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8
votes
3answers
279 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 ...
2
votes
1answer
108 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 ...
3
votes
0answers
78 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 ...
3
votes
1answer
192 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 ...
8
votes
0answers
210 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.) ...
9
votes
0answers
171 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 ...
3
votes
0answers
97 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 ...
5
votes
1answer
181 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 ...
9
votes
0answers
124 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 ...
0
votes
1answer
144 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 ...
-2
votes
1answer
195 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 ...
4
votes
1answer
121 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
319 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
207 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
170 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
321 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
202 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
441 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
148 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 ...
6
votes
3answers
308 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
142 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
368 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 ...
7
votes
1answer
278 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
335 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
225 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
156 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
97 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
382 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
329 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
252 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
211 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
499 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
216 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
238 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
226 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 ...
6
votes
1answer
139 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
111 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
771 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 $\mathcal{T}$ ...
2
votes
0answers
112 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
350 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
235 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
177 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 : ...
3
votes
1answer
445 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 $\mu(X)=1$. ...
10
votes
1answer
384 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
325 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
309 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 $x_0$ ...
10
votes
0answers
576 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
157 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
880 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 ...
10
votes
0answers
516 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 ...