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Given the relation $$\mathcal{R}=\{(1,1), (1,2), (2,2), (2,3), (3,3), (3,1)\}$$ the problem is to determine whether this relation is reflexive/symmetric/antisymmetric/transitive or not.

I understand why this relation is reflexive, and also why it is not symmetric and transitive.

But the textbook says it is NOT antisymmetric whereas I think it is antisymmetric.

My idea:

Def. of antisymmetry: $x\mathcal{R}y$ and $y\mathcal{R}x \implies x=y$

For $(a,b)$ in $\mathcal{R}$, there are only 3 elements in $\mathcal{R}$ that satisfy the condition "$x\mathcal{R}y$ and $y\mathcal{R}x$"; $(1,1), (2,2), (3,3)$. So the condition "$x=y$" trivially holds for those elements. Also, there is NO counterexample: that is, there is no $(2,1), (3,2)$ and $(1,3)$ in $\mathcal{R}$. So I think it is antisymmetric, but the book says it is NOT antisymmetric and there is no explanation.

Am I wrong? Please give your help!

mlc
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keon kwan
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    To me, this definitely looks antisymmetric by the definition you have given in your post. The textbook must be wrong or something. – Sarvesh Ravichandran Iyer Jul 30 '17 at 09:29
  • You are right.. Maybe it's a typo in your book? – bof Jul 30 '17 at 09:54
  • The relation is antisymmetric, but it is not asymmetric. Is is possible that your book meant asymmetric instead of antisymmetric? – mlc Jul 30 '17 at 10:02
  • Good. Thank you for your helps! – keon kwan Jul 30 '17 at 10:05
  • Other point: the relation is only reflexive under the extra condition that it is a is a relation on ${1,2,3}$. – drhab Jul 30 '17 at 10:26
  • This is the "$a$ either beats or draws against $b$ in a game of rock-paper-scissors" relation – Akiva Weinberger Jul 30 '17 at 10:56
  • Note that without knowing on which set $\mathscr R $ is a relation we can actually not tell whether it is reflexive. If it is supposed to be a relation on ${ 1, 2, 3, 4 } $ then it's not reflexive since it doesn't include $(4,4) $. – md2perpe Jul 30 '17 at 11:13

1 Answers1

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Your argument is perfectly fine, there must be a mistake in the book. You can also use the equivalent form of antysimmetry: If $ x \mathcal{R} y $ and $ x \neq y $ then $(y,x) \not \in \mathcal{R} $ which is the second part of your argument

Joaquin San
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