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$A = \{1, 2, 3, 4\}, R \subset A \times A$

Why is $\{(1,1),(2,2),(3,3)\}$ an asymmetric relation?

$(a,b)$ where $a=b$ must come under symmetric relation.

George
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  • Are you asking whether your relation is asymmetric or symmetric? – Mankind Apr 01 '15 at 06:20
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    @HowDoIMath I am confused why the above relation is called an asymmetric relation. – George Apr 01 '15 at 06:22
  • What is the definition of "asymmetric relation"? – bof Apr 01 '15 at 06:24
  • @George I get your confusion then. It also doesn't look asymmetric to me, given the usual definition: If $xRy$ then it is not true that $yRx$. Is this your definition as well? – Mankind Apr 01 '15 at 06:25
  • @bof If (a,b) is present then (b,a) must not be present. – George Apr 01 '15 at 06:26
  • @HowDoIMath If (a,b) is present then (b,a) must not be present. – George Apr 01 '15 at 06:28
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    @George in that case it isn't asymmetric. For instance, you definition fails, if I let $a=1$ and $b=1$, because then $(a,b)$ is present, but so is $(b,a)$. – Mankind Apr 01 '15 at 06:30
  • @HowDoIMath Peter Woolfitt says my definition is correct. Why does it fail? – George Apr 01 '15 at 06:43
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    @George no, your definition doesn't fail in the sense that it is incorrect. :) Your definition is correct, but the definition shows that your given relation is not asymmetric. This is what I mean, when I write that your definition fails for your relation: The definition is not fulfilled in this specific case. – Mankind Apr 01 '15 at 06:43
  • @HowDoIMath Thank you for the help and sorry for misunderstanding your comment. – George Apr 01 '15 at 06:47

1 Answers1

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The relation you give is in fact not actually an asymmetric relation. The definition of an asymmetric relation is that if $(a,b)\in R$, then $(b,a)\not\in R$. This does not hold because $(a,a)\in R$.

On the other hand, the given relation is an antisymmetric relation. The definition of antisymmetric is that if $(a,b)\in R$ and $(b,a)\in R$, then $a=b$. Notice that our set satisfies this definition. The definition may not be completely clear, and the way I like to think of it is that if one of $(a,b)$ or $(b,a)$ is in the set (for $a\ne b$ !) then the other isn't.

It is important to note that neither asymmetric nor antisymmetric is the opposite of symmetric - indeed the relation you give is both symmetric and antisymmetric, and we can look at the relation $\{(a,a),(a,b)\}$ to see that a relation may not be symmetric and not be asymmetric at the same time.

Peter Woolfitt
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  • Okay. My definition is wrong. But is asymmetric and antisymmetric the same thing? – George Apr 01 '15 at 06:31
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    @George I'm sorry, I was completely wrong. I misread your original post as antisymmetric as this is a sort of question which gets asked often. You are indeed correct that the relation you have given is not asymmetric, and you are correct that the definition of asymmetric is that if $(a,b)$ is present then $(b,a)$ is not present. I will delete this post in a couple minutes (so you have time to see this comment). – Peter Woolfitt Apr 01 '15 at 06:36
  • Okay. Thank you. So number of asymmetric relations can be counted by subtracting symmetric ones from the total? – George Apr 01 '15 at 06:41
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    @George That's not quite right, you can have relations which are neither symmetric nor asymmetric ${(a,a),(a,b)}$ is one such relation. – Peter Woolfitt Apr 01 '15 at 06:44
  • And don't delete it if possible. Just edit in the comment. It made me aware of antisymmetric. – George Apr 01 '15 at 06:44
  • Alright, I'll update it – Peter Woolfitt Apr 01 '15 at 06:46