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Questions tagged [ultrafilters]

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Can there be a p-point ultrafilter that is 'aggressively non-Ramsey'?

These are fairly standard terms, but for the sake of completeness: An ultrafilter $\mathcal{U}$ on $\omega$ is a p-point if whenever $(A_n)_{n<\omega}$ is a partition of $\omega$ such that $A_n \...
James Hanson's user avatar
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5 votes
1 answer
198 views

Infinite tensor/Fubini product of ultrafilters

Given an infinite family $\{\mathcal{F}_{\lambda}$, $\lambda <\kappa\}$, $\kappa \geq \omega_0$, of (ultra)filters of a set $X$, how it is defined the infinite tensor/Fubini product $$\bigotimes_{\...
BTN's user avatar
  • 53
5 votes
0 answers
84 views

Comparing Mathias forcing notions relative to various filters

Let $\mathcal F$ be a (non-principle, non trivial, ...) filter on $\omega$. The Mathias Forcing relative to $\mathcal F$ is the forcing notion $\mathbb M(\mathcal F)$ consisting of pairs $(s, X)$ with ...
Corey Bacal Switzer's user avatar
12 votes
2 answers
743 views

Ultrafilter lemma for arbitrary lattice

Can someone kindly confirm whether the ultrafilter lemma for arbitrary (i.e., not necessarily Boolean) bounded lattices is equivalent to Zorn's lemma? To be precise, if $\mathbf{L} = (L, \leq, \land, \...
Menander I's user avatar
4 votes
1 answer
241 views

Supremum of infimum of measure of members of a free ultrafilter

For a set $A\subseteq \omega$ we let the upper density of $A$ be defined as $d^+(A) := \lim\sup_{n\to\infty}\frac{|A\cap(n+1)|}{n+1}$. Let $\text{FrU}(\omega)$ be the collection of free ultrafilters ...
Dominic van der Zypen's user avatar
2 votes
0 answers
113 views

An ultrafilter on $\omega_1$ with a nice Fubini product with an ultrafilter on $\omega$

Fix an ultrafilter $U$ on $\omega$ (that is, $U$ is an ultrafilter on the Boolean algebra of all subsets of $\omega$). Let $(f_\alpha \mid \alpha < \omega_1)$ be an increasing sequence in $\mathbb{...
Gawr Gura's user avatar
  • 153
4 votes
1 answer
136 views

Decomposition of an ultrafilter on the fibers of a map

Short version: If I have a map $f:Y \to I$, and $\mu$ an ultrafilter on $Y$, under what condition can $\mu$ be written as a limit/sum/integral of ultrafilters on the fibers of $f$ along the ...
Simon Henry's user avatar
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3 votes
0 answers
130 views

A p-point game with infinitely many ultrafilters

The following game-theoretic characterization of p-points is well known: Theorem A. An ultrafilter $D$ on the set $\omega$ of natural numbers is a p-point if and only if player I does not have a ...
Goldstern's user avatar
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12 votes
1 answer
618 views

Ultrafilter subtraction and "zero"

This is related to a couple recent MO/MSE questions of mine, namely 1,2. Belatedly, I've tweaked this post to remove an overly-ambitious secondary question; see the edit history if interested. Let $\...
Noah Schweber's user avatar
4 votes
0 answers
229 views

ultrapower(ultrapower)=ultrapower

Is there a nonprincipal ultrafilter $\omega$ on $\mathbb N$ such that for any metric space $M$ there is an isometry $$(M^\omega)^\omega\to M^\omega?$$ (In other words, the $\omega$-power of $\omega$-...
Anton Petrunin's user avatar
5 votes
1 answer
238 views

Ultrafilters of closed sets

The following definition should be standard, but let me state it just in case there is some ambiguity: If $\mathscr{L}$ is a set of subsets of a set $X$ that is closed under finite unions and ...
Gro-Tsen's user avatar
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5 votes
1 answer
227 views

Silver-like forcing preserves p-points (reference request)

A Silver forcing "below $2^n$" is defined e.g. in Definition 7.4.11 of [Tomek Bartoszyński and Haim Judah, Set Theory: on the structure of the real line, A. K. Peters, Wellesley, 1995.]. It ...
Andrzej's user avatar
  • 233
3 votes
2 answers
131 views

Is $(\omega+1)^\omega/{\cal U}$ complete for ${\cal U}$ free ultrafilter?

Let ${\cal U}$ be a free ultrafilter on $\omega$. Is the linearly ordered set $(\omega+1)^\omega/{\cal U}$ complete?
Dominic van der Zypen's user avatar
30 votes
3 answers
2k views

Are all free ultrafilters 'the same' in some sense?

Consider the set of ultrafilters $\beta(\mathbb N)$ on $\mathbb N$. Any function $f\colon\mathbb N\to\mathbb N$ extends to a function $\beta f\colon \beta \mathbb N \to \beta\mathbb N$. We say that ...
Squala's user avatar
  • 954
5 votes
1 answer
461 views

Can there be no "surprisingly averageable" second-order sentences?

