Are there any examples of reflexive, symmetric and intransitive relations on sets containing more than one element (not explicitly defined by their elements? (edited due to imprecision, I apologize.)
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Have you tried to find some example on sets with 2 or 3 elements for example? – Arnaud D. Nov 21 '19 at 17:12
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Possible duplicate : https://math.stackexchange.com/questions/268726/are-there-real-life-relations-which-are-symmetric-and-reflexive-but-not-transiti – Arnaud D. Nov 21 '19 at 17:15
2 Answers
Sure. How about $R=\{(1,1),(2,2),(3,3), (1,2), (2,1), (2,3), (3,2)\}$ on $\{1,2,3\}$?
It's reflexive because of $(1,1), (2,2), $ and $(3,3)$, and it's symmetric,
but it's not transitive, because $(1,2),(2,3)\in R$ but $(1,3)\not\in R$.
Addendum in response to edit of the question:
If you want such a relation defined with some properties,
rather than explicitly in terms of the elements,
define $(x,y)\in R\;$ by $\;|x-y|\le1$.

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W.Tanner, I forgot to ask (precisely), my big mistake, if there exist such relations defined with some operation or number properties. – PinkyWay Nov 21 '19 at 17:19
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1@VerkhotsevaKatya: you mean like $(x,y)\in R$ defined by $|x-y|\le1$ ? – J. W. Tanner Nov 21 '19 at 17:57
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For a big class of examples, consider an undirected graph $G$ on a set of vertices $V$, and say two vertices $v_1, v_2 \in V$ are related if $v_1=v_2$ or there is an edge from $v_1$ to $v_2$ in $G$.
This defines a reflexive, symmetric relation on $V$, but will not typically be transitive. (It will be transitive if $G$ is complete, or a union of complete subgraphs, but not in general.)

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Thank you @Clive Newstead! I even forgot to write I was looking for relations that are not necessarily defined by their elements counted, but you gave me an interesting answer anyway! – PinkyWay Nov 21 '19 at 17:23