Victor Porton's Math Blog http://portonmath.wordpress.com Math research of Victor Porton Sat, 07 Nov 2009 21:09:38 +0000 http://wordpress.com/ en hourly 1 http://www.gravatar.com/blavatar/f5dd7d3b4b95f8bca977e752fe1d88fe?s=96&d=http://s.wordpress.com/i/buttonw-com.png Victor Porton's Math Blog http://portonmath.wordpress.com Exposition: Complementive filters are complete lattice http://portonmath.wordpress.com/2009/11/02/exposition-complementive-filters/ http://portonmath.wordpress.com/2009/11/02/exposition-complementive-filters/#comments Sun, 01 Nov 2009 22:40:00 +0000 porton http://portonmath.wordpress.com/?p=217 ]]>

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. In fact this is equivalent to our conjecture:

Proposition 4 Our main conjecture is equivalent to join-closedness of the sublattice {Z (D \mathcal{A})} of the lattice {\mathfrak{F}}.

Proof: The reverse implication is trivial. Let’s prove the direct implication: 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 X} 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 X}.

    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})} is equivalent to 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.

    ]]> http://portonmath.wordpress.com/2009/11/02/exposition-complementive-filters/feed/ 0 porton Filter objects http://portonmath.wordpress.com/2009/10/31/filter-objects/ http://portonmath.wordpress.com/2009/10/31/filter-objects/#comments Sat, 31 Oct 2009 20:33:49 +0000 porton http://portonmath.wordpress.com/?p=208 ]]>

    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.)

    ]]> http://portonmath.wordpress.com/2009/10/31/filter-objects/feed/ 1 porton Principal filters are center – solved http://portonmath.wordpress.com/2009/10/31/principal-filters-are-center-solved/ http://portonmath.wordpress.com/2009/10/31/principal-filters-are-center-solved/#comments Sat, 31 Oct 2009 19:47:58 +0000 porton http://portonmath.wordpress.com/?p=198 ]]>

    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

    ]]>     http://portonmath.wordpress.com/2009/10/31/principal-filters-are-center-solved/feed/ 1 porton     Are principal filters the center of the lattice of filters? http://portonmath.wordpress.com/2009/10/31/are-principal-filters-the-center-of-the-lattice-of-filters/ http://portonmath.wordpress.com/2009/10/31/are-principal-filters-the-center-of-the-lattice-of-filters/#comments Sat, 31 Oct 2009 16:38:55 +0000 porton http://portonmath.wordpress.com/?p=193 ]]>

    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.

    ]]>     http://portonmath.wordpress.com/2009/10/31/are-principal-filters-the-center-of-the-lattice-of-filters/feed/ 2 porton     Collaborative math research – a real example http://portonmath.wordpress.com/2009/10/25/collaborative-research-of-filters/ http://portonmath.wordpress.com/2009/10/25/collaborative-research-of-filters/#comments Sat, 24 Oct 2009 21:36:37 +0000 porton http://portonmath.wordpress.com/?p=62 ]]>

    There were much talking about writing math research articles collaboratively but no real action. I present probably the first real example of a research manuscript ready to be written collaboratively.

    I wrote the draft Filters on Posets and Generalizations which I present to the online mathematical society to finish writing it as a collaborative project.

    This blog post will discuss both general aspects of collaboratively writing math research and particular aspects of this “Filters on Posets and Generalizations” manuscript.

    I just finished to copy from my old document to the wiki site. Now it is ready for finishing it writing collaboratively.

    Licensing

    I would enjoy to license my manuscript under GFDL or an other free license.

    But in the future I will need to publish my manuscript or fragments thereof in a peer-reviewed math journal (to be able to properly refer to it from my future manuscripts and from Wikipedia and PlanetMath). I’m afraid that the journal may not accept an article licensed under GFDL.

    So currently I require that every submitter would surrender his copyright for me. In exchange I promise only to acknowledge significant contributions.

    This situation is clearly unacceptable. We should use free licenses for collaborative math research. But this depends on journals not to me alone. Excuse me for borrowing your copyright for now. Hopefully I will indeed re-license it under a free license for unlimited copying, distribution, and modification by other authors.

    We need a reform of the journal publishing to support free licenses.

    Choice of the wiki engine and site

    Probably the best choice for the wiki site to do research writing online would be Wikiversity. But they require Creative Commons Attribution-ShareAlike license. This seems not acceptable for me because of the above mentioned reasons.

