# Difference between revisions of "Funcoid bases/Proving existence of funcoid through lattice Gamma"

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\in S' \right\} = \\ \left\{ \langle X \rangle x \mid X | \in S' \right\} = \\ \left\{ \langle X \rangle x \mid X | ||

\in S' \right\}$ | \in S' \right\}$ | ||

+ | |||

+ | (taken into account that $\supfun{X_i}x$ is a principal filter). | ||

It follows $\langle Y \rangle x \in \left\{ \langle X \rangle x \mid X \in S' \right\}$ that is | It follows $\langle Y \rangle x \in \left\{ \langle X \rangle x \mid X \in S' \right\}$ that is |

## Revision as of 20:39, 13 April 2017

**Statement** Under the conditions "1" or "2" of our conjecture, we have that $S=\operatorname{up}\bigwedge^{\mathsf{FCD}} S=\operatorname{up}\bigwedge^{\mathsf{FCD}} S'$.

**Proof**
Denote $S' = S \cap \Gamma$. It's easy to show that $S'$ is a filter on
$\Gamma$ (in my book it's proved that $\Gamma$ is a sublattice of
$\mathsf{FCD}$).

For every ultrafilter $x$ we have

$\operatorname{up} \langle Y \rangle x \subseteq \operatorname{up} \left\langle \bigwedge^{\mathsf{FCD}} S' \right\rangle x = \\ \operatorname{up} \bigwedge_{X \in S'} \langle X \rangle x = \\ \left\{ \langle X_0 \rangle x \sqcap \ldots \sqcap \langle X_n \rangle x \mid i = 0, \ldots, n, X_i \in S' \right\} = \\ \left\{ \langle X \rangle x \mid X \in S' \right\}$

(taken into account that $\supfun{X_i}x$ is a principal filter).

It follows $\langle Y \rangle x \in \left\{ \langle X \rangle x \mid X \in S' \right\}$ that is

$\langle Y \rangle x = \langle X \rangle x$ for some $X \in S'$.

The next question is whether we can find such $X \in S$ (not necessarily $X
\in S'$!) that $\langle Y \rangle x = \langle X \rangle x$ for **all**
ultrafilters $x$. Or is there a counter-example for existence of such
$X$?