Filters on Posets and Generalizations: Submitted to another math journal

February 3, 2010 by porton

I submitted to “Topology” math journal by email the manuscript Filters on Posets and Generalizations. In the email I asked them to confirm receipt of the email as soon as they receive it. Until now there were no response from “Topology” math journal. So I count them unresponsive and submitted the same work to an other journal (also by Elsevier publisher) “Topology and its Applications” 26 Jan 2010, this time through an automated submission tracking system of Elsevier (EES). Until now my manuscript has “Submitted to Journal” status. So I await when they will take an action on my manuscript.

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Filters on Posets and Generalizations – preprint

January 23, 2010 by porton

I submitted a preprint of the “Filters on Posets and Generalizations” article for peer-review and publication in Topology journal. The current version of this article is located at this URL. Read the rest of this entry »

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Co-separability of filter objects – solved

January 19, 2010 by porton

I solved a problem earlier formulated in this blog post.

A solution (of a slightly more general problem) can be found at this wiki page.

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Chain-meet-closed sets on complete lattices

December 12, 2009 by porton

Let \mathfrak{A} is a complete lattice. I will call a filter base a nonempty subset T of \mathfrak{A} such that \forall a,b\in T\exists c\in T: (c\le a\wedge c\le b). I will call a chain (on \mathfrak{A}) a linearly ordered subset of \mathfrak{A}.

Now as a part my research of filters I attempt to solve this problem (the problem seems not very difficult and I hope to prove it today or tomorrow, however who knows how difficult it may be):

Definition A subset S of a complete lattice \mathfrak{A} is chain-meet-closed iff for every non-empty chain T\in\mathscr{P}S we have \bigcap T\in S.

Conjecture A subset S of a complete lattice \mathfrak{A} is chain-meet-closed iff for every filter base T\in\mathscr{P}S we have \bigcap T\in S.

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Open problem: co-separability of filter objects

November 29, 2009 by porton

Conjecture Let a and b are filters on a set U. Then

a\cap b = \{U\} \Rightarrow \\ \exists A\in a,B\in b: (\forall X\in a: A\subseteq X \wedge \forall Y\in b: B\subseteq Y \wedge A \cup B = U).

This conjecture can be equivalently reformulated in terms of filter objects:

Conjecture Let \mathcal{A} and \mathcal{B} are filter objects on a set U. Then

\mathcal{A}\cup^{\mathfrak{F}}\mathcal{B}=U \Rightarrow \exists A,B\in\mathcal{P}U: A\subseteq\mathcal{A} \wedge B\subseteq\mathcal{B} \wedge A\cup B=U.

(where \mathfrak{F} is the set of filter objects on U).

It is simple to show (by applying the above conjecture twice) that it is equivalent to a yet simpler conjecture

Conjecture Let \mathcal{A} and \mathcal{B} are filter objects on a set U. Then

\mathcal{A}\cup^{\mathfrak{F}}\mathcal{B}=U \Rightarrow \exists A\in\mathcal{P}U: A\subseteq\mathcal{A} \wedge A\cup^{\mathfrak{F}}\mathcal{B}=U.

Put in yet simpler words, this conjecture can be formulated: the filtrator of filters on a set is with co-separable center.

Looking innocent? Indeed I currently don’t know how to attack this looking simple problem. Maybe you will help?

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Open problem: Pseudodifference of filters

November 24, 2009 by porton

Let {U} is a set. A filter {\mathcal{F}} (on {U}) is a non-empty set of subsets of {U} such that {A, B \in \mathcal{F} \Leftrightarrow A \cap B \in \mathcal{F}}. Note that unlike some other authors I do not require {\emptyset \notin \mathcal{F}}.

I will call the set of filter objects the set of filters ordered reverse to set theoretic inclusion of filters, with principal filters equated to the corresponding sets. See here for the formal definition of filter objects. I will denote {(\mathrm{up} a)} the filter corresponding to a filter object {a}. I will denote the set of filter objects (on {U}) as {\mathfrak{F}}.

I will denote {(\mathrm{atoms} a)} the set of atomic lattice elements under a given lattice element {a}. If {a} is a filter object, then {(\mathrm{atoms} a)} is essentially the set of ultrafilters over {a}.

