Questions tagged [ultrafilters]

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Points in the Stone Cech compactification are intersection of open sets

Let $\beta \mathbb{N}$ be the Stone Cech compactification of the natural numbers and let $ x\in \beta \mathbb{N}$. Is it true that there exists a sequence of open sets $\{U_n\}_{n=1}^\infty$ in $\beta ...
Serge the Toaster's user avatar
12 votes
1 answer
472 views

Is the Tukey order well-founded

Consider the Tukey order restricted to directed orders of the form $(U,\supseteq)$, where $U$ is an ultrafilter on $\omega$. It is defined as follows: For two ultrafilters $U,W$ on $\omega$, we say ...
Tom Benhamou's user avatar
4 votes
1 answer
166 views

Maximal intersecting families on $\omega$ that are not ultrafilters

A family ${\cal S}\subseteq{\cal P}(\omega)$ is intersecting if any two members of ${\cal S}$ have non-empty intersection. Zorn's Lemma implies that every intersecting family is contained in a maximal ...
Dominic van der Zypen's user avatar
5 votes
1 answer
100 views

Are sequences going to +infty along a ultrafilter on $\omega$ essentialy increasing?

The question is essentially in the title: suppose you have a non-principal ultrafilter $p$ on $\omega$ and a sequence $(u_n)_{n\in \omega}$ of elements of $\omega$ such that the $p$-limit of $(u_n)$ ...
AntoninG's user avatar
2 votes
1 answer
151 views

Ultralimit of $w^*$-continuous maps

Let $\omega$ be a free ultrafilter on $\mathbb N.$ Let $(\mathcal M_n)$ be a sequence of finite von Neumann algebras. Let $\mathcal N$ be another finite von Neumann algebra and we have maps $\phi_n:\...
A beginner mathmatician's user avatar
0 votes
1 answer
92 views

A question about the Stone-Čech compactification and ultrafilter

Let $X$ be a Tychonoff space and let $\beta X$ is the Stone-Čech compactification of $X$. Assume $f:X\longrightarrow \mathbb{R}$ is a bounded function. Then there exists a function $f^{\beta }:\beta X\...
Mehmet Onat's user avatar
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4 votes
1 answer
222 views

Mysior plane is not realcompact

Let $X = \mathbb{R}^2$ with $(x, y)\in X$ for $y\neq 0$ isolated and $(x, 0)$ having neighbourhood basis of the form $$U_n(x) = \{(x, y) : y\in (-1/n, 1/n)\}\cup \{(x+y+1, y) : 0 < y < 1/n\}\cup ...
Jakobian's user avatar
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1 vote
0 answers
58 views

Must an ultrafilter orthogonal to all ultrafilters containing an element $a$ contains $a^\perp$?

Let $L$ be an orthocomplemented lattice. We may consider the collection $U$ of ultrafilters on $L$. We say two elements $a, b \in L$ are orthogonal to each other, written as $a \perp b$, if $a \leq b^\...
David Gao's user avatar
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-1 votes
1 answer
58 views

The existence of a maximal “cross-sectional” filter on the Boolean algebra of measurable subsets of [0, 1] modulo almost everywhere equivalence

Let $\mathcal{B}([0, 1])$ be the Boolean algebra of measurable subsets of $[0, 1]$ modulo almost everywhere equivalence, i.e., two measurable sets which differ only by a Lebesgue null set are ...
David Gao's user avatar
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167 views

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
261 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
104 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
794 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
264 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
122 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
142 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
150 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
625 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
246 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
7 votes
1 answer
302 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
  • 30.2k
5 votes
1 answer
254 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
140 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
31 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
  • 964
5 votes
1 answer
559 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
17 votes
0 answers
1k 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
261 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
310 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
259 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
5 votes
1 answer
247 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
  • 1,823
4 votes
1 answer
319 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
  • 1,133
9 votes
2 answers
925 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
263 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
201 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
187 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
141 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
237 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
174 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
221 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
158 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
199 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
  • 10.3k
2 votes
1 answer
245 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
148 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
  • 1,161
12 votes
1 answer
638 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
  • 1,216
2 votes
1 answer
213 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
321 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
226 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
  • 4,969
6 votes
1 answer
157 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
130 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
  • 40.9k
0 votes
0 answers
45 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
334 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

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