The ultrafilters tag has no usage guidance.

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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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votes

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### What are the minimal requirements for the definable hyperreal field plus transfer?

It is interesting that to prove the transfer principle for the definable hyperreal field, one requires no more choice than for proving, for instance, the countable additivity of the Lebesgue measure. ...

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121 views

### Can There be Rudin-Keisler Immediate Sucessors?

There are several well-studied orderings on the set $\omega^*$ of ultrafilters on the natural numbers. Three popular ones are $\le_i$ for $i = 1,2,3$. We define $\mathcal U \le_i \mathcal V$ to mean ...

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328 views

### Description of $\big(\ell^\infty(\mathbb N)\big)^{\!*}$ via ultrafilters

Let $\beta\mathbb N$ is the set of ultrafilters on $\mathbb N$ and $\mathscr F\in\beta\mathbb N$. Assume that $l_{\mathscr F}\in\big(\ell^\infty(\mathbb N)\big)^{\!*}$ is the functional which assigns ...

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120 views

### Representation of the elements of $c_0^\perp$ as integrals over ultrafilters

Let
$$
X=\big\{\varphi\in\ell_\infty^{\,*}(\mathbb N) : \varphi(\{a_n\})=0\,\,\text{whenever $a_n\to 0$}\big\}.
$$
If $\varphi_{\mathscr F}(\{a_n\})$ is the limit of $\{a_n\}$ with respect to the non-...

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votes

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94 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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votes

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225 views

### When can you canonically extend an ultrafilter after forcing?

Suppose that $V$ is a model of $\sf ZFC$, and fix some regular $\kappa$, say $\omega_1$ for practical purposes.
Let $\cal U$ be an ultrafilter on $\omega_1$ in $V$ which is non-principal and even ...

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223 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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186 views

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

Let us consider $\beta\mathbb{N}$, the Stone-Czech 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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### 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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votes

**1**answer

188 views

### Comparing cardinalities of the spectrum of two masas in $B(H)$

If I imagine that (the self-adjoint part of) a C*-algebra $A$ represents the algebra of observables of some quantum system, then certain perspectives on algebraic quantum theory would ask me to ...

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130 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\...

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151 views

### Lowering from filters to ultrafilters for an infinitary relation

Let $U$ be a set. Let $N$ be a (possibly infinite) index set. Let $f$ be an $N$-ary relation on $U$ (that is $f$ is a set of functions $N\rightarrow U$).
I denote $\mathcal{L}\in \upuparrows f \...

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206 views

### Expressing a value related to an infinitary relation through ultrafilters

Let $U$ be a set. I denote $\mathfrak{A}$ the lattice of filters on $U$ ordered reverse to set theoretic inclusion of filters. I denote $\bigvee$ and $\bigwedge$ correspondingly the supremum and ...

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123 views

### Ultrafilters of weight $\aleph_2$ in Sacks model

It is well-known that in Sacks model there are P-points and even Ramsey ultrafilters, but what the usual (i.e. findable in the literature) proofs for these facts do is proving that ground model P-...

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328 views

### Ultraproduct of Forcing Extensions & Forcing Extension of Ultraproduct

Notation:
$M[{\mathbb{P}}:G]$ denotes the forcing extension of $M$ by $\mathbb{P}$-generic filter $G$.
$\prod_{\mathcal{F}}\langle M_i~|~i\in I\rangle$ denotes the ultraproduct of models using the ...

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236 views

### Preservation of ultrafilters by Sacks forcing

It is well-known that, if $p$ is a Ramsey (selective) ultrafilter on $\omega$, then after adding a Sacks real $p$ remains an ultrafilter (well, it's really the upwards closure of $p$ the one that's an ...

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votes

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179 views

### How many elementary equivalent models are unifiable by ultrapower?

Definition. A class $\mathcal{C}$ of pairwise elementary equivalent $\mathcal{L}$-structures is unifiable by ultrapower if there is an index set $I$ and an ultrafilter $F$ on it such that $\forall M,N\...

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vote

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326 views

### Can the structure of an ultrafilter determine the structure of its ultrapower?

Usually we work with ultrafilters as pure sets without any structure.
Q1. Is there any important notion of structure on an ultrafilter?
Q2. Is there any non-trivial notion of structure on ...

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vote

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210 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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451 views

### Commutative algebras whose bidual is commutative

Let $k$ be a commutative ring and $A$ a commutative $k$-algebra. Call $D(A) := \mathrm{Hom}_k(A,k)$ the dual of $A$ as a $k$-module, and $DD(A) := \mathrm{Hom}_k(D(A),k)$ the dual of the latter. Let ...

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### Shelah's proof that proper forcing preserves P-points

In Proper and Improper forcing, VI.5; Claim 5.1 part 1 is the following:
If $F$ is a P-point in $V$, $P$ is a proper forcing notion and
$\Vdash_P `` F$ generates an ultrafilter"
Then the ultrafilter ...

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votes

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322 views

### In $L$, does there exist a definable non-principal ultrafilter on $\mathbb{N}$

The axiom of constructibility $V=L$ leads to some very interesting consequences, one of which is that it becomes possible to give explicit constructions of some of the "weird" results of AC. For ...

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152 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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373 views

### Ultrafilter-based Fourier-Walsh-like Functions

Here is a (little wild) question about Boolean functions with countably many variables and a wild analog for Fourier-Walsh functions and analysis based on them.
Let $x_1,x_2,\dots,x_n,\dots$ be ...

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291 views

### A special c.c.c forcing notion and adding minimal generic reals

This question is related to my question "Forcing with c.c.c forcing notions, Cohen reals and Random reals".
A natural way to answer Prikry's conjecture is to build a c.c.c. forcing notion which adds ...

