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I was studying relations and functions and transitive relations were defined as :

$$R=\{(x,y)\} ; xRy ~\text{and}~ yRz \implies xRz$$

But what about a single element $(a,b)$?

For example, if the relation was

$$R = \{(1,2),(2,1),(1,1),(2,2)\}$$

Then clearly, it is transitive, but if the relation were : $$R = \{(1,2),(2,1),(1,1),(2,2),(5,6)\}$$

Or simply :

$$R=\{(5,6)\}$$

Then are these relations transitive ?

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  • $\begingroup$ You might want to change the question in the title to the one actually being asked. (First and foremost, replace "set" by "relation".) $\endgroup$ – martin.koeberl May 10 '17 at 1:34
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Yup, that's transitive! For every triple of elements $a, b, c$, if $aRb$ and $bRc$ then $aRc$. It doesn't matter that there are some elements - namely, $5$ and $6$ - which can't be "extended" to form a triple. There's no more to transitivity than the definition you wrote above.

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  • $\begingroup$ Then what is '$c$' here ( in my example)? Is it defined to be transitive or there is any explanation? $\endgroup$ – Jaideep Khare May 10 '17 at 1:23
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    $\begingroup$ @JaideepKhare You're misunderstanding the definition. The definition just says whenever you have such a triple $a, b, c$, then [stuff]. It doesn't say that every element is part of some triple. The only way a relation $S$ is non-transitive is if there are three elements $a, b, c$ with $aSb$, $bSc$, but not $aSc$ - can you find such a triple of elements in this case? $\endgroup$ – Noah Schweber May 10 '17 at 1:24
  • $\begingroup$ Oh, now I understand, I think I misunderstood 'if ...(..)' as 'iff...(..)'. Right? $\endgroup$ – Jaideep Khare May 10 '17 at 1:26
  • $\begingroup$ @JaideepKhare Yeah, it's just a one-way implication. In particular, the empty relation is transitive! $\endgroup$ – Noah Schweber May 10 '17 at 1:41
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    $\begingroup$ @BrianTung It is a mystery to me that what do you mean by 'Not quite', btw thanks for your explanation! $\endgroup$ – Jaideep Khare May 10 '17 at 1:48

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