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Questions tagged [filters]

Filters (and ultrafilters) are used in various areas of mathematics, e.g. general topology, set theory, boolean algebras, model theory. In topology they can be used to study convergence in a more general way than just convergence of sequences. For questions about filters in the sense of signal processing, do not use this tag and instead use (signal-processing) or another appropriate tag.

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weakly normal filters

Kanamori (Ultrafilters over Uncountable Cardinals) in his Phd Thesis defines a filter $\mathcal F$ as weakly normal whenever every function $f$ such that $\{\xi<\kappa\mid f(\xi)<\xi\}\in\...
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Is there a proof of the proposition that $\equiv$ is an equivalence relation?

Would anyone happen to have a proof that the following use of $\equiv$ is an equivalence relation: Let $\mathcal{F}(\mathbb{N})$ be an ultrafilter on $\mathbb{N}$, constructed by choosing subsets of $...
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1answer
38 views

Are ultrafilters unique?

I'm trying to get a feel for what an ultrafilter is by looking at some finite examples. Firstly, I took a look at this question: Example of a filter on a set, and it stoked the question: can sets such ...
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$Z[(I, J)]$ is the set of all $Z_1 \cap Z_2 $, where $Z_1 \in Z[I]$ and $Z_2 \in Z[J]$

Definition 1: $Z(f)= \{ x \in X : f(x) = 0 \}$ is a zero-set.($f \in C(X)$) $Z(X) =\{ Z(f): f \in C (X)\}$ ($C(X)$ is the ring of real continuous function on $X$.) Definition 2: A ...
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1answer
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Covariance in and input-output filter [Stationary Stochastic Processes]

The weakly stationary processes $X_t$, $\;t=0$, $\,\pm$$1$, $\,\pm$$2$,$\,\ldots$ and $Y_t$,$\;$ $t=0$, $\,\pm$$1$, $\,\pm$$2$,$\ldots$ are input and output of a linear filter according to $...
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1answer
29 views

Limit point compact uniform space

I'm working on a theorem on compactness for uniform spaces. Here are the definitions I'm using: $X$ is compact if every open cover of $X$ reduces to a finite subcover. $X$ is filter-compact ...
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1answer
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Determining elements of a Boolean algebra by a set of ultrafilters

Let $A$ be a Boolean algebra and let $Ult(A)$ be its Stone space. Let us say that a set $U\subseteq Ult(A)$ determines an element $a\in A$ if there exists $V\subseteq U$ such that $$\big\{b\in A\!: (\...
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Proof: $U$ is an ultrafilter of a Boolean algebra $B$ if and only if for all $x$ in $U$ exactly one of $x$,$x^*$ belongs to $U$.

I have been stuck with this problem for a while now. I have a proof that letting $U$ be an ultrafilter, exactly one of $x,x^*$ belongs to $U$ for all $x$ in $B$, I did this by showing that both belong ...
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1answer
66 views

Sequential compactness and filters

I'm trying to work with the many equivalent definitions of compactness for a topological space $X$; in particular, Every (proper) filter on $X$ has a (proper) convergent refinement. I'll ...
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1answer
26 views

$(f,g) = C(X) \Longleftrightarrow Z(f) \cap Z(g) = \emptyset$

Theorem: a: If $I$is an ideal in $C(X)$ , then the family $Z[I] = \{ Z(f) : f \in I \} $ is a $z$-filter on $X$. b: if $\mathbf{F}$ is a $z$-filter on $X$, then the family $Z^{-1} [ \mathbf{F} ] = ...
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Proving that functions send ultrafilter basis to ultrafilter basis

I'm currently revisiting a proof of Tychonoff's theorem via ultrafilters. The definitions we were working with are as follows, $\mathcal{B}$ is a basis for a filter $\mathcal{F}$ on a set $X$ if $\...
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About isomorphic of a boolean algebra to bolean algebra

