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Let $R$ be a relation that is transitive. Does complementary of $R$ ($\overline R$) is transitive?($\overline R$is hold transitive)

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You need more hypotheses, because as it stands, the statement is false: $\varnothing$ and $A\times A$ are complementary, transitive relations on $A$. –  Brian M. Scott Jan 3 '13 at 15:28
    
Sorry.I think new question is correct. –  geni Jan 3 '13 at 15:39

2 Answers 2

This isn't true, even for nontrivial relations.

For instance, consider $R \subseteq \{ 1,2,3 \}^2$ given by $$R = \{ (1,2),(1,3),(2,3) \}$$ Then $R$ is transitive, and $$\bar R = \{(1,1), (2,1), (2,2), (3,1), (3,2), (3,3) \}$$ which is also transitive.


Edit: Even with the question edit, the result still doesn't hold. For instance, let $R = \{ (1,3) \}$ on the set $\{1,2,3\}$. Then $(1,2) \in \bar R$ and $(2,3) \in \bar R$ but $(1,3) \not \in \bar R$, so $\bar R$ is not transitive, even though $R$ is transitive.

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Or more generally $<$ and $\ge$, where $<$ is any linear order on the underlying set. –  Brian M. Scott Jan 3 '13 at 15:36
    
Sorry,I edited body of question. –  geni Jan 3 '13 at 15:44
    
@geni: It still doesn't hold; see my edit. –  Clive Newstead Jan 3 '13 at 15:49
    
(I may have missed something though; I don't know what 'hold transitive' means.) –  Clive Newstead Jan 3 '13 at 15:50

Knowing only that a relation is transitive doesn't allow you to conclude anything about the transitivity of its complement.

For example the standard order relation $<$ on the integers is transitive, but its complement $\geq$ is also transitive.

On the other hand, the equality relation $=$ on the integers is transitive, but its complement $\neq$ is not.

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