Problem Which of the following expressions are pairwise equal for all {a, b \in \mathfrak{F}} for each set {U}? (If some are not equal, provide counter-examples.)

  1. {\bigcap^{\mathfrak{F}} \left\{ z \in \mathfrak{F} | a \subseteq b \cup^{\mathfrak{F}} z \right\}};
  2. {\bigcup^{\mathfrak{F}} \left\{ z \in \mathfrak{F} | z \subseteq a \wedge z \cap^{\mathfrak{F}} b = \emptyset \right\}};
  3. {\bigcup^{\mathfrak{F}} (\mathrm{atoms} a \setminus \mathrm{atoms} b)};
  4. {\bigcup^{\mathfrak{F}} \left\{ a \cap^{\mathfrak{F}} (U\setminus B) | B \in \mathrm{up} b \right\}}.

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Exposition: Complementive filters are complete lattice

November 2, 2009 by porton

(In a past version of this article I erroneously concluded that our main conjecture follows from join-closedness of {Z (D \mathcal{A})}.)

Let {U} is a set. A filter {\mathcal{F}} (on {U}) is a non-empty set of subsets of {U} such that {A, B \in \mathcal{F} \Leftrightarrow A \cap B \in \mathcal{F}}. Note that unlike some other authors I do not require {\emptyset \notin \mathcal{F}}. I will denote {\mathfrak{f}} the lattice of all filters (on {U}) ordered by set inclusion.

Conjecture 1 Let {\mathcal{A} \in \mathfrak{f}} is some (fixed) filter. Let {D = \left\{ \mathcal{X} \in \mathfrak{f} \hspace{1em} | \hspace{1em} \mathcal{X} \supseteq \mathcal{A} \right\}}. (Obviously {D} is a bounded lattice.) Then the set of complemented elements of the lattice {D} (ordered by set inclusion) is a complete lattice.

This conjecture was first formulated in this blog post and suggested as a polymath project in this blog post.

1. Filter objects

(Borrowed from this blog post)

For greater clarity I will use {filter objects} instead of filters. Below I will describe the properties of filter objects without exact definition and the proofs. You can look here for the formalistic behind.

I will denote the set of all filters objects on {U} as {\mathfrak{F}}. Filter objects are bijectively related with filters by the bijection “{\mathrm{up}}” from the set of filter objects to the set of filters. A filter object corresponding to principal filter generated by a set {A} is equal to {A}. (Thus the set of subsets of {U} is a subset of {\mathfrak{F}}.)

For formal definition of filter objects in the framework of ZF see here. Below we will not need the exact definition of filter objects, but only the facts that “{\mathrm{up}}” is a bijection from filter objects to filters and that a filter object corresponding to principal filter generated by a set {A} is equal to {A}.

I will define the order on the set of filter objects by the formula {\mathcal{A} \subseteq \mathcal{B} \Leftrightarrow \mathrm{up} \mathcal{A} \supseteq \mathrm{up} \mathcal{B}} for every filter objects {\mathcal{A}} and {\mathcal{B}}. This order well-agrees with the order of sets on {U}.

{\mathfrak{F}} is a complete lattice. (See here for a proof.)

2. Second definition

Let {\mathfrak{A}} is a bounded distributive lattice. Let {a \in \mathfrak{A}}. I will denote {D a = \left\{ x \in \mathfrak{A} \hspace{1em} | \hspace{1em} x \subseteq a \right\}}. Obviously {D a} is a bounded lattice.

The center of a bounded distributive lattice {\mathfrak{A}} is by definition the set {Z (\mathfrak{A})} of all complemented elements of {\mathfrak{A}}. It is a well known fact that the center is a boolean lattice.

I will call {Z (D \mathcal{A})} the elements {complementive} to {\mathcal{A}}.

Now our conjecture can be equivalently reformulated:

Conjecture 2 {Z (D \mathcal{A})} is a complete lattice whenever {\mathcal{A}} is an element of the lattice {\mathfrak{F}}.

If the conjecture is true it may be generalized for the case of {\mathfrak{F}} being the set of filter objects on some lattice instead of our less general case of filters on a set.