Let $\mathcal A$ be a boolean algebra. It`s non-empty subset $\mathcal F$ is called a filter if $∅ \notin \mathcal F$, for all $A ∈ \mathcal A$, $B ∈ \mathcal F$ from $B ⊂ A$ follows $A ∈ \mathcal F$, ...
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3answers
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find a free ultrafilter on $\Bbb N$

Suppose $(x_n)_n$ is a bounded sequence of complex numbers, there must exist a accumulation point, say $x_0$, thus we can find a free ultrafilter $\mathcal{F}$ on $\Bbb N$ such that $\lim_{\mathcal{F}}...
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intersection of elements of $\beta\Bbb N\setminus \Bbb N$ [duplicate]

$\omega \in \beta \Bbb N\setminus \Bbb N$ is a free ultrafilter on $\Bbb N$.My question is :Is $\cap_{\omega\in \beta \Bbb N \setminus \Bbb N}=\Bbb N $?
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1answer
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intersection of sets corresponding to free ultrafiler

If $\omega \in \beta \Bbb N\setminus \Bbb N$,we define $S_{\omega}=\{(x_n) \in \prod M_n(\Bbb C):lim_{n \to \omega}tr_n(x_n)=0\}$ Is the intersection $\cap_{\omega \in \beta \Bbb N \setminus \Bbb N}...
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1answer
39 views

difference between convergence along free ultrafilter and common convergence

Suppose $\mathcal{F}$ is any free ultrafilter on $\beta\mathbb{N}\setminus\mathbb{N}$,$(x_n)$ is a sequence of complex numbers. My question is:What is the deference between the limit along $\mathcal{...
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1answer
116 views

Club filter of $\kappa$ is $\kappa$-complete

I'm trying to show that club filter of $\kappa$ is $\kappa$-complete for uncountable regular cardinal $\kappa$: Let $\kappa$ be uncountable regular cardinal, let $C(\kappa)$ be the club filter ...
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2answers
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Let $A$ be a Boolean algebra and $F\subseteq A$ be a filter on $A$. Why are the following properties equivalent?

Let $\mathcal{A}$ be a Boolean algebra and $F\subseteq \mathcal{A}$ be a filter on $\mathcal{A}$. Why are the following properties equivalent? $$(1)\,\,\,A\land B\in F\Rightarrow A,B\in F$$ $$(2)\,\,\...
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Countably closed ultrafilters on incomplete Boolean algebras

Suppose that $B$ is a Boolean algebra. Say that an ultrafilter, $U$, on $B$ is countably closed iff whenever $X\subseteq U$ is countable and the meet $\bigwedge X$ exists, $\bigwedge X\in U$. I ...
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1answer
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free ultrafilter

If $\omega$ is a free ultrafilter on $\mathbb{N}$,$(x_n)$ is a sequence of complex numbers,what is the precise definition of "$lim_{\omega}(x_n)$ does not converge to $x$"?
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How many ultrafilters there are in an infinite space?

I'm stuck with the next exercise from the book Rings of Continuous Functions by Gillman. If $X$ is infinite, there exist $2^{2^{|X|}}$ ultrafilters on $X$ all of whose members are of cardinal $X$. ...
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1answer
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What does the word “extend” mean in the context of model theory?

Consider the following two problems: (1) Let $L=\{E\}$ be a language consisting one binary relation symbol. Let $T$ be the $L$-theory saying that $E$ is an equivalence relation with infinitely many ...
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1answer
21 views

Is the algebra map of the ultrafilter monad continuous?

