Talk:Relation is Connected iff Union with Inverse and Diagonal is Trivial Relation

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I might be being dense but in the definition of a connected relation I can't see that connected relations are necessarily reflexive. If they are not necessarily reflexive then I do not see how line 3 of the theorem section of this page that: $\mathcal R \cup \mathcal R ^{-1} = S \times S$

You are correct. It goes wrong at the point 'Since $\mathcal R$ is connected, either $(a,b) \in \mathcal R$ or $(b,a)\in\mathcal R$' where it is implicitly assumed that $a \ne b$.
The correct equivalent statement is that $S \times S \setminus \Delta \left({S}\right) \subseteq \mathcal R \cup \mathcal{R}^{-1}$. Here, $\Delta$ is the diagonal mapping. --Lord_Farin 07:21, 23 June 2012 (UTC)
Okay, leave it with me - I'll sort it out. --prime mover 08:21, 23 June 2012 (UTC)
I had confused Definition:Connected Relation with Definition:Total Relation, not having noticed the need for $a$ and $b$ to be distinct in the former for the condition to hold. (And I keep getting the words "connected" and "complete" confused, which is just senility.) --prime mover 12:44, 23 June 2012 (UTC)