappa$, ...
Dominic van der Zypen's user avatar
40 votes
2 answers
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 ...
Adam P. Goucher's user avatar
4 votes
1 answer
117 views

The example of the idempotent filter or subsets family with finite intersections property

From the answer of Andreas Blass and comments of Ali Enayat on my question Selective ultrafilter and bijective mapping it became clear that for any free ultrafilter $\mathcal{U}$ we have $\mathcal{U}\...
ar.grig's user avatar
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4 votes
1 answer
202 views

Nonexistence of a 'product universal' compact Hausdorff pseudotopological space?

I'm largely following the definitions of this paper, but I will replicate the relevant ones here. I'm taking a pseudotopological space to be a set $X$ together with a relation $\rightarrow$ on the ...
James Hanson's user avatar
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6 votes
1 answer
284 views

On infinite combinatorics of ultrafilters

Let $\mathcal{U}$ be (selective) ultrafilter and $\omega = \coprod_{i<\omega}S_i$ be partition of $\omega$ with small sets $S_i\notin\mathcal{U}$. All $S_i$ are infinite. Does there exist a system ...
ar.grig's user avatar
  • 1,133
12 votes
1 answer
487 views

Selective ultrafilter and bijective mapping

For arbitrary (selective) ultrafilter $\mathcal{F}$ does there exist bijection $\phi:[\omega]^2\to\omega$ with the property: $\forall B\in\mathcal{F} : \phi([B]^2)\in\mathcal{F}$ ?
ar.grig's user avatar
  • 1,133
3 votes
0 answers
68 views

A notion of largeness somewhere between $\mathrm{IP}_+$ and $\mathrm{IP}_+^*$

It is a well known fact that a set $A \subset \mathbb{N}$ is $\mathrm{IP}$ if and only if there exists an idempotent $p \in \beta \mathbb{N}$ (i.e., $p+p=p$) such that $A \in p$. Similarly, $B \subset ...
Jakub Konieczny's user avatar
6 votes
1 answer
330 views

Valuation Rings and Ultrafilters

I notice there is a certain similarity between the definition of a valuation ring and the definition of an ultrafilter. To begin, take a field $K$ and let $\mathcal{A}$ be the set of subrings of $K$. ...
user avatar
4 votes
1 answer
140 views

Complexity of ultrafilter limits

Let $\mathscr{F}$ be a free ultrafilter on $\mathbf{N}$ and, for each $A\subseteq \mathbf{N}$ and $n \in \mathbf{N}$, define $$ d_n(A):=\frac{|A\cap [1,n]|}{n}. $$ Question. Considering $\mathcal{P}...
Paolo Leonetti's user avatar
3 votes
1 answer
116 views

Ultrafilters preserving infinite joins

A filter $U$ over a boolean algebra (isomorphic to a powerset) $A$ "preserves" a join $a = \bigcup_{i\in I}a_i$, if $a\in U$ implies $a_i\in U$ for some $i\in I$. (A join $a$ is infinite if $I$ is.) ...
Michal Walicki's user avatar
15 votes
1 answer
699 views

Completeness number of ultrafilters

In what I write below, by "ultrafilter" I mean a non-principal ultrafilter. Given an ultrafilter $U$ on some set $S$, let $\mu$ be the least cardinal such that $U$ is $\mu$-complete but not $\mu^+$-...
almost nowhere's user avatar
9 votes
3 answers
487 views

A property of an ultrafilter

Let $\mathcal U$ be a free ultrafilter on a set $X$. For $n\in\mathbb N$ let $\mathcal F$ be a family of $n$-element subsets of $X$ such that $\bigcup\mathcal F\in\mathcal U$. Question. Is there a ...
Taras Banakh's user avatar
  • 40.9k
2 votes
1 answer
140 views

Are separability and ccc equivalent for closed subspaces of $\beta N$?