Say that a second-order sentence $\varphi$ is averageable iff there exists some infinite cardinal $\kappa$ and some nonprincipal ultrafilter $\mathcal{U}$ on $\kappa$ such that for every $\kappa$-...
Noah Schweber's user avatar
16 votes
0 answers
848 views

Non-rigid ultrapowers in $\mathsf{ZFC}$?

Originally asked and bountied at MSE: Question: Can $\mathsf{ZFC}$ prove that for every countably infinite structure $\mathcal{A}$ in a countable language there is an ultrafilter $\mathcal{U}$ on $\...
Noah Schweber's user avatar
6 votes
1 answer
255 views

Are these two definitions of $\mathcal{U}$-Ramsey set equivalent?

Let $\mathcal{U}$ be an ultrafilter over $\omega$, and let $\mathcal{X} \subseteq [\omega]^\omega$. In two separate texts, there are two possible interpretations of a $\mathcal{U}$-Ramsey set, as ...
Clement Yung's user avatar
6 votes
1 answer
281 views

Ramsey ultrafilters on partial order

$\newcommand{\U}{\mathcal{U}}$ $\newcommand{\P}{\mathbb{P}}$ $\newcommand{\Q}{\mathbb{Q}}$ $\newcommand{\F}{\mathcal{F}}$ Recall the following equivalent definitions of a Ramsey ultrafilter over $\...
Clement Yung's user avatar
3 votes
2 answers
239 views

Is the set of $\kappa$-complete ultrafilters closed in $\beta X$?

Given an arbitrary set $X$, let $\beta X$ be the set of all ultrafilters over $X$. Consider endowing $\beta X$ with a topology consisting of the following open sets: $$ \{\mathcal{U} \in \beta X : A \...
Clement Yung's user avatar
4 votes
1 answer
201 views

Can Tychonoffs theorem for a countable number of spaces be proven with ZF plus the axiom of (countable) dependent choice?

It can be proven without any form of infinite choice that the product of two compact spaces (and thus any finite product) is compact, while on the other hand, it is well known that the general form of ...
saolof's user avatar
  • 1,733
4 votes
1 answer
298 views

On a completeness property of hyperreals

Let $\mathbb{R}_*=\mathbb{R}^\omega/\mathcal U$ for some ultrafilter $\cal U$. In the definitions of this question and assuming ZFC + CH there are only three types of cuts in $\mathbb{R}_*$: $(\omega,\...
ar.grig's user avatar
  • 1,135
9 votes
2 answers
881 views

SPOT as a conservative extension of Zermelo–Fraenkel

In Infinitesimal analysis without the Axiom of Choice, Hrbacek and Katz have shown that it is possible to formulate an axiomatic theory which provides a formalisation of calculus procedures which make ...
Hollis Williams's user avatar
3 votes
1 answer
240 views

"Good limit" of an uncountable sequence of elements of an ultrafilter

Let $U$ be an ultrafilter on $\mathcal{P}(\omega)$ and $\langle \sigma _\alpha \mid \alpha < \omega_1 \rangle$ be a sequence of elements of $U$. I know that the limit sup of $\sigma _\alpha$'s ($= \...
Gawr Gura's user avatar
  • 153
3 votes
1 answer
183 views

Grigorieff forcing and destruction of ultrafilters

I was interested in the Grigorieff forcing (you can read the definition here: How "much" does (Grigorieff) forcing destroy an ultrafilter?) I couldn't prove that it destroys ultrafilters, ...
D. Hershko's user avatar
3 votes
1 answer
176 views

${\frak b}$ and ${\frak d}$ in the Rudin-Keisler preordering

If $(Q,\leq)$ is any preordered set (that is, $\leq$ is a reflexive and transitive, but not necessarily anti-symmetric relation), then we say that $S\subseteq Q$ is unbounded if for all $q\in Q$ ...
Dominic van der Zypen's user avatar
1 vote
1 answer
135 views

Is the Rudin-Keisler ordering a continuous relation?

If $X, Y$ are topological, and $R\subseteq X\times Y$ we say that $R$ is continuous (from $X$ to $Y$) if for every $V\subseteq Y$ with $V$ open, we have $$R^{-1}(V) = \{u\in U: \exists v\in V:(u,v)\...
Dominic van der Zypen's user avatar
3 votes
2 answers
223 views

"Completion property" in $(\beta\omega,+)$

Let $\beta\omega$ be collection of all ultrafilters on $\omega$ (principal and non-principal). We endow $\beta\omega$ with an operation $+$ in the following way. For ${\bf a}, {\bf b}\in \beta\omega$, ...
Dominic van der Zypen's user avatar
2 votes
1 answer
164 views

Minimal components of the translation action on the Stone–Čech compactification

$\newcommand\Cb{C^\text b}$Let $\Cb(\mathbb R)$ be the C*-algebra formed by all bounded, continuous, complex valued functions on $\mathbb R$. Consider the action $\tau $ of $\mathbb R$ on $\Cb(\...
Black's user avatar
  • 471
4 votes
1 answer
213 views

