Union of two partial orderings

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As exitingcorpse remarks, antisymmetry may fail.

Transitivity can be a problem too, for example on domain $\{a,b,c\}$, with $S=\{(a,a), (b,b), (a,b),(c,c)\}, R= \{(a,a),(b,c),(c,c),(b,b)\}$. Then the union $S \cup R$ is not transitive: $(a,c)$ should be in it as $(a,b)$ and $(b,c)$ are...

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Heisenberg
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Heisenberg

Updated on June 11, 2020

Comments

  • Heisenberg
    Heisenberg over 3 years

    Suppose S and R are partial orderings. Does is necessarily mean that $R \cup S$ (union) is a partial ordering? If not what conditions would have to be met for it to be a partial ordering?

    • Cameron Buie
      Cameron Buie over 10 years
      A sufficient (but perhaps not necessary) condition is that the domains of $S$ and $R$ be disjoint.
    • citedcorpse
      citedcorpse over 10 years
      take the set to be $\{a,b\}$ with $S = \{(a,a), (a,b), (b,b)\}$ and $R = \{(a,a),(b,a),(b,b)\}$. Then the union fails to be antisymmetric.
    • Git Gud
      Git Gud over 10 years
  • Lord_Farin
    Lord_Farin over 10 years
    Hello, welcome to Mathematics.SE and thank you for your answer! I've enhanced the typesetting of your answer. For some basic information about writing maths at this site see e.g. here, here, here and here.