3. Special case of {\mathcal{A}} being a set

I almost believe that our conjecture is true in general, because it is true in the special case when {\mathcal{A}} is a set. Trueness of this special case of our conjecture trivially follows from the following theorem:

Theorem 3 {Z (D A) = \mathscr{P} A} if {A \in \mathscr{P} U}.

Proof: It follows from this theorem. \Box

4. Ways to attack our conjecture

4.1. Calculating meets and joins

To prove that a lattice is complete is enough to prove one of the following two statements: 1. it has all joins; or 2. it has all meets.

Directly calculating meets and joins for the lattice {Z (D \mathcal{A})} seems a complicated problem.

It could be simplified if meets or joins on {Z (D \mathcal{A})} coincide with meets or joins on {\mathcal{\mathfrak{F}}}.

For meets it is not the case. (A counter-example is simple to find in the case {\mathcal{A} = U} when {U} is any infinite set.)

For joins it seems true that joins on {Z (D \mathcal{A})} coincide with joins on {\mathcal{\mathfrak{F}}}. However I failed to prove it.

Proposition 4 If our main conjecture is true, then of the sublattice {Z (D \mathcal{A})} of the lattice {\mathfrak{F}} join-closeded.

Proof: Let {S \in \mathscr{P} Z (D \mathcal{A})}. Let our main conjecture is true and thus {\bigcup^{Z (D \mathcal{A})} S} exists. We need to prove that {\bigcup^{Z (D \mathcal{A})} S = \bigcup^{\mathfrak{F}} S}. Because {Z (D \mathcal{A})} is a sublattice of {\mathfrak{F}} it’s enough to prove that for any {\mathcal{F} \in D \mathcal{A}}

\displaystyle \forall K \in S : K \subseteq \mathcal{F} \Rightarrow \mathcal{F} \supseteq \bigcup {\nobreak}^{Z (D \mathcal{A})} S.

Really: Let {C \in \mathrm{up}^{D \mathcal{A}} \mathcal{F}} and {\forall K \in S : K \subseteq \mathcal{F}}. Then {C \supseteq \mathcal{F}}; {C \supseteq \bigcup {\nobreak}^{Z (D \mathcal{A})} S}; {C \in \mathrm{up}^{D \mathcal{A}} \bigcup {\nobreak}^{Z (D \mathcal{A})} S}. Thus {\mathcal{F} \supseteq \bigcup {\nobreak}^{Z (D \mathcal{A})} S}. \Box

4.2. Intersection with a set

An other way to characterize elements of {Z (D \mathcal{A})} is:

Theorem 5 {\mathcal{X} \in Z (D \mathcal{A}) \Leftrightarrow \exists X \in \mathscr{P} U : \mathcal{X} = X \cap^{\mathfrak{F}} \mathcal{A}} for every{\mathcal{X} \in \mathfrak{F}}.

Proof:

  • Reverse implication Let {\mathcal{X} = X \cap^{\mathfrak{F}} \mathcal{A}}. Let also {\mathcal{Y} = (U \setminus X) \cap^{\mathfrak{F}} \mathcal{A}}. Then {\mathcal{X} \cap^{\mathfrak{F}} \mathcal{Y} = \emptyset} and {\mathcal{X} \cup^{\mathfrak{F}} \mathcal{Y} = (X \cup (U \setminus X)) \cap^{\mathfrak{F}} \mathcal{A} = U \cap^{\mathfrak{F}} \mathcal{A} = \mathcal{A}}. So {\mathcal{X} \in Z (D \mathcal{A})}.
  • Direct implication Let {\mathcal{X} \in Z (D \mathcal{A})}. Then exists {\mathcal{Y} \in Z (D \mathcal{A})} such that {\mathcal{X} \cap^{\mathfrak{F}} \mathcal{Y} = \emptyset} and {\mathcal{X} \cup^{\mathfrak{F}} \mathcal{Y} = \mathcal{A}}. Then exists {X \in \mathscr{P} U} such that {X \in \mathrm{up} \mathcal{X}} and {X \cap^{\mathfrak{F}} \mathcal{Y} = \emptyset}. Obviously {( \mathcal{X} \cup^{\mathfrak{F}} \mathcal{Y}) \cap^{\mathfrak{F}} X = X \cap^{\mathfrak{F}} \mathcal{A}}; {\mathcal{X} \cup^{\mathfrak{F}} (X \cap^{\mathfrak{F}} \mathcal{Y}) = X \cap^{\mathfrak{F}} \mathcal{A}}; {\mathcal{X} = X \cap^{\mathfrak{F}} \mathcal{A}}. \Box

    We may attempt to attack our conjecture by proving that meets or joins of filter objects of the form {K \cap^{\mathfrak{F}} \mathcal{A}} are also of the form {K \cap^{\mathfrak{F}} \mathcal{A}}.