    Finally, I put my manuscript on the wikidot.com site. Why this choice of the wiki site?

    I only found good enough inline math formulas support in the following freely available wiki sites:

    With Wikia for me there was the problem that I was unable to login to that site. As the only remaining choice I knew I picked wikidot.com. I think wikidot.com is a good choice for collaboratively writing math manuscripts.

    However wikidot.com has some short comings:

    • The free account can host at most 5 wiki sites.
    • Storage per site is limited to 300 Mb (seems enough for most math books however).
    • Wrong vertical alignment of math formulas.
    • No PDF export, no LaTeX export.
    • No support for LaTeX preamble and no way to defined LaTeX macroses.

    So, yes, we need a better math wiki. But for now we can use wikidot. We already can use it, not waiting when a suitable software will appear.

    BTW, somebody may setup MediaWiki or wikidot.org wiki on their own site specifically for collaboratively edited scientific research drafts. This may probably require having a dedicated server (or a virtual server).

    Now you may see my wiki for writing Filters on Posets and Generalizations collaboratively on wikidot.com.

    Relation with other math projects on the Web

    I has proposed finishing writing Filters on Posets and Generalizations collaboratively as a Polymath project. I hope that Polymath administrators will pick this my project.

    Work already done

    Previously I wrote an old version of the manuscript in PDF form using TeXmacs math text editor.

    Then I manually converted the text and formulas to the wikidot wiki. (In process of copying I also did some actual math rewriting so that it was also a research by the way.)

    Now it is ready for editing by everyone who wants to join the site.

    In the future when writing will be finished, I will convert back to TeXmacs file format and then export to LaTeX in order to be published with a conventional publisher.

    The structure of my wiki site

    Instead of reading this about the structure of the site, you could browse the actual site.

    However, my wiki consists (among other) of:

    • the main page;
    • the TODO page which lists things yet to do in the project;
    • other pages, particularly the manuscript itself.

    Among other my manuscripts contain some unsolved problems. Welcome, collaborators!

    About collaborative research in general

    So in the face of my project we have a test bed for the future of math research, doing research collaboratively.

    In the same venue as open source software reviolutionarized software industry, open research would a lot benefit for the progress of mathematical research.

    Last words

    Welcome, collaborators!

    ]]>     http://portonmath.wordpress.com/2009/10/25/collaborative-research-of-filters/feed/ 0 porton     Complete lattice generated by a partitioning – finite meets http://portonmath.wordpress.com/2009/10/20/generated-lattice-finite-meets/ http://portonmath.wordpress.com/2009/10/20/generated-lattice-finite-meets/#comments Tue, 20 Oct 2009 15:14:04 +0000 porton http://portonmath.wordpress.com/?p=169 ]]>

    I conjectured certain formula for the complete lattice generated by a strong partitioning of an element of complete lattice. Now I have found a beautiful proof of a weaker statement than this conjecture. (Well, my proof works only in the case of distributive lattices, but the case of non-distributive lattices is outside of my research area.)

    Let’s denote R = \left\{ \bigcup{}^{\mathfrak{A}}X | X\in\mathscr{P}S \right\} where S is a strong partitioning an element of the complete lattice \mathfrak{A}. Our conjecture is trivially equivalent to the statement that R is closed under arbitrary meets and joins.

    That R is closed regarding any joins is obvious. To finish proving the conjecture we need to show that R is closed under arbitrary meets. In this post I prove weaker result that R is closed under finite meets.

    I hope this finite case may serve as a model for the general infinite case. However it seems that generalizing it to infinite case is non-trivial.

    Theorem Let \mathfrak{A} is a distributive complete lattice and S is a strong partitioning of some element of this lattice. Then R is closed under finite meets.

    Proof Let X,Y\in\mathscr{P}S.

    Then \bigcup^{\mathfrak{A}} X \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y = \bigcup^{\mathfrak{A}} ((X \cap Y) \cup (X \setminus Y)) \cap^{\mathfrak{A}}\bigcup^{\mathfrak{A}} Y = ( \bigcup^{\mathfrak{A}} (X \cap Y)\cup^{\mathfrak{A}} \bigcup^{\mathfrak{A}} (X \setminus Y))\cap^{\mathfrak{A}} \bigcup Y = ( \bigcup^{\mathfrak{A}} (X \cap Y)\cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y) \cup ( \bigcup^{\mathfrak{A}} (X\setminus Y) \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y) = (\bigcup^{\mathfrak{A}} (X \cap Y) \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}}Y) \cup^{\mathfrak{A}} 0 = \bigcup^{\mathfrak{A}} (X \cap Y)\cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y.

