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I've seen a few answers to this, like here and but they are not satisfying to me (possibly too advanced).

The definitions in my book are as follows:

  • A binary relation $\mathrel{R}$ on two sets $A$ and $B$ is a subset of $A \times B$, whose elements can be written $a \mathrel{R} b$.
  • When we say $\mathrel{R}$ is binary relation on $A$, we mean that $R$ is a subset of $A \times A$.
  • The relation $R$ is transitive if $a \mathrel{R} b$ and $b \mathrel{R} c$ imply $a \mathrel{R} c$, for all $a,b,c \in A$.
  • The relation $R$ is symmetric if $a \mathrel{R} b$ implies $b \mathrel{R} a$.

Browsing Math Stack it appears those definitions are standard. Consider the following question: if a nonempty relation is symmetric and transitive, is it also reflexive?

I say yes. But in a discussion with a peer, they provide the example: consider the relation $R$ on $A$ where $A = \{0,1,2\}$ but $R = \{(0,1), (0,2), (1,0), (2,0), (2,1), (1,2)\}$. They claim this relation is transitive but I say no, because in order for it to be so we need $0 \mathrel{R} 1$ and $1 \mathrel{R} 0$ to imply $0 \mathrel{R} 0$, but clearly $(0, 0) \notin R$.

Who's right? And is it possible to generate such a nonempty relation?

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  • $\begingroup$ No. Any one of the three is independent of the other two. $\endgroup$ – Brevan Ellefsen Jan 19 at 21:24
  • $\begingroup$ The reflexivity of the relation is here to ensure that any element is in relation with another one. It makes perfect sense from a linguistic point of view right ? If someone has relation with no one, then maybe he should no be part of the group. $\endgroup$ – J.F Jan 19 at 21:40
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You can produce relation that is both transitive and symetric but not reflexive b considering $R=\{(0,1), (1,0), (1,1), (0,0)\}$ on the set $X=\{0,1,2\}$.

(the problem here is that $(2,2)\not\in R$)

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    $\begingroup$ Exactly this. Transitivity does not imply that every element of the set belongs to the relation- only reflexivity guarantees this property. $\endgroup$ – LuuBluum Jan 19 at 21:28
  • $\begingroup$ @Ispil 's comment is what answered my question. But let me just make sure. So, just because (w.r.t this answer post) there does not exist some $x \mathrel{R} 2$ does not mean that $R$ is not transitive? And if that is true, doesn't that mean that my definition of transitivity is off "... for all $a, b, c \in A$"? $\endgroup$ – Zduff Jan 19 at 22:14
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    $\begingroup$ The definition for transitivity is an implication: that if $aRb$ and $bRc$ then $aRc$. If there's no $aRb$ and $bRc$, then the implication is true regardless of the truth of $aRc$. $\endgroup$ – LuuBluum Jan 20 at 5:32
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The example by your peer is indeed not transitive, as you pointed out. A correct counterexample would be $\{(0,0),(0,1),(1,0),(1,1)\}$ on the set $\{0,1,2\}$.

More can be said, though. Let $R$ be a transitive symmetric relation on $A$. Then for all $a\in A$ such that $aRb$ for some $b\in A$, we have $aRa$. Indeed, by symmetry $aRb$ and $bRa$ and by transitivity $aRa$.

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  • $\begingroup$ Is my definition of transitivity incorrect then, since it states "... for all $a,b,c \in A$"? $\endgroup$ – Zduff Jan 19 at 22:16
  • $\begingroup$ I don't understand why you suddenly think your definition is incorrect. $\endgroup$ – SmileyCraft Jan 19 at 22:17
  • $\begingroup$ I don't understand why you think the thought was "sudden". $\endgroup$ – Zduff Jan 19 at 23:17
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a) your peer is incorrect with their example, as also stated by you and SmileyCraft, because $(0,0)\notin R$.

generalizing the example given by GF.:

b) still, not every binary symmetric relation $R$ on $A$ that is symmetric and transitive is also reflexive. The relation $R$ would be reflexive if $(X,X)\in R$ for all $X\in A$. Symmetry and transitivity only guarantee that IF $(X,Y) \in R$ for a given $X\in A$ THEN also $(Y,X) \in R$ by symmetry and $(X,X) \in R$ by transitivity. However, there can be some $X\in A$ for which there is no $(X,Y) \in R$ or $(Y,X) \in R$ for any $Y\in A$.

symmetry, transitivity and reflexivity together are the defining properties of an equivalence class.

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