# Questions tagged [ultrafilters]

The ultrafilters tag has no usage guidance.

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### 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 ...

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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 $\beta\mathbb{Z}\setminus\mathbb{Z}$? ...

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### 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 ...

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### 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\...

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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 ...

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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 ...

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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.) (...

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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 ...

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### 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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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 ...

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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 ...

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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 ...

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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 ...

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### 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 ...

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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 ...

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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 ...

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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 "...

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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 $...

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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 ...

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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 ...

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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 ...

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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 ...

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### 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 ...

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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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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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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 ...

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### 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\...