Let $\beta$ be the ultrafilter functor from Sets to Sets, which sends a set $X$ to the set of all ultrafilters on the powerset of $\mathcal{P}(X)$ equipped with its Boolean algebra structure. Then $\...
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1answer
29 views

Image of a $z-$ultrafilter is contained in a unique $z-$ultrafilter

I understand that for a continuous map $f:X\longrightarrow Y$, if $\mathcal{F}$ is a $z-$ultrafilter on $X$, then $f(\mathcal{F})=\{A\in Z(Y):f^{-1}(A)\}$ is a prime $z-$filter on $Y$ and may not be a ...
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1answer
28 views

Ultrafilters preserving infinite joins

A filter $U$ over a boolean algebra $A$ (isomorphic to a powerset algebra) "preserves" a join $a = \bigcup_{i\in I}a_i$, if $a\in U$ implies $a_i\in U$ for some $i\in I$. A join $a$ is infinite if $I$ ...
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1answer
37 views

Colimits where maps in are determined by maps into a component?

I am curious if there is a class of colimits $\mathsf{colim} D$ where maps $A \to \mathsf{colim} D$ must factor through the colimit cocone $D(X) \to \mathsf{colim}D$ for some $X$? For example, would ...
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1answer
39 views

When is every principal filter an intersection of ultrafilters?

The question is in the title: what property does a lattice need to have such that for every element of the lattice $x$, there exists a set of ultrafilters in the lattice such that the intersection of ...
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1answer
52 views

Show that a prime closed filter is not always a closed ultrafilter

I still don't understand how is that true. Here are the definitions: Let P be a class of closed subsets of a topological space X which is closed under finite intersections and finite unions. A closed ...
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$f : \omega \to \omega$ unbounded on every ultrafilter element, must $f$ be strictly increasing some ultrafilter element?

Consider an ultrafilter $\mathcal{U}$ and $f : \omega \to \omega$ such that $f|_{S}$ is unbounded for every $S \in \mathcal{U}$. Must there exist $S \in \mathcal{U}$ such that $f|_{S}$ is strictly ...
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1answer
36 views

Nonprincipal ultrafilters over $\mathbb{N}$

So, I'm given $\mathcal{A}\subseteq \mathcal{P}(\mathbb{N})$ that has the property that for any $\mathcal{A}_0\subseteq \mathcal{A}$ finite, $\cap\mathcal{A}_0$ is infinite. I have to show that there ...
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1answer
42 views

Definition of trace in Bourbaki

Bourbaki, General Topology, p. 61 (1966) What is the definition of trace in the following Proposition? Proposition 8. Let $\mathcal{F}$ be a filter on a set $X$ and $A$ a subset of $X$. Then the ...
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1answer
48 views

limit of a sequence along ultrafilter

If $X$ is a compact and Hausdorff topological space,$(x_n)_{n}$ is a sequence in $X$, for any ultrafilter $\mathcal{F}$ on $\mathbb{N}$, I know the fact that $\lim_{\mathcal{F}}x_n$ exists and is ...
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1answer
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free ultrafilters on $\mathbb{N}$

Let $\mathbb{N}$ be the set of natural numbers, I know the fact that the number of free ultrafilters on $\mathbb{N}$ is uncountable. I have two questions: 1.How to construct these uncountable ...
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1answer
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Filter convergence in discrete space.

Which filters converge in a discrete topological space? Is this correct: Let $(X,\tau)$ discrete topological space and $\cal{F}$ a filter and $\cal{U}_x$ the neighborhood system of $x$. Then $\{x\}$ ...
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1answer
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What is the difference between a directed set and a filtered category?

This may seem like a stupid question, but these two concepts seem to be identified so often that it's just a detail I've overlooked. Apparently a filtered category is a generalisation of a directed ...
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Topological “filters” and convolutions, are they related or is it just a coincidence of name choice?