    Let {S \in \mathscr{P} Z (D \mathcal{A})}. Then {S = \left\{ X \cap^{\mathfrak{F}} \mathcal{A} \hspace{1em} | \hspace{1em} X \in T \right\}} where {T \in \mathscr{P} \mathscr{P} U}.

    How to calculate {\bigcup^{Z (D \mathcal{A})} \left\{ X \cap^{\mathfrak{F}} \mathcal{A} \hspace{1em} | \hspace{1em} X \in T \right\}} and {\bigcap^{Z (D \mathcal{A})} \left\{ X \cap^{\mathfrak{F}} \mathcal{A} \hspace{1em} | \hspace{1em} X \in T \right\}}?

    I conjecture:

    {\bigcup^{Z (D \mathcal{A})} \left\{ X \cap^{\mathfrak{F}} \mathcal{A} \hspace{1em} | \hspace{1em} X \in T \right\} = \mathcal{A} \cap^{\mathfrak{F}} \bigcup T} and {\bigcap^{Z (D \mathcal{A})} \left\{ X \cap^{\mathfrak{F}} \mathcal{A} \hspace{1em} | \hspace{1em} X \in T \right\} = \mathcal{A} \cap^{\mathfrak{F}} \bigcap T}.

    It probably may be helpful if we would calculate {\bigcup^{\mathfrak{F}} S} or {\bigcap^{\mathfrak{F}} S}. (However it simple to show that in general {\bigcap^{\mathfrak{F}} S \neq \bigcap^{Z (D \mathcal{A})} S}.)

    {\bigcup^{\mathfrak{F}} S = \bigcup^{\mathfrak{F}} \left\{ X \cap^{\mathfrak{F}} \mathcal{A} \hspace{1em} | \hspace{1em} X \in T \right\}}. Sadly there are no distributive law we could apply here.

    {\bigcap^{\mathfrak{F}} S = \bigcap^{\mathfrak{F}} \left\{ X \cap^{\mathfrak{F}} \mathcal{A} \hspace{1em} | \hspace{1em} X \in T \right\} = \mathcal{A} \cap^{\mathfrak{F}} \bigcap^{\mathfrak{F}} T}. Sadly {\bigcap^{\mathfrak{F}} T} is not necessarily a set.

    5. Consequences of our conjecture

    5.1. Closedness of joins on {Z (D \mathcal{A})}

    Above it was proved that closedness of joins of {Z (D \mathcal{A})} follows from our main conjecture.

    5.2. Coinciding pseudodifference and second pseudodifference

    It seems that using our conjecture we can prove that quasidifference and second quasidifference coincide for the lattice of filter objects.

    See this and this wiki pages for details about the current state of the problem about coinciding quasidifference and second quasidifference.

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    Filter objects

    October 31, 2009 by porton

    Let U is a set. A filter (on U) \mathcal{F} is by definition a non-empty set of subsets of U such that A,B\in\mathcal{F} \Leftrightarrow A\cap B\in\mathcal{F}. Note that unlike some other authors I do not require \varnothing\notin\mathcal{F}.

    For greater clarity I will use filter objects instead of filters. Below I will describe the properties of filter objects without exact definition and the proofs. You can look here for the formalistic behind.

    I will denote the set of all filters objects on a set U as \mathfrak{F}. Filter objects are bijectively related with filters by the bijection “\mathrm{up}” from the set of filter objects to the set of filters. A filter object corresponding to principal filter generated by a set A is equal to A. (Thus the set of subsets of U is a subset of \mathfrak{F}.)

    Formal definition of filter objects in the framework of ZF is given here. We will not need the exact definition of filter objects, but only the facts that “\mathrm{up}” is a bijection from filter objects to filters and that a filter object corresponding to principal filter generated by a set A is equal to A.

