Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have proved the below conjecture for the special cases $n\in\{0,1,2\}$. The cases $n\ge 3$ (finite and infinite) are unknown.

If the following conjecture is true, I don't expect that you will be able to prove it, because I myself failed (and I am not a moron).

But maybe it has a counterexample? Can you provide a counter-example? (maybe for the case $n=3$?) I was too obsessed with the idea that my conjecture is true too much believing my intuition. Now I humble down and ask to try to find a counter-example.

Or maybe I overestimate my skill to solve this kind of problems and someone can nevertheless prove it? (Even the special case $n=3$ would be interesting.)

Definition 1 A filtrator is a pair $\left( \mathfrak{A}; \mathfrak{Z} \right)$ of a poset $\mathfrak{A}$ and its subset $\mathfrak{Z}$.

Having fixed a filtrator, we define:

Definition 2 $\mathrm{up}\,x = \left\{ Y \in \mathfrak{Z} \hspace{0.5em} | \hspace{0.5em} Y \geqslant x \right\}$ for every $X \in \mathfrak{A}$.

Definition 3 $E^{\ast} K = \left\{ L \in \mathfrak{A} \hspace{0.5em} | \hspace{0.5em} \mathrm{up}\,L \subseteq K \right\}$ (upgrading the set $K$) for every $K \in \mathscr{P} \mathfrak{Z}$.

Definition 4 A free star on a join-semilattice $\mathfrak{A}$ with least element $0$ is a set $S$ such that $0 \not\in S$ and

$\displaystyle \forall A, B \in \mathfrak{A}: \left( A \cup B \in S \Leftrightarrow A \in S \vee B \in S \right) .$

Definition 5 Let $\mathfrak{A}$ be a family of posets, $f \in \mathscr{P} \prod \mathfrak{A} (\prod \mathfrak{A}$ has the order of function space of posets), $i \in \mathrm{dom}\,\mathfrak{A}$, $L \in \prod \mathfrak{A}|_{\left( \mathrm{dom}\,\mathfrak{A} \right) \setminus \left\{ i \right\}}$. Then

$\displaystyle \left( \mathrm{val}\,f \right)_i L = \left\{ X \in \mathfrak{A}_i \hspace{0.5em} | \hspace{0.5em} L \cup \left\{ (i ; X) \right\} \in f \right\} .$

Definition 6 Let $\mathfrak{A}$ is a family of posets. A multidimensional funcoid (or multifuncoid for short) of the form $\mathfrak{A}$ is an $f \in \mathscr{P} \prod \mathfrak{A}$ such that we have that:

$\left( \mathrm{val} f \right)_i L$ is a free star for every $i \in \mathrm{dom} \mathfrak{A}$, $L \in \prod \mathfrak{A}|_{\left( \mathrm{dom} \mathfrak{A} \right) \setminus \left\{ i \right\}}$ and $f$ is an upper set.

$\mathfrak{A}^n$ is a function space over a poset $\mathfrak{A}$ that is $a\le b\Leftrightarrow \forall i\in n:a_i\le b_i$ for $a,b\in\mathfrak{A}^n$.

Conjecture Let $\mho$ be a set, $\mathfrak{F}$ be the set of filters on $\mho$ ordered reverse to set theoretic inclusion, $\mathfrak{P}$ be the set of principal filters on $\mho$, let $n$ be an index set. Consider the filtrator $\left( \mathfrak{F}^n ; \mathfrak{P}^n \right)$. If $f$ is a multifuncoid of the form $\mathfrak{P}^n$, then $E^{\ast} f$ is a multifuncoid of the form $\mathfrak{F}^n$.

My solution for the cases $n\in\{0,1,2\}$

share|improve this question
    
Would you please make an effort to think of a more substantive subject line for your post? –  MJD Apr 30 '12 at 19:49
    
@Mark Dominus: Done. –  porton Apr 30 '12 at 19:51
    
What does f.o. abbreviate? Never mind: from your paper I see that it must be filters. –  Brian M. Scott Apr 30 '12 at 20:36
    
@Brian M. Scott: Oh, I removed this unconventional term. Now I say: "the set of filters on ℧ ordered reverse to set theoretic inclusion." –  porton Apr 30 '12 at 20:39
    
By upper set do you mean a set that if it contains $y$ then it contains all $x$ such that $x\geq y$? –  Apostolos Apr 30 '12 at 22:28

1 Answer 1

up vote -4 down vote accepted

Oh, I unexpectedly found a really simple proof for this conjecture.

The proof is presented in this my draft article: http://www.mathematics21.org/binaries/nary.pdf

Also my blog post: http://portonmath.wordpress.com/2012/05/01/upgrading-conjecture-proved/

share|improve this answer
3  
You should expand a bit on the details, so that the answer is helpful to others. After some time, you can accept this answer. –  Mariano Suárez-Alvarez May 2 '12 at 6:19
    
I should admit that I can't understand why these posts are downvoted. As downvote needs 125 reps they are from good users. It may be a policy or.. I don't know..However I hardly believe it's natural.. –  user59671 Feb 16 '13 at 0:43
1  
"Oh, I UNEXPECTEDLY found a really simple proof for this conjecture." ROFL, obviously trying to promote his "research" –  Alexander Grothendieck Nov 16 '13 at 23:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.