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Please, can you help a beginner mathematician with the following problem?

Is there a binary relation that is reflexive, symmetric, antisymmetric but not transitive?

Definitions:

Relation

Let be two sets $A$,$B$ $\neq$$\emptyset$. A relation $\mathscr{R}$ of $A$ to $B$ is the ordered triple ($A$,$B$,$\mathscr{R}$) where $\mathscr{R}$ $\subset$ $A$$\times$$B$, $A$ is called input set, $B$ is called output set and $\mathscr{R}$ is called matching rule or graphic.

Note: A particular case of relation is when the input set and output set are equal i.e. $A$=$B$.
Let $A$ $\neq$$\emptyset$. Hereinafter, we say that $\mathscr{R}$ it is a relation of $A$ to $A$. Furthermore, $(a,b)$$\in$ $\mathscr{R}$, then we will denote $a$ $\mathscr{R}$ $b$.

Reflexive

A relation $\mathscr{R}$ is called reflexive iff: $\forall x\in A:$ $x$ $\mathscr{R}$ $x$.

Symmetric

A relation $\mathscr{R}$ is called symmetric iff: $\forall x,y\in A:$ $x$ $\mathscr{R}$ $y$ $\Rightarrow$ $y$ $\mathscr{R}$ $x$.

Transitive

A relation $\mathscr{R}$ is called transitive iff: $\forall x,y,z\in A:$ $x$ $\mathscr{R}$ $y$ $\wedge$ $y$ $\mathscr{R}$ $z$ $\Rightarrow$ $x$ $\mathscr{R}$ $z$.

Antisymmetric

A relation $\mathscr{R}$ is called antisymmetric iff: $\forall x,y\in A:$ $x$ $\mathscr{R}$ $y$ $\wedge$ $y$ $\mathscr{R}$ $x$ $\Rightarrow$ $x$ $=$ $y$

If the answer is true, then please show me a couple of examples.
Thank you.

Quote of the day:

"There is no branch of mathematics, however abstract, which may not some day be applied to phenomena of the real world".

Nicolai Ivanovitch Lobachevsky
1792-1856

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  • $\begingroup$ As I told you, it can be only isolated points. $\endgroup$
    – L F
    Nov 1, 2016 at 22:58
  • $\begingroup$ Oh, I understand. Please, could you explain with a couple of examples? $\endgroup$
    – Oromion
    Nov 1, 2016 at 23:06
  • $\begingroup$ The answer of Alephnull shows that a relation with your characteristics is nessesary identity, which is transitive. So you can't ask for non transivity $\endgroup$
    – L F
    Nov 1, 2016 at 23:13
  • $\begingroup$ Pretty sure Nicolai Lobachevsky is wrong... $\endgroup$ Nov 2, 2016 at 2:02
  • $\begingroup$ See also: Examples and Counterexamples of Relations which Satisfy Certain Properties $\endgroup$ Oct 25, 2021 at 13:14

2 Answers 2

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Assume we have such a relation. It is symmetric so xRy implies yRx. It is antisymmetric so xRy and yRx implies x=y. But putting this together we get xRy implies x=y. Thus our relation is the identity function over some set. But the identity function is transitive vacuously. This is a contradiction.

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  • $\begingroup$ i'll study your proof. Thank so much.. $\endgroup$
    – Oromion
    Nov 1, 2016 at 22:24
  • $\begingroup$ Thank you. I've already understood. $\endgroup$
    – Oromion
    Jul 10, 2017 at 20:16
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I don't know if this is correct, but it seems that if the relation is symmetric then x R y -> y R x. Hence, x R y ^ y R x -> x = y. So, if you have x R y, y R z, since x = y -> x R z, so the relation is necessarily transitive.

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    $\begingroup$ When you don't know if your answer is correct, it's best to leave your "answer" as a comment below Oromion's post. $\endgroup$
    – amWhy
    Nov 1, 2016 at 22:10
  • $\begingroup$ oh ok gotcha. thanks. $\endgroup$
    – green frog
    Nov 1, 2016 at 22:11
  • $\begingroup$ i'll study your proof too. Thank you. $\endgroup$
    – Oromion
    Nov 1, 2016 at 22:25

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