Questions tagged [ultrafilters]
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217
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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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Infinite tensor/Fubini product of ultrafilters
Given an infinite family $\{\mathcal{F}_{\lambda}$, $\lambda <\kappa\}$, $\kappa \geq \omega_0$, of (ultra)filters of a set $X$, how it is defined the infinite tensor/Fubini product $$\bigotimes_{\...
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Can there be no "surprisingly averageable" second-order sentences?
Say that a second-order sentence $\varphi$ is averageable iff there exists some infinite cardinal $\kappa$ and some nonprincipal ultrafilter $\mathcal{U}$ on $\kappa$ such that for every $\kappa$-...
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Extending ground model ultrafilters
Work over a model of $\sf GCH$. Suppose that $\kappa$ is some regular cardinal, and $U$ is a uniform ultrafilter on $\kappa^+$.
Does $U$ have some canonical extension after forcing with $\...
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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" ...
user36136
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Normal measures on $P_{\kappa }(\lambda )$ extend the club filter
This is (a variation on) exercise 20.4 in Jech's "Set Theory." Let $j : V \to M$ witness $\lambda$-supercompactness of $\kappa$, and consider the normal measure $U$ on $P_{\kappa}(\lambda)$ ...
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Does a generic normal measure extend the club filter?
This question is related to this one. The setup is as follows:
In $V$, $\kappa$ is supercompact and $\mathbb{P} = \mathrm{Coll}(\kappa, \aleph_2)$ is the Levy collapse. $G$ is $(V,P)$-generic, and ...
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Comparing Mathias forcing notions relative to various filters
Let $\mathcal F$ be a (non-principle, non trivial, ...) filter on $\omega$. The Mathias Forcing relative to $\mathcal F$ is the forcing notion $\mathbb M(\mathcal F)$ consisting of pairs $(s, X)$ with ...
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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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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 ...
- 3,138
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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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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 ...
- 2,781
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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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Multiplicative and additive groups of the field $(\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z})/\simeq_{\cal U}$
Let ${\cal U}$ be a non-principal ultrafilter on $\omega$, and for each $n\in\omega$, let $p_n$ denote the $n$th prime, that is $p_0 = 2, p_1=3, \ldots$
Next we introduce the following standard ...
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Models of $\mathsf{ZFC}$ with neither $P$- nor $Q$-points
A $P$-point is an ultrafilter $\scr U$on $\omega$ such that for every function $f:\omega\to\omega$ there is $x\in {\scr U}$ such that the restriction $f|_x$ is either constant, or finite-to-one.
A $Q$...
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Maximal intersecting families on $\omega$ that are not ultrafilters
A family ${\cal S}\subseteq{\cal P}(\omega)$ is intersecting if any two members of ${\cal S}$ have non-empty intersection. Zorn's Lemma implies that every intersecting family is contained in a maximal ...
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Questions on ultrafilters
Hi, I know that there are models in ZFC where don't exist p-points, but I can't find (neither on internet) a proof that I understand, some of you could help me please?
Another question: I know that ...
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Cardinality of a set of pairwise non-order-isomorphic ultrafilters on $\omega$
It is well known that there are $2^{2^{\aleph_0}}$ many non-principal ultrafilters on $\omega$. Is there a set ${\frak U}$ of non-principal ultrafilters on $\omega$ with $|{\frak U}| = 2^{2^{\aleph_0}}...
- 45.5k
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Supremum of infimum of measure of members of a free ultrafilter
For a set $A\subseteq \omega$ we let the upper density of $A$ be defined as $d^+(A) := \lim\sup_{n\to\infty}\frac{|A\cap(n+1)|}{n+1}$. Let $\text{FrU}(\omega)$ be the collection of free ultrafilters ...
- 45.5k
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Addition and Rudin-Keisler ordering in $\beta \omega$
$\DeclareMathOperator{\RK}{\mathrm{RK}}$Let $\beta\omega$ be the Stone-Cech compactification of $\omega$ with the discrete topology. We can endow $\beta\omega$ with an addition operation that extends ...
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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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Minimal cardinality of a filter base of a non-principal uniform ultrafilters
Let $\kappa$ be an infinite cardinal. An ultrafilter ${\cal U}$ on $\kappa$ is said to be uniform if $|R|=\kappa$ for all $R\in{\cal U}$. If ${\cal U}$ is a non-principal ultrafilter on $\kappa$, ...
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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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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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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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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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On a completeness property of hyperreals
Let $\mathbb{R}_*=\mathbb{R}^\omega/\mathcal U$ for some ultrafilter $\cal U$. In the definitions of this question and assuming ZFC + CH there are only three types of cuts in $\mathbb{R}_*$: $(\omega,\...
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Are there q-filters which are not ultrafilters?
I have just read that a selective ultrafilter must necessary be an ultrafilter. Is this also true for q-filters?
Im not sure if using CH for instance, we can follow the construction of a q-point but ...
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"Gaps" in the Rudin-Keisler ordering
If $(P,\leq)$ is a poset and $p\in P$, then we say that $p$ is the lower part of a gap there is $q \in P$, $q>p$ such that $[p,q] = \{p,q\}$. (This is equivalent to the statement that $(\uparrow p) ...
