Questions tagged [ultrafilters]

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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 ...
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4 votes
1 answer
161 views

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

A Silver forcing "below $2^n$" is defned 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 ...
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3 votes
2 answers
114 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?
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29 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 ...
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  • 802
4 votes
1 answer
354 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 $\kappa$-...
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16 votes
0 answers
658 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 $\...
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6 votes
1 answer
235 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 ...
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6 votes
1 answer
260 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 $\...
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3 votes
2 answers
224 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 \...
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3 votes
1 answer
168 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 ...
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4 votes
1 answer
262 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,\...
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  • 1,095
8 votes
2 answers
843 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 ...
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3 votes
1 answer
192 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 ($= \...
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3 votes
1 answer
168 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, ...
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3 votes
1 answer
158 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 $q\in Q$ ...
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1 vote
1 answer
124 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)\...
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3 votes
2 answers
208 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 ${\bf a}, {\bf b}\in \beta\omega$, ...
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2 votes
1 answer
158 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(\...
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  • 441
4 votes
1 answer
206 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 ...
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4 votes
0 answers
148 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 ...
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6 votes
1 answer
186 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 $\...
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2 votes
1 answer
202 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 ...
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1 vote
1 answer
132 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{...
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12 votes
1 answer
584 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 $M$ ...
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2 votes
1 answer
201 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 ...
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4 votes
2 answers
264 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 ...
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2 votes
0 answers
125 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.: ...
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6 votes
1 answer
151 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 $i_0$...
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2 votes
1 answer
110 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 ...
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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
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5 votes
1 answer
267 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 ...
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0 votes
1 answer
200 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, ...
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7 votes
1 answer
261 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 ...
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9 votes
1 answer
446 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 ...
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6 votes
1 answer
242 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<\...
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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 ...
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7 votes
1 answer
388 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 ...
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  • 1,095
3 votes
1 answer
210 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 ...
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5 votes
1 answer
183 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?
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  • 1,095
3 votes
1 answer
186 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 $\Gamma$ (...
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1 vote
1 answer
123 views

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, ...
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12 votes
0 answers
184 views

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 ...
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1 vote
1 answer
109 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{...
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5 votes
1 answer
317 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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  • 996
7 votes
0 answers
163 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 ...
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  • 32.9k
5 votes
1 answer
196 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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5 votes
0 answers
286 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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  • 2,573
2 votes
1 answer
84 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 $\...
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  • 1,095
4 votes
1 answer
312 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 $Q$...
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3 votes
1 answer
128 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 $\mathcal{F_1}=\{A\subset\omega~|~\rho(A)=1\}$. ...
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