File:PartialOrders redundencies.pdf

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Summary

Description
English: Connection between strict and non-strict partial orders established by converse (cnv), reflexive closure (cls), and irreflexive kernel (ker). The mappings cls and ker (on the domain of irreflexive and reflexive relations, respectively) are inverse to each other, and cnv is inverse to itself, and commutes with cls and with ker. Therefore, we have a commutative diagram.

For illustration, we use four example relations on the set {1,2,3,4,5}. Each relation table shows a "*" whenever (x,y)∈R holds, where x and y corresponds to the row and column, respectively, like this:

Each of the four relations corresponds, with some (own) grain of salt, to the one Hasse diagram in the image center: Since a Hasse diagram cannot show if an element is related to itself, such relations must be tacitly understood to be absent for the strict versions, and present for the non-strict ones. For the ">" and the "≥" relation, the diagram must be read upside-down.

Moreover, if cpl denotes the complement of a relation, then, for every non-strict partial order R, one has

  1. cnv(ker(R)) ⊆ cpl(R), and
  2. equality holds iff R is a non-strict total order.
Proof of (1): If (x,y) ∈ cnv(ker(R)), then (y,x) ∈ ker(R), hence (y,x) ∈ ker(R) and yx, hence (x,y) ∉ R, since else x=y by asymmetry, hence (x,y) ∈ cpl(R).
Proof of (2): "If": Let R be connected, let (x,y) ∈ cpl(R), then (x,y) ∉ R, hence (y,x) ∈ R, which also implies that xy, hence (y,x) ∈ ker(R), hence (x,y) ∈ cnv(ker(R)).
"Only if": Assume for contradiction equality for a non-connected partial order R. Let (x,y), (y,x) ∉ R, then (x,y) ∈ cpl(R), hence (x,y) ∈ cnv(ker(R)), hence (y,x) ∈ ker(R), hence (y,x) ∈ R, contradicting the assumption.
Date
Source Own work
Author Jochen Burghardt
Other versions File:PartialOrders redundencies.pdf * File:PartialOrders redundencies svg.svg
LaTeX source code
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% foreground
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% y -->  1    2    3    4   5
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