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

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(Created page with "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 ultrafilt...")
 
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$\Gamma$ (in my book it's proved that $\Gamma$ is a sublattice of
 
$\Gamma$ (in my book it's proved that $\Gamma$ is a sublattice of
 
$\mathsf{FCD}$).
 
$\mathsf{FCD}$).
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 +
FIXME: Write what we should prove and then make blogpost at the discussion thread.
  
 
For every ultrafilter $x$ we have
 
For every ultrafilter $x$ we have

Revision as of 11:34, 13 April 2017

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 $\mathsf{FCD}$).

FIXME: Write what we should prove and then make blogpost at the discussion thread.

For every ultrafilter $x$ we have

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

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 $X$?