Addition and Rudin-Keisler ordering in $\beta \omega$

$\DeclareMathOperator{\RK}{\mathrm{RK}}$Let $\beta\omega$ be the Stone-Cech compactification of $\omega$ with the discrete topology. We can endow $\beta\omega$ with an addition operation that extends ...
Dominic van der Zypen's user avatar
4 votes
0 answers
154 views

Finite pre-orders embeddable in the Rudin-Keisler ordering

$\DeclareMathOperator{\NPU}{\operatorname{NPU}}\DeclareMathOperator{\RK}{\,\mathrm{RK}}$A pre-ordered set is a pair $(P, \leq)$ where $P$ is a set and $\leq\subseteq P\times P$ is a reflexive and ...
Dominic van der Zypen's user avatar
6 votes
1 answer
195 views

Topological complexity of ultrafilters in $2^\kappa$ for uncountable $\kappa$

It is a well known fact that if $\mathcal{F}$ is a non-principal ultrafilter on $\omega$, then the set $\{ \alpha \in 2^\omega : \alpha \in \mathcal{F}\}$ (conflating binary strings with subsets of $\...
James Hanson's user avatar
  • 8,036
2 votes
1 answer
230 views

Unbounded $\omega_1$-sequence in $^*\mathbb{N}$

Let $\mathcal{F}$ be a non-principal ultrafilter on $\omega$. Let $^*\mathbb{N}$ = $\mathbb{N}^\omega/\mathcal{F}$ be an ultrapower. Let $\{n_\alpha\}_{\alpha\in\omega_1}$ be a strictly increasing ...
Sergey Grigoryants's user avatar
1 vote
1 answer
143 views

Regularity and ultrafilter

I read the following result in an article. Let $X$ be a regular space. Let $\mathcal{M}$ be free closed ultrafilter on $X$. Set $\mathcal{U=}\left\{ U:U\text{ is open and there exists a }F\in \mathcal{...
Mehmet Onat's user avatar
12 votes
1 answer
612 views

When do two ultrafilters yield isomorphic ultrapowers?

Fix a cardinal $\lambda$$\newcommand{\cU}{\mathcal U}\newcommand{\cV}{\mathcal V}$. Consider the equivalence relation on $\beta\lambda$ given by $\cU\sim \cV$ when for all first-order structures $M$ ...
tomasz's user avatar
  • 1,070
2 votes
1 answer
211 views

External ultrafilters definitions

I am reading a paper by Goldberg, in which he defines ultrafilter over a model of set theory (not transitive). These are the definitions: I get the definition of an M-ultrafilter, it is a real subset ...
D. Hershko's user avatar
4 votes
2 answers
275 views

Multiplicative and additive groups of the field $(\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z})/\simeq_{\cal U}$

Let ${\cal U}$ be a non-principal ultrafilter on $\omega$, and for each $n\in\omega$, let $p_n$ denote the $n$th prime, that is $p_0 = 2, p_1=3, \ldots$ Next we introduce the following standard ...
Dominic van der Zypen's user avatar
2 votes
0 answers
181 views

Ultralimit of metric spaces vs. inductive limits of underlying topological spaces

Let $\{(X_n,d_n)\}_{n =1}^{\infty}$ be a sequence of bounded metric spaces such that: $X_n \subseteq X_{n+1}$ is a metric subspace of $X_n$. Let $\omega$ denote a non-principal ultrafilter (i.e.: ...
ABIM's user avatar
  • 4,883
6 votes
1 answer
153 views

Regular limit points of possible cofinalities

Let $A$ be a non-empty set of regular cardinals such that $\vert A\vert <\text{min}\ A$, and $\{\nu_i\mid i<i_0\}\subseteq \text{pcf}\ A$ be a strict increasing sequence having limit length $i_0$...
Sho Banno's user avatar
2 votes
1 answer
120 views

What is the smallest cardinality of a base of an ultrafilter on $\omega$ related to an almost disjoint family of cardinality $\mathfrak c$?

Let $(A_\alpha)_{\alpha\in\mathfrak c}$ be an almost disjoint family of infinite subsets of $\omega$. The almost disjointness of the family means that $A_\alpha\cap A_\beta$ is finite for any ordinals ...
Taras Banakh's user avatar
  • 38.7k
0 votes
0 answers
43 views

Notation of $P^+$-families - bibliography searching

have you ever met with notation of $P^+$-families in other papers than Iian B. Smythe "A local Ramsey theory for block sequences" and his phd? Thank you in advance
Andrzej Starosolski's user avatar
5 votes
1 answer
295 views

NCF, P-points, weak P-points, and cardinalities

The post is a bit long, but all the questions are similar or concern the same topic. Let $\omega^*=\beta\omega\setminus\omega$. A well-known topological definition of a P-point (on $\omega$) is as ...
Damian Sobota's user avatar
0 votes
1 answer
230 views

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
  • 59
7 votes
1 answer
270 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
9 votes
1 answer
484 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
248 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
119 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
  • 38.7k
7 votes
1 answer
415 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
  • 1,135
3 votes
1 answer
220 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
200 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
  • 1,135
3 votes
1 answer
214 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
  • 939

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