    Applying the formula \bigcup^{\mathfrak{A}} X \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y = \bigcup^{\mathfrak{A}} (X \cap Y) \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y twice we get

    \bigcup^{\mathfrak{A}} X \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y = \bigcup^{\mathfrak{A}} (X \cap Y) \cap^{\mathfrak{A}} \bigcup (Y \cap (X \cap Y)) = \bigcup^{\mathfrak{A}} (X \cap Y) \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} (X \cap Y) = \bigcup^{\mathfrak{A}} (X \cap Y).

    But for any A,B\in R exist X,Y\in S such that A=\bigcup^{\mathfrak{A}}X and B=\bigcup^{\mathfrak{A}}Y. So A\cap^{\mathfrak{A}}B = \bigcup^{\mathfrak{A}} X \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y = \bigcup^{\mathfrak{A}} (X \cap Y) \in R.

    ]]>     http://portonmath.wordpress.com/2009/10/20/generated-lattice-finite-meets/feed/ 0 porton     Complete lattice generated by a partitioning of a lattice element http://portonmath.wordpress.com/2009/10/20/complete-lattice-generated-by-a-partitioning-of-a-lattice-element/ http://portonmath.wordpress.com/2009/10/20/complete-lattice-generated-by-a-partitioning-of-a-lattice-element/#comments Mon, 19 Oct 2009 23:02:25 +0000 porton http://portonmath.wordpress.com/?p=138 ]]>

    In this post I defined strong partitioning of an element of a complete lattice. For me it was seeming obvious that the complete lattice generated by the set S where S is a strong partitioning is equal to \left\{ \bigcup{}^{\mathfrak{A}}X | X\in\mathscr{P}S \right\}. But when I actually tried to write down the proof of this statement I found that it is not obvious to prove. So I present this to you as a conjecture:

    Conjecture The complete lattice generated by a strong partitioning S of an element of a complete lattice \mathfrak{A} is equal to \left\{ \bigcup{}^{\mathfrak{A}}X | X\in\mathscr{P}S \right\}.

    Proposition Provided that this conjecture is true, we can prove that the complete lattice [S] generated by a strong partitioning S of an element of a complete lattice is a complete atomic boolean lattice with the set of its atoms being S (Note: So [S] is completely distributive).

    Proof Completeness of [S] is obvious. Let A\in[S]. Then exists X\in\mathscr{P}S such that A=\bigcup{}^{\mathfrak{A}}X. Let B=\bigcup{}^{\mathfrak{A}}(S\setminus X). Then B\in[S] and A\cap B=0. A\cup B=\bigcup{}^{\mathfrak{A}}S is the biggest element of [S]. So we have proved that [S] is a boolean lattice.

    Now let prove that [S] is atomic with the set of atoms being S. Let z\in S and A\in[S]. If A\neq z then either A=0 or x\in X where A=\bigcup^{\mathfrak{A}}X, X\in\mathscr{P}S and x\neq z. Because S is a strong partitioning, \bigcup^{\mathfrak{A}}(X\setminus\{z\})\cap^{\mathfrak{A}}z=0 and \bigcup^{\mathfrak{A}}(X\setminus\{z\})\neq 0. So A=\bigcup^{\mathfrak{A}}X=\bigcup^{\mathfrak{A}}(X\setminus\{z\})\cup^{\mathfrak{A}}z\nsubseteq z.

    Finally we will prove that elements of [S]\setminus S are not atoms. Let A\in[S]\setminus S and A\neq 0. Then A\supseteq x\cup^{\mathfrak{A}}y where x,y\in S and x\neq y. If A is an atom then A=x=y what is impossible. QED

    The above conjecture as a step to solution to the original conjecture may also be considered for the polymath research problem. Or maybe we should research both these two problems in a single polymath set, as the solution of one of them may inspire the solution of the other of these two problems.

    ]]>     http://portonmath.wordpress.com/2009/10/20/complete-lattice-generated-by-a-partitioning-of-a-lattice-element/feed/ 1 porton     Partitioning elements of distributive and finite lattices http://portonmath.wordpress.com/2009/10/18/partitioning-distributive-finite-lattices/ http://portonmath.wordpress.com/2009/10/18/partitioning-distributive-finite-lattices/#comments Sun, 18 Oct 2009 15:45:58 +0000 porton http://portonmath.wordpress.com/?p=120 ]]>

    I proposed this open problem for the next polymath project. Now I will consider some its special simple cases.