    I will define the order on the set of filter objects by the formula \mathcal{A}\subseteq\mathcal{B} \Leftrightarrow \mathrm{up} \mathcal{A} \supseteq \mathrm{up} \mathcal{B} for every filter objects \mathcal{A} and \mathcal{B}. This order well-agrees with the order of sets on U.

    \mathfrak{F} with the above defined order is a complete lattice. (See this draft article for a proof.)

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    Principal filters are center – solved

    October 31, 2009 by porton

    I have proved this conjecture:

    Theorem 1 If {\mathfrak{F}} is the set of filter objects on a set {U} then {U} is the center of the lattice {\mathfrak{F}}. (Or equivalently: The set of principal filters on a set {U} is the center of the lattice of all filters on {U}.)

    Proof: I will denote {Z (\mathfrak{F})} the center of the lattice {\mathfrak{F}}. I will denote {\mathrm{atoms}^{\mathfrak{A}} a} the set of atoms of a lattice {\mathfrak{A}} under its element {a}.

    Let {\mathcal{X} \in Z (\mathfrak{F})}. Then exists {\mathcal{Y} \in Z (\mathfrak{F})} such that {\mathcal{X} \cap^{\mathfrak{F}} \mathcal{Y} = \emptyset} and {\mathcal{X} \cup^{\mathfrak{F}} \mathcal{Y} = U}. Consequently, there are {X \in \mathrm{up} \mathcal{X}} such that {X \cap^{\mathfrak{F}} \mathcal{Y} = \emptyset}; we have also {X \cup^{\mathfrak{F}} \mathcal{Y} = U}. Suppose {X \supset \mathcal{X}}. Then (because for {\mathfrak{F}} is true the disjunct propery of Wallman, see [1]) exists {a \in \mathrm{atoms}^{\mathfrak{F}} X} such that {a \notin \mathrm{atoms}^{\mathfrak{F}} \mathcal{X}}. We can conclude also {a \notin \mathrm{atoms}^{\mathfrak{F}} \mathcal{Y}}. Thus {a \notin \mathrm{atoms}^{\mathfrak{F}} ( \mathcal{X} \cup^{\mathfrak{F}} \mathcal{Y})} and consequently {\mathcal{X} \cup^{\mathfrak{F}} \mathcal{Y} \neq U} what is a contradiction. We have {\mathcal{X} = X \in \mathscr{P} U}.

    Let now {X \in \mathscr{P} U}. Then {X \cap (U \setminus X) = 0} and {X \cup (U \setminus X) = U}. Thus {X \cap^{\mathfrak{F}} (U \setminus X) = \bigcap^{\mathfrak{F}} \left\{ X \cap (U \setminus X) \right\} = \emptyset}; {X \cup^{\mathfrak{F}} (U \setminus X) = \bigcap^{\mathfrak{F}} (\mathrm{up} X \cap \mathrm{up} (U \setminus X)) = \bigcap^{\mathfrak{F}} \left\{ U \right\} = U} (used formulas from [1]). We have shown that {X \in Z (\mathfrak{F})}. \Box

    This theorem may be generalized for a wider class of filters on lattices than only filters on lattices of a subsets of some set.

    [1] Victor Porton. Funcoids and Reloids. http://www.mathematics21.org/binaries/set-filters.pdf

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    Are principal filters the center of the lattice of filters?

    October 31, 2009 by porton

    This conjecture has a seemingly trivial case when \mathcal{A} is a principal filter. When I attempted to prove this seemingly trivial case I stumbled over a looking simple but yet unsolved problem:

    Let U is a set. A filter (on U) \mathcal{F} is by definition a non-empty set of subsets of U such that A,B\in\mathcal{F} \Leftrightarrow A\cap B\in\mathcal{F}. Note that unlike some other authors I do not require \varnothing\notin\mathcal{F}. I will denote \mathscr{F} the lattice of all filters (on U) ordered by set inclusion. (I skip the proof that \mathscr{F} is a lattice).

    Conjecture The set of principal filters on a set U is the center of the lattice of all filters on U.

    Note that by center of a (distributive) lattice I mean the set of all its complemented elements.

    I did a little unsuccessful attempt to solve this problem before I’ve put it into this blog. I will think about this more. You may also attempt to solve this open problem for me.

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