Being an engineer, I have sometimes heard of Topological filters and in particular ultrafilters. They seem to have some meaning regarding partial ordering of sets but be of a more pure math origin ...
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1answer
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Trouble understanding filter and ultrafilter

A filter F on S is a collection of subsets of S in which two conditions hold: If A and B belong to the collection F then A∩B also belongs to the collection. If A belongs to the collection F and A is ...
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1answer
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Norm on hyperreals ${^*\mathbb R}$ using the ultrafilter construction

Suppose we construct the hyperreals by fixing a free ultrafilter $\mathcal F$ – formalizing the idea of "large subsets of $\mathbb N$" – and defining an equivalence relation between two real sequences ...
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1answer
41 views

Ultrafilters on finite boolean algebras

I am asked to prove a special case of Stone duality, namely that $B\cong \mathcal{P}(\text{Ult}(B))$ by the map $\phi:B\to \mathcal{P}(\text{Ult}(B))$ given by the homomorphism $$ \phi(x)=\{V\in \text{...
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2answers
115 views

Example of a free ultrafilter on natural numbers

We know that free ultrafilter exist on natural numbers. I would like to see an example of a free ultrafilter on natural numbers.
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1answer
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Showing that a net is universal

Let $\mathcal{U}$ be an ultrafilter on $\mathbb{N}$ and let $x:\mathcal{U} \rightarrow \mathbb{N}$ be a net such that $x(U) \in U, \forall U \in \mathcal{U}$. Show that $x$ is universal. I tried some ...
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Infinite family $\mathscr{A}\subseteq P(\omega)$ with criteria

First part: Prove that there's an infinite family $\mathscr{A}\subseteq P(\omega)$ such that: $X \in \mathscr{A} \Rightarrow |X|=\aleph_0$ $(X,Y\in \mathscr{A} \wedge X\ne Y)\Rightarrow |X \cap Y|&...
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1answer
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A strictly decreasing nested sequence of non-empty compact subsets of S has a non-empty intersection with empty interior.

S is an Hausdorff topological space. A decreasing nested sequence of non-empty compact subsets of S has a non-empty intersection. In other words, supposing $C_{k}$ is a sequence of non-empty, compact ...
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1answer
69 views

Ultrafilter on a boolean algebra containing $u$ but not $v$

I was reading stone's representation theorem and a part of the proof was omitted and i wanted to check if i got it right: So assume that $B$ is a boolean algebra and $u,v \in B \wedge u\neq v$ and we ...
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1answer
69 views

If every finite subset of a theory has an infinite model then the theory has an infinite model

Consider the famous compactness theorem, one of the first theorems in model theory: Let $T$ be a theory, i.e., a set of first-order sentences. If every finite subset of $T$ has a model, then the whole ...
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1answer
26 views

Is every z-filter $\mathscr{F}$ the intersection of all z-ultrafilters refining it?

Every z-filter $\mathscr{F}$ on a topological space $X$ is the intersection of all z-primefilters refining it. Is it also true that it is the intersection of all z-ultrafilters refining it? Can you ...
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$f(\mathcal F):=\{f(F):F \in \mathcal F\}$ is filter-basis and $G = \{G \subset N: f^{-1}(G) \in \mathcal F\}$

Let $\mathcal F$ be a filter on $M$ and $f:N \rightarrow N$. I want to show that $f(\mathcal F):=\{f(F):F \in \mathcal F\}$ is a filter-basis of a filter $\mathcal G$ on N and that $\mathcal G = \{G \...
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1answer
35 views

Why does a filter base uniquely define a filter?

Why does a filter base uniquely define a filter? We define the filter base $\mathcal B$ of the filter $\mathcal F$as: $\forall F \in \mathcal F \ \exists \ B \in \mathcal B: B\subset F$ So why can $...
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Can we add a ordinal bigger than every other ordinal and end up with a transitive model?

Given some (set sized) transitive model of ZFC, $M$ we can construct Hyper-$M$ as follows. We construct an ultrafilter $U$ on $Ord_M$ such that $(\exists \alpha.S = \{ \beta : \beta > \alpha \}) \...
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2answers
133 views

Meaning behind Filter in Set Theory

In a course in logic and set theory, we studied the concept of a Filter. We defined a filter $F \in P(S)$ on $S$ an equivalent of the following definition from Jech's Introduction to Set Theory: (a) $...