Let $\beta \mathbb N$ be the Stone-Cech compactification of the integers. Then $\beta \mathbb N\setminus \mathbb N$ is non-separable because if fails the ccc condition, that is, it has an uncountable ...
user129208's user avatar
8 votes
5 answers
463 views

Ideals on $\mathbb N$ and large sets that have small intersection

Let $\mathcal I$ be a (non-principal) ideal of subsets of $\mathbb N$. Suppose that every family $\mathcal{A} \subset \wp(\mathbb N)\setminus \mathcal I$ with the following property is countable: $$A,...
Tomasz Kania's user avatar
  • 11.3k
3 votes
0 answers
206 views

Identification of ultrafilters with measures

We know that each ultrafilter $p$ on $\mathbb{N}$ can be identified with a finitely additive $\{0,1\}$-valued probablity measure $\mu_{p}$ on the power set of $\mathbb{N}$. Now my question is which ...
mahdi meisami's user avatar
4 votes
0 answers
115 views

Selectors for bases of ultrafilters

If $\mathcal{U}$ is a selective (Ramsey) ultrafilter on $\omega$ and $\mathcal{B}$ is its base, then for every sequence $A_0\supsetneq A_1\supsetneq A_2\supsetneq\ldots$ in $\mathcal{B}$ there exists $...
Damian Sobota's user avatar
3 votes
2 answers
365 views

Selective ultrafilter on $\omega$ is normal. Clear proof

In this question I had asked about proof of the property of selective ultrafilter. As was answered, the proof is trivial if we know that ultrafilter is selective iff it is Ramsey ultrafilter. The ...
ar.grig's user avatar
  • 1,133
3 votes
2 answers
521 views

What are the components of the Stone-Cech Remainder?

Suppose $X = \displaystyle\bigsqcup_{i \in I} X_i$ is the disjoint union of infinitely many continua. The components of the Stone-Cech remainder $X^*$ can be described as follows. Treat $I$ as a ...
Daron's user avatar
  • 1,761
4 votes
0 answers
220 views

Set of subsequences with the same ultrafilter limit of the original sequence

Let $\mathscr{U}$ be a free ultrafilter on the positive integers $\mathbf{N}$ and fix $U \in \mathscr{U}$ such that $U$ is not cofinite (thanks J.D.Hamkins for the correction.) Consider the natural ...
Paolo Leonetti's user avatar
22 votes
2 answers
1k views

Ultrafilters and diagonal arguments

Is there a diagonal argument to show that if $x$ is infinite then ${\cal P}(x)$ (the power set of $x$) is smaller than $\beta x$ (the set of ultrafilters on $x$)? (Added later. I tried commenting ...
Thomas Forster's user avatar
8 votes
0 answers
226 views

Is there a 'local' version of Near Coherence of Filters?

The axiom Near Coherence of Filters (NCF) is known to be independent of ZFC. Axiom (NCF I): For any two free ultrafilters $\mathcal D$ and $\mathcal E$ on $\mathbb N$, there exist finite-to-one ...
Daron's user avatar
  • 1,761
6 votes
1 answer
474 views

Permutation on $\omega$ and Ramsey ultrafilter

let $\pi:\omega\to\omega$ be permutation and $\mathcal{F}$ is Ramsey selective ultrafilter on $\omega$. There are uncountable many increasing subsequences of $\pi$. Can one proof that one of them has ...
ar.grig's user avatar
  • 1,133
7 votes
0 answers
205 views

Possible cofinalities of cuts of ultraproducts

Suppose $\kappa$ is a regular cardinal, $\bar{\mu}=(\mu_i: i<\kappa)$ is an increasing sequence of regular cardinals ($>\kappa$) and set $pcut(\bar \mu)=\{ (\lambda_1, \lambda_2):$ for some ...
Mohammad Golshani's user avatar
3 votes
1 answer
142 views

The Wallman and interval topologies on non-principal ultrafilters with the Rudin-Keisler preorder

If $(P,\leq)$ is a pre-odered set (that is, $\leq$ is a reflexive and transitive relation) and $x\in P$, we set $(\uparrow_{\leq} x) = \{p\in P: p\geq x\}$ and $(\downarrow_{\leq} x) = \{p\in P: p\leq ...
Dominic van der Zypen's user avatar
4 votes
1 answer
175 views

"Gaps" in the Rudin-Keisler ordering

If $(P,\leq)$ is a poset and $p\in P$, then we say that $p$ is the lower part of a gap there is $q \in P$, $q>p$ such that $[p,q] = \{p,q\}$. (This is equivalent to the statement that $(\uparrow p) ...
Dominic van der Zypen's user avatar
2 votes
1 answer
272 views

Infima 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 (\exists f:...
Dominic van der Zypen's user avatar