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345 views

### Does an ultrapower of an Aronszajn tree have an $\omega_{1}$-branch?

Throughout this question, I shall let $A^{\mathcal{U}}$ denote the ultrapower of a structure $A$ by an ultrafilter $\mathcal{U}$. Suppose that $T$ is an Aronszajn tree and $\mathcal{U}$ is an ...

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237 views

### Character of normal ultrafilters

The character of an ultrafilter $U$, denoted $\chi(U)$, is the minimal size of an $A \subseteq U$ such that $(\forall x \in U ) (\exists y \in A) y \subseteq x$. This cardinal characteristic has been ...

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159 views

### Is there a truth approximation on a cumulative hierarchy?

Note to the following well known theorem:
Theorem (1): If $\kappa$ be a "measurable" cardinal and $\mathcal{F}$ be a "non-principal $\kappa$-complete normal" ...

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101 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 ...

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387 views

### Is the product of ultrafilters cancellative?

Suppose that $\mathcal{U},\mathcal{V}$ are ultrafilters on sets. Recall that $\mathcal{U}\leq_{RK}\mathcal{V}$ (here we say $\mathcal{U}$ is Rudin-Keisler less than or equal to $\mathcal{V}$) iff for ...

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367 views

### Stone-Cech compatification and ultrafilter [closed]

I have been studding about compatification of a topological space $X$. But I have low understanding about the Stone-Cech compatification, specially construction of the Stone-Cech compatification on ...

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275 views

### Idempotent ultrafilters and the Rudin-Keisler ordering

Short version: what can we say about the place of idempotent ultrafilters in the Rudin-Keisler ordering?
Longer version:
If $U$, $V$ are (nonprincipal) ultrafilters on $\omega$, then we write $U\ge_{...

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215 views

### ultrafilter characterisation of huge cardinals

A cardinal $\kappa$ is huge iff there is $\lambda>\kappa$ and a $\kappa$-complete normal ultrafilter on
$P_{\leq \kappa}(\lambda)$, or, equivalently, on the set of families of subsets of $\lambda$ ...

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523 views

### Why does CH imply that there is a unique ultrapower of $\mathbb{N}$?

I've read these words: "How many ultra products $∏_Uℕ$ exist up to isomorphism, where $U$ is a non-principal ultrafilter over $ℕ$? If continuum hypothesis(CH) holds, then obviously just one ..."
i ...

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219 views

### When can we “displace” an ultrafilter limit with another limit?

Let $\cal A$ be a Banach algebra, $\cal U$ be a free ultrafilter, and $\phi$ be a character. Let ${(w_{\alpha})}_{\alpha}$ be a net in $(\cal A)_{\cal U}$, and suppose that for every $(a_i)\in (\cal A)...

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244 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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235 views

### an elementary substructure of a natural numbers ultrapower

Hi
I'm looking for an elementary substructure of a natural numbers ultrapower with a free ultrafilter over a numerable set also must not be isomorphism between the elementary substructure and any ...

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142 views

### Are irrational multiples of central sets again central?

Let me begin by giving the relevant definitions. A set $A \subset \mathbb{N}$ is said to be central if and only if there exists a topological system $(X,T)$ (with $X$ a compact metric space, $T$ a ...

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### 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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### Stone–Čech Compactification of $\mathbb{Z}$ with Fürstenberg Topology

The Stone–Čech Compactification of $\mathbb{N}$ as a discrete space has been extensively studied and can be represented using ultrafilters.
Consider $X=(\mathbb{Z},\mathcal{T})$, where $\mathcal{T}$ ...

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### 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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355 views

### Kadison-Singer problem in exotic Hilbert spaces

The Kadison-Singer problem is considered in relation to the separable Hilbert space:
KS: Does every pure state on the diagonal (atomic) masa of $B(\ell_2)$ has a unique extension to $B(\ell_2)$?
...

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239 views

### Extending complete filters

Suppose $\kappa$ is a measurable cardinal and let $\mathcal{F}\subset\wp(\kappa)$ be a $\kappa$-complete non-principal filter. Can we extend $\mathcal{F}$ to a $\kappa$-complete ultrafilter?
My ...

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### Dimension of An Ultraproduct Field as a Vector Space over Another Ultraproduct Field

Suppose $F \subseteq K$ are fields with $G$ an ultrafilter on an infinite set $X$. If $F^{\ast}$ and $K^{\ast}$ represent the ultraproducts respectively of $F$ and $K$, it is easy to see that $[K : F]...

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448 views

### special extremally disconnected spaces with only finite isolated points

We Know that a cardinal $\kappa$ is measurable if there is a set $X$ with cardinal $\kappa$ and a {0,1}-measure $\mu: P(X) \rightarrow ${$0,1$} so that for all $x \in X$, $\mu(x)=0$ and $\mu(X)=1$. ...

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444 views

### Ultralimit versus partial limit

Let $\omega$ be a nonprincipal ultrafilter on $\mathbb N$.
A standard construction gives an $\omega$-limit, say $x_\omega$, for any bounded sequence $(x_n)$ of real numbers.
Namely, there is unique ...

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331 views

### How much $\beta \mathbb{N}$ is homogenous?

Let $p,q\in \beta \mathbb{N}\setminus \mathbb{N}$. Must always the spaces $\beta \mathbb{N}\setminus \{p\}$ and $\beta \mathbb{N}\setminus \{q\}$ be homeomorphic? If no, can we for each point $p\in \...

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319 views

### Determinacy and definable ultrafilters

It is a simple consequence of AD that there are no non-principal ultrafilters on $\omega$: for $U$ an ultrafilter on $\omega$, consider the game $G_U$ where players I and II play natural numbers $x_0$ ...

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