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Large filters of quasiorder lattices can be generated by few elements

Gábor Czédli

Abstract

For a poset $(P;\leq)$, the quasiorders (AKA preorders) extending the poset order "$\leq$" form a complete lattice $F$, which is a filter in the lattice of all quasiorders of the set $P$. We prove that if the poset order "$\leq$" is small, then $F$ can be generated by few elements.

Large filters of quasiorder lattices can be generated by few elements

Abstract

For a poset , the quasiorders (AKA preorders) extending the poset order "" form a complete lattice , which is a filter in the lattice of all quasiorders of the set . We prove that if the poset order "" is small, then can be generated by few elements.
Paper Structure (5 sections, 3 theorems, 61 equations, 2 figures)

This paper contains 5 sections, 3 theorems, 61 equations, 2 figures.

Key Result

Lemma 1.1

If $X$ is an at least two-element set of an accessible cardinality and this cardinality is different from $4$, then the complete lattice $\textup{Quo}(X)$ is $4$-generated as a complete lattice. For $|X|\geq 3$, $\textup{Quo}(X)$, as a complete lattice, is not $3$-generated. For $|X|=4$, $\textup{Qu

Figures (2)

  • Figure 1: A forest ${\mathbb P}$; see Example \ref{['exmpl:BlTrsst']}.
  • Figure 2: A forest ${\mathbb P}$; see Example \ref{['exmpl:jtKgmbRnk']}.

Theorems & Definitions (17)

  • Lemma 1.1: Czédli czg2017fourgen, Czédli and Kulin czgkulin
  • Definition 2.1
  • Definition 2.2
  • Definition 2.3
  • Theorem 2.4
  • Remark 2.5
  • Example 2.6
  • Example 2.7
  • Remark 2.8
  • proof : Proof of Theorem \ref{['thmmain']}
  • ...and 7 more