Questions tagged [filters]

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Milnor and Moore paper "On the structure of Hopf algebra" proposition 1.7 , filtration and grading

In the Milnor and Moore paper, "On the structure of Hopf algebras" proposition 1.7 said the following: $A$ a connected $K$-algebra. $N$ a left $A$ module that is connected as a $K$-graded ...
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The existence of a maximal “cross-sectional” filter on the Boolean algebra of measurable subsets of [0, 1] modulo almost everywhere equivalence

Let $\mathcal{B}([0, 1])$ be the Boolean algebra of measurable subsets of $[0, 1]$ modulo almost everywhere equivalence, i.e., two measurable sets which differ only by a Lebesgue null set are ...
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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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Are arbitrary nonempty intersections of principal filters principal?

Suppose $\langle L,\leq\rangle$ is a lattice with join $\sqcup$. Let $F_1$ and $F_2$ be principal filters on $L$. Thus, for $i\in I=\{1,2\}$ there are $x_i\in L$ so that $F_i=\{y\in L:x_i\leq y\}$. In ...
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1 vote
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What's the autocovariance of 2 concurrent hidden states of a stochastic gaussian walk, using the Kalman filter?

A latent variable evolves as $x_t = x_{t-1} + e_t$ where $e_t$ has a gaussian distribution with 0 mean and a variance which defines the volatility of the overall process. Let $o_t$ be the noisy ...
1 vote
33 views

Posets which extend centered sets to filters

(Post cross-posted from math.se.) Suppose $(\mathcal O, \leq)$ is an arbitrary poset. Let us say that $\mathcal O$ is compact if every $\mathcal C\subseteq\mathcal O$ which is centered (any finite ...
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Without Choice: Are there filters of cardinality continuum?

Is it provable, in ZF (without Choice), that every filter can be extended to one of cardinality continuum? The extended filter is not requested to be an ultrafilter.
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Is certain topology-related set a distributive lattice?

In my research the following problem appeared (and if it is true, this solves positively several my conjectures): Let $U$ be a fixed set (usually $U$ is infinite). Let $n$ be a fixed index set ($n$ ...
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$\kappa$-translatability

I asked the following on MSE a few weeks ago but I did not get any answer : https://math.stackexchange.com/questions/1039593/carlsons-translatability-are-theses-characterisations-equivalent Reference ...
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Why the intersection of a scott open (or \w the relatively compactness property) filter on a topology of a sober (and 2nd countable) space is compact?

Definitions and notations. Let $\mathcal{P}(X)$ the power set of $X$. Let $\tau_X\subseteq\mathcal{P}(X)$ a topology on X. We call $A$ irreducible if every time $A=B\cup C$ with $B,C$ closed set ...
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Intersections of open sets and $\alpha$-favorable spaces

I would like to ask about some classes of topological spaces whether they have been studied, what are they called (if they have a name) and whether they have some interesting properties. For the ...
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Analogous filter to Kalman filter that maximized mode instead (as opposed to minimizing variance)

I may have a potential application where maximizing the mode (as opposed to typically minimizing the variance) would be useful for state estimates. The situation may arise from skewed distributions ...
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A characterization of the poset of filters on a set

For the lattices of all subsets of a given set, an axiomatic characterization is known: A poset is isomorphic to a set of all subsets of some set iff it is a complete atomic boolean algebra. The ...
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1 vote
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About ordering and equivalence of filters

Let $U$ is a set. I will speak about filters on this set. If $f$ is a function and $a$ is a filter then I define $f \left[ a \right]$ as the filter whose base is $\lbrace f[A] | A \in a \rbrace$. I ...
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This question arose more from curiosity than from an actual problem. There are situations when you embed some space $X$ in a set of filters on $X$, which inherits properties of $X$ or has even better ...