    First, it is quite obvious that every strong partitioning is a weak partitioning.

    If (finite) meets of our lattice are distributive over arbitrary joins (or, equivalently, our lattice is brouwerian) then the reverse implication holds, that is a weak partitioning is a strong partitioning.

    Proof: Let S is a weak partitioning, let A,B\in\mathscr{P}\mathfrak{A} and A\cap B=0. Then

    \bigcup^{\mathfrak{A}} A \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} B =\bigcup^{\mathfrak{A}} \left\{ \left(\bigcup^{\mathfrak{A}} A \right) \cap^{\mathfrak{A}} Y | Y \in B \right\} \subseteq\bigcup^{\mathfrak{A}} \left\{ \bigcup^{\mathfrak{A}} (S \setminus Y) \cap^{\mathfrak{A}} Y | Y \in B \right\} =\bigcup^{\mathfrak{A}} \left\{ 0 \right\} = 0
    It was taken into account that S\setminus Y\supseteq A.

    So S is a strong partitioning. QED

    This does not solve the problem for me, because I want to prove it for the cases when our complete lattice is not brouwerian but is dual that is co-brouwerian. Co-brouwerian lattices appear as lattices of filter objects in my manuscript about filters and as lattices of funcoids in my texts about funcoids and reloids; it is simple to prove that for infinite set it is not brouwerian.

    Further: it is trivial that every finite distributive lattice is brouwerian. From this follows that weak and strong partitioning coincide for finite distributive lattices.

    A natural question to ask: Are there non-distributive finite lattices for which weak and strong partitioning do not coincide?

    Solving this question (what I have not yet attempted) would involve some combinatorics. If that shall go complex we may use a computer experiment enumerating many lattices. But I hope that the solution to this tiny question is simpler.

    ]]>     http://portonmath.wordpress.com/2009/10/18/partitioning-distributive-finite-lattices/feed/ 0 porton     Proposal: Partitioning a lattice element http://portonmath.wordpress.com/2009/10/17/proposal-partitioning/ http://portonmath.wordpress.com/2009/10/17/proposal-partitioning/#comments Sat, 17 Oct 2009 19:51:24 +0000 porton http://portonmath.wordpress.com/?p=117 ]]>

    I’ve given two different definitions for partitioning an element of a complete lattice (generalizing partitioning of a set). I called them weak partitioning and strong partitioning.

    The problem is whether these two definitions are equivalent for all complete lattices, or if are not then under which additional conditions these are equivalent. (I suspect these may be equivalent under the additional condition that our lattice is distributive.)

    I may seem greedy producing now already second or third (dependently on how to count) polymath proposal about my problems, but I just want to present all of my proposal for (hopefully) fair judges Terence Tao and Tim Gowers. The more the better.

    ]]>     http://portonmath.wordpress.com/2009/10/17/proposal-partitioning/feed/ 2 porton     Partitioning of a lattice element: a conjecture http://portonmath.wordpress.com/2009/10/17/partitioning-lattice-element/ http://portonmath.wordpress.com/2009/10/17/partitioning-lattice-element/#comments Sat, 17 Oct 2009 16:49:54 +0000 porton http://portonmath.wordpress.com/?p=102 ]]>

    Let \mathfrak{A} is a complete lattice. Let a\in\mathfrak{A}.

    I will call weak partitioning of a a set S\in\mathscr{P}\mathfrak{A}\setminus\{0\} such that

    \bigcup{}^{\mathfrak{A}}S = a \text{ and } \forall x\in S: x\cap^{\mathfrak{A}}\bigcup{}^{\mathfrak{A}}(S\setminus\{x\}) = 0.

    I will call strong partitioning of a a set S\in\mathscr{P}\mathfrak{A}\setminus\{0\} such that

    \bigcup{}^{\mathfrak{A}}S = a \text{ and } \forall A,B\in\mathscr{P}S: (A\cap B=0\Rightarrow \bigcup{}^{\mathfrak{A}}A\cap{}^{\mathfrak{A}}\bigcup{}^{\mathfrak{A}}B = 0).

    Question: Do exist complete lattices for which weak partitioning and strong partitioning are not the same?

    If this conjecture (that it is the same for arbitrary complete lattices) is indeed false for arbitrary complete lattices, we must find the cases when it is true. (I strongly suspect that it is true for distributive complete lattices.)

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