# A counter-example  for a set-theoretic problem?

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 $n\in\{0,1,2\}$

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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
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## 1 Answer

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/

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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.. – CutieKrait Feb 16 at 0:43
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