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3
votes
1answer
99 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 ...
5
votes
1answer
274 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 ...
7
votes
1answer
159 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 ...
2
votes
1answer
129 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 ...
1
vote
3answers
294 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 ...
1
vote
0answers
178 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 $S^*_\kappa$, ...
9
votes
1answer
414 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 ...
5
votes
1answer
119 views

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 ...
5
votes
3answers
233 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 ...
7
votes
0answers
111 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 ...
3
votes
2answers
348 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 ...
8
votes
1answer
212 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 ...
14
votes
3answers
300 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 ...
7
votes
1answer
198 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 ...
5
votes
1answer
143 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" ...
2
votes
0answers
66 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 ...
7
votes
3answers
363 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 ...
0
votes
1answer
186 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 ...
10
votes
1answer
210 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 ...
3
votes
1answer
197 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 $\lambda$ ...
13
votes
2answers
395 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 ...
1
vote
1answer
197 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 ...
5
votes
0answers
200 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 ...
1
vote
1answer
175 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 ...
3
votes
0answers
103 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 ...
4
votes
0answers
83 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 ...
9
votes
2answers
677 views

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 $\mathcal{T}$ ...
1
vote
0answers
105 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 $D$. $\phi(x)$ ...
4
votes
1answer
309 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 $B(\ell_2)$? ...
4
votes
1answer
199 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 ...
2
votes
1answer
159 views

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 : ...
2
votes
1answer
408 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 $\mu(X)=1$. ...
7
votes
1answer
261 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 ...
5
votes
1answer
304 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 ...
7
votes
1answer
280 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 $x_0$ ...
8
votes
0answers
507 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 $\beta\mathbb{Z}\setminus\mathbb{Z}$? ...
2
votes
1answer
147 views

Whether the result that an ultraproduct which satisfies ACCP is automatically a field generalizes to ultrafilters on larger indexing sets

Let $F$ be an ultrafilter on some set $X$, $R$ an integral domain and $R_F$ the resulting ultraproduct ring. For an element $(a)$ in the product ring of $R$ indexed by $X$, denote its equivalence ...
11
votes
1answer
728 views

Characterization of Stone-Cech compactifications

Suppose I have an infinite discrete topological space $X$ of cardinality $\kappa$. Then I know some things about the Stone-Cech compactification, $\beta X$: it is Hausdorff and compact but not ...
9
votes
0answers
469 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 ...
4
votes
1answer
282 views

Lattice of differences between ultrafilters

Consider two ultrafilters, $U$ and $V$, on the same cardinal $\kappa$. Let $D(U, V)=\lbrace X\subseteq \kappa: X\in U-V\rbrace$; clearly $D(U, V)$ is a lattice under $\subseteq, \cap, \cup $ since the ...
1
vote
1answer
164 views

A Q-point not Ramsey

May I ask where can I read the example of a q-point which is not Ramsey. I'm especially looking for a coloring of the set of two element subsets of $\mathbb{N}$ without a homogeneous set in the ...
5
votes
0answers
514 views

Mathias forcing with Ramsey ultrafilters, and Cohen reals [closed]

Edit/update: One reason this question never received an answer is because it was founded on a faulty premise! The Blaszczyk-Shelah paper I mentioned below does not prove that Mathias forcing with a ...
2
votes
2answers
204 views

Ultrafilter and contracting maps

I was trying to construct some element with specific properties in an ultraproduct and it boils down to a question which seems relatively natural but leaves me perfectly clueless. ...
5
votes
1answer
424 views

An Extender is a Generalization of an Ultrafilter?

I'm not sure whether this belongs here or on math stackexchange, but I'll give a try here. I've heard that an extender is a generalization of an ultrafilter. This is not to say that an ultrafilter ...
4
votes
1answer
426 views

Does ultrafilter have measure one?

Define a new product measure on cantor space as follows:u({0})=a,u({1})=1-a,where a$\in$(0,1/2]. Does any ultrafiter U hasn't measure one? When a=1/2,I know U hasn't measue one.I guess neither when ...
1
vote
2answers
282 views

Ultrafilters over vector spaces

Perhaps my question is naive, but let me try. Take a (real or complex) vector space $V$ and consider an ideal $\mathcal{I}$ of subsets of $V$ with the following property (call it (*)): for each ...
3
votes
1answer
170 views

closed set and z-ultrafilter on normal space

Let $X$ be a completely regular, Hausdorff topological space and let $\cal F$ be a $z$-ultrafilter on $X$. Then for each zero set $W$ in $X$, either $W\in \cal F$ or there exists $Z\in \cal F$ such ...
0
votes
2answers
262 views

ultrafilters' succession

hi I'n looking for a increasing and bounded ultrafilters' succession in natural numbers with Rudin-Keisler order, actually I need to prove there is that succession the idea is $U_1,U_2,....$ with ...
5
votes
0answers
312 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 ...
6
votes
1answer
438 views

Existence of non-principal ultrafilters on sets

Is it known to be consistent with ZF that there is no non-principal ultrafilter on any infinite set? (Feel free to use your favorite interpretation of "infinite" in this context. If infinite just ...