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Questions tagged [ultrafilters]

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12
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0answers
629 views

Existence (or non) of “definable” ultrafilters

This is a question which I suspect has an absurdly easy answer, but I'm not seeing it. Let $\langle\cdot,\cdot\rangle:\omega^2\rightarrow\omega$ be your favorite pairing map (for me, this is the ...
11
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0answers
650 views

A basic question on Stone-Cech compactification of $\mathbb{Z}$

Can the identity isomorphism on the additive group $\mathbb{Z}$ be extended to a non-identity semigroup isomorphism on $\beta\mathbb{Z}$, and still preserves $\beta\mathbb{Z}\setminus\mathbb{Z}$? ...
9
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0answers
304 views

Embedding $\beta\mathbb{N}$ into a product of Cantor sets

Let us consider $\beta\mathbb{N}$, the Stone-Čech compactification of the natural numbers (where we do not take $0$ to be a natural number, so the only idempotent elements are nonprincipal ...
9
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0answers
163 views

Is it possible that all ultrafilters are determined by the meet-semilattice of sub-ultrapowers?

Suppose that $\mathcal{Z}$ is a filter on a set $X$. Let $\Pi(X)$ denote the lattice of all partitions of the set $X$. Then $(\Pi(X),\wedge)$ is a meet-semilattice where $P\wedge Q=\{R\cap S|R\in P,S\...
8
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0answers
204 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 ...
8
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0answers
135 views

Non-wellorderable ultrafilters with wellorderable bases

There are some models in which $2^\omega$ is not wellorderable but there is a free ultrafilter over $\omega$. What about the consistency of: $2^\omega$ is not wellorderable + AC for countable sets of ...
8
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0answers
316 views

What is the name for a Banach space property closed under ultraproducts?

In Banach space theory, a super-property is a property of a Banach space that is preserved under ultrapowers. (Update (2015-09-28): The property must also be closed under isometric embeddings.) (...
8
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196 views

How much do idempotent ultrafilters generate in terms of semigroups?

It is known that the set of ultrafilters on, say, the natural numbers $\mathbb{N}$, can naturally be endowed with the structure of a compact topological left semigroup (which fails to be anything ...
7
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0answers
130 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 ...
7
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0answers
180 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 ...
7
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0answers
197 views

Extending a Ramsey filter

We take filter to mean filter on $\omega$ containing all cofinite sets. We say a filter $F$ is Ramsey if ZERO does not have a winning strategy in the following infinite game between the two players ...
6
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0answers
154 views

Does Łoś's theorem imply choice given a free ultrafilter?

In the paper "Łoś's theorem and the boolean prime ideal theorem imply axiom of choice" Howard has shown that Łoś's theorem and the boolean prime ideal theorem imply axiom of choice. At the end of the ...
5
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0answers
191 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 ...
5
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0answers
141 views

Preservation of Baumgartner's I-ultrafilters under various forcings

For $I\subset \mathcal{P}(2^\omega)$, an ultrafilter $U$ on $\omega$ is said to be an I-ultrafilter if for all $f:\omega \to 2^\omega$, there exists $A\in U$ such that $f''[A]\in I$ [Baumgartner]. In ...
5
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0answers
278 views

Double ultrapower of the hyperfinite $II_1$-factor

Let $\omega$ be a free ultrafilter on the natural numbers and $R$ be the hyperfinite $II_1$-factor (the definition of $R$ is recalled in the comments). Question: Does there exist another free ...
5
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0answers
373 views

Boolean Prime Ideal Theorem and non-principal ultrafilters

Somewhat related to my other question Existence of non-principal ultrafilters on sets, is it known whether it is consistent with ZF that every infinite set has a free (non-principal) ultrafilter, but ...
4
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0answers
171 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 "...
4
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0answers
99 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 $...
4
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0answers
188 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 ...
4
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0answers
177 views

Are ultralimits the Gromov-Hausdorff limits of a subsequence?

Let $(M_i,p_i)$ be a sequence of $n$-dimensional Riemannian manifolds with lower Ricci curvature bound $-1$. Fix a non-orincipal ultrafilter and let X be the ultralimit of the sequence. Does there ...
3
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0answers
57 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 ...
3
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0answers
124 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 ...
3
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0answers
125 views

What is known about the krull dimension of an ultrapower ring?

Let $R$ be a ring, $F$ a free ultrafilter on a set $X$ which is not countably complete, and $R_F$ the ultrapower of $R$ with $R \not\cong R_F$. The following two results are from a masters thesis ...
3
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0answers
120 views

Compactness-like property for universal generalization?

Hi all! I have the following problem. Suppose I have a sequence of models $M_1,M_2,...$, all of which have the same countable domain (call it $D$). $x_1,x_2,...$ is a well-ordering of $D$. $\phi(x)$ ...
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0answers
230 views

Ultrapowers of ultrapowers

Suppose that you have some structure $S$, and you want to construct an ultrapower of cardinality $\kappa$ to obtain $S^*_\kappa$. Then, say you want to construct a new ultrapower from $S^*_\kappa$, ...
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0answers
227 views

Defining filters in closure algebras: reference request

A closure algebra C is a boolean algebra B together with a unary closure operator, and additional axioms, the Kuratowski axioms, that the closure operator must satisfy. (The Wikipedia article prefers ...
0
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0answers
58 views

Existance of bijective function which maps tensor product of subsets of a selective ultrafilter into the ultrafilter

In the answer on this question Andreas Blass had shown that for any selective ultrafilter $\scr{U}$ on $\omega$ and for any free subfilter $\scr{F}\subset{U}$ doesn't exist bijection $\varphi:\omega^2\...