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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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Decomposition of an ultrafilter on the fibers of a map
Short version: If I have a map $f:Y \to I$, and $\mu$ an ultrafilter on $Y$, under what condition can $\mu$ be written as a limit/sum/integral of ultrafilters on the fibers of $f$ along the ...
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Nonexistence of a 'product universal' compact Hausdorff pseudotopological space?
I'm largely following the definitions of this paper, but I will replicate the relevant ones here.
I'm taking a pseudotopological space to be a set $X$ together with a relation $\rightarrow$ on the ...
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Notion of non-selectivity of ultrafilters
A non-principal ultrafilter $\mathcal{U}$ on $\omega$ is called selective (or Ramsey) if for every partition $\mathcal{P}$ of $\omega$ disjoint with $\mathcal{U}$ there is $A\in\mathcal{U}$ such that $...
- 1,153
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The property of the dense subfilter of a selective ultrafilter
Let us define the density of subset $A\subset\omega$ :
$$\rho(A)=\lim_{n\to\infty}\frac{|A\cap n|}{n}$$
if the limit exists. Let $\mathcal{F_1}=\{A\subset\omega~|~\rho(A)=1\}$. $\mathcal{F_1}$ is the ...
- 1,133
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Lattice of differences between ultrafilters
Consider two ultrafilters, $U$ and $V$, on the same cardinal $\kappa$. Let $D(U, V)=\lbrace X\subseteq \kappa: X\in U-V\rbrace$; clearly $D(U, V)$ is a lattice under $\subseteq, \cap, \cup $ since the ...
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Does ultrafilter have measure one?
Define a new product measure on cantor space as follows:u({0})=a,u({1})=1-a,where a$\in$(0,1/2].
Does any ultrafiter U hasn't measure one?
When a=1/2,I know U hasn't measue one.I guess neither when ...
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closed set and z-ultrafilter on normal space
Let $X$ be a completely regular, Hausdorff topological space and let $\cal F$ be a $z$-ultrafilter on $X$. Then for each zero set $W$ in $X$, either $W\in \cal F$ or there exists $Z\in \cal F$ such ...
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The example of the idempotent filter or subsets family with finite intersections property
From the answer of Andreas Blass and comments of Ali Enayat on my question Selective ultrafilter and bijective mapping it became clear that for any free ultrafilter $\mathcal{U}$ we have $\mathcal{U}\...
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Complexity of ultrafilter limits
Let $\mathscr{F}$ be a free ultrafilter on $\mathbf{N}$ and, for each $A\subseteq \mathbf{N}$ and $n \in \mathbf{N}$, define
$$
d_n(A):=\frac{|A\cap [1,n]|}{n}.
$$
Question. Considering $\mathcal{P}...
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cardinality of local bases in the non-standard reals
Given a index set $S$ together with a ultrafilter $\mu$ on $S$ (such that no set of cardinality $< |S|$ has measure $1$). Let the ordered field $\mathbb{R}(S,\mu)$ denote the ultrapower of $\mathbb{...
- 10.3k
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Mysior plane is not realcompact
Let $X = \mathbb{R}^2$ with $(x, y)\in X$ for $y\neq 0$ isolated and $(x, 0)$ having neighbourhood basis of the form $$U_n(x) = \{(x, y) : y\in (-1/n, 1/n)\}\cup \{(x+y+1, y) : 0 < y < 1/n\}\cup ...
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ultrapower(ultrapower)=ultrapower
Is there a nonprincipal ultrafilter $\omega$ on $\mathbb N$ such that for any metric space $M$ there is an isometry
$$(M^\omega)^\omega\to M^\omega?$$
(In other words, the $\omega$-power of $\omega$-...
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Finite pre-orders embeddable in the Rudin-Keisler ordering
$\DeclareMathOperator{\NPU}{\operatorname{NPU}}\DeclareMathOperator{\RK}{\,\mathrm{RK}}$A pre-ordered set is a pair $(P, \leq)$ where $P$ is a set and $\leq\subseteq P\times P$ is a reflexive and ...
- 45.5k
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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 $...
- 1,153
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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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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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How much are reduced powers different?
Given two infinite sets $X$ and $I$, and a filter ${\cal F}$ on $I$, one defines as usual the equivalence relation $\approx_{\cal F}$ on $X^I$ and obtains the reduced power $Y = X^I / \approx_{\cal F}$...
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"Completion property" in $(\beta\omega,+)$
Let $\beta\omega$ be collection of all ultrafilters on $\omega$ (principal and non-principal). We endow $\beta\omega$ with an operation $+$ in the following way. For ${\bf a}, {\bf b}\in \beta\omega$, ...
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Is the set of $\kappa$-complete ultrafilters closed in $\beta X$?
Given an arbitrary set $X$, let $\beta X$ be the set of all ultrafilters over $X$. Consider endowing $\beta X$ with a topology consisting of the following open sets:
$$
\{\mathcal{U} \in \beta X : A \...
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Is $(\omega+1)^\omega/{\cal U}$ complete for ${\cal U}$ free ultrafilter?
Let ${\cal U}$ be a free ultrafilter on $\omega$. Is the linearly ordered set $(\omega+1)^\omega/{\cal U}$ complete?
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