Publish here. Donate

Difference between revisions of "Funcoid bases/Reduce to meets of applying funcoids to ultrafilters"

From Virtual scientific conference
Jump to: navigation, search
(math typo)
(Undo revision 75 by Victor Porton (talk))
Line 5: Line 5:
 
Let $X_i\in S$. We need to prove for every binary relations $Q \in \operatorname{up} (X_0 \sqcap^{\mathsf{FCD}}
 
Let $X_i\in S$. We need to prove for every binary relations $Q \in \operatorname{up} (X_0 \sqcap^{\mathsf{FCD}}
 
\ldots \sqcap^{\mathsf{FCD}} X_n)$ that is $\langle Q \rangle x
 
\ldots \sqcap^{\mathsf{FCD}} X_n)$ that is $\langle Q \rangle x
\subseteq \langle X_0 \rangle x \sqcap \ldots \sqcap \langle X_n \rangle x$
+
\sqsupseteq \langle X_0 \rangle x \sqcap \ldots \sqcap \langle X_n \rangle x$
 
that is $\operatorname{up} \langle Q \rangle x \sqsupseteq \left\{ K_0 \sqcap \ldots
 
that is $\operatorname{up} \langle Q \rangle x \sqsupseteq \left\{ K_0 \sqcap \ldots
 
\sqcap K_n \mid \forall i : K_i \in \operatorname{up} \langle
 
\sqcap K_n \mid \forall i : K_i \in \operatorname{up} \langle

Revision as of 19:14, 11 April 2017

The idea behind this attempted proof is to reduce behavior of funcoids $\langle f\rangle$ with better known behavior of filters $\langle f\rangle x$ for an arbitrary ultrafilter $x$ (I remind that knowing $\langle f\rangle x$ for all ultrafilters $x$ on the domain, it's possible to restore funcoid $f$) and then to replace $\langle X_0 \sqcap^{\mathsf{FCD}} \ldots \sqcap^{\mathsf{FCD}} X_n\rangle x$ with $\langle X_0 \rangle x \sqcap \dots \sqcap \langle X_n \rangle x$.

The below is my attempted proof (trying to reduce the statement for $n$ to the statement for $n-1$):

Let $X_i\in S$. We need to prove for every binary relations $Q \in \operatorname{up} (X_0 \sqcap^{\mathsf{FCD}} \ldots \sqcap^{\mathsf{FCD}} X_n)$ that is $\langle Q \rangle x \sqsupseteq \langle X_0 \rangle x \sqcap \ldots \sqcap \langle X_n \rangle x$ that is $\operatorname{up} \langle Q \rangle x \sqsupseteq \left\{ K_0 \sqcap \ldots \sqcap K_n \mid \forall i : K_i \in \operatorname{up} \langle X_i \rangle x \right\}$ that $Q \in S$.

The below is a chain of equalities attempting the proof:

$\operatorname{up} \langle Q \rangle x \subseteq \left\{ K_0 \sqcap \ldots \sqcap K_n \mid \forall i : K_i \in \operatorname{up} \langle X_i \rangle x \right\} = \\ \left\{ P \sqcap K_n \mid P \in \operatorname{up} (\langle X_0 \rangle x \sqcap \ldots \sqcap \langle X_{n - 1} \rangle x) \wedge K_n \in \operatorname{up} \langle X_n \rangle x \right\} = \\ \left\{ P \sqcap K_n \mid P \in \operatorname{up} \langle X_0 \sqcap^{\mathsf{FCD}} \ldots \sqcap^{\mathsf{FCD}} X_{n - 1} \rangle x \wedge K_n \in \operatorname{up} \langle X_n \rangle x \right\} = \\ \left\{ P \sqcap K_n \mid K \in \operatorname{up} (X_0 \sqcap^{\mathsf{FCD}} \ldots \sqcap^{\mathsf{FCD}} X_{n - 1}), P \in \operatorname{up} \langle K \rangle x \wedge K_n \in \operatorname{up} \langle X_n \rangle x \right\} = \\ \bigcup_{K \in \operatorname{up} (X_0 \sqcap^{\mathsf{FCD}} \ldots \sqcap^{\mathsf{FCD}} X_{n - 1})} \left\{ P \sqcap K_n \mid P \in \operatorname{up} \langle K \rangle x \wedge K_n \in \operatorname{up} \langle X_n \rangle x \right\} = \\ \bigcup_{K \in \operatorname{up} (X_0 \sqcap^{\mathsf{FCD}} \ldots \sqcap^{\mathsf{FCD}} X_{n - 1})} \operatorname{up} (\langle K \rangle x \sqcap \langle X_n \rangle x) = \\ \bigcup_{K \in \operatorname{up} (X_0 \sqcap^{\mathsf{FCD}} \ldots \sqcap^{\mathsf{FCD}} X_{n - 1})} \operatorname{up} \langle K \sqcap^{\mathsf{FCD}} X_n \rangle x \subseteq \bigcup_{K \in \operatorname{up} (X_0 \sqcap^{\mathsf{FCD}} \ldots \sqcap^{\mathsf{FCD}} X_{n - 1})} Q_K$

because $\langle K \rangle x \sqcap \langle X_n \rangle x = \langle K \sqcap^{\mathsf{FCD}} X_n \rangle x \sqsupseteq Q_K$ for $Q_K \in S$

The remaining attempt to prove our conjecture is stalled.