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Take the set $A=\{1,4,5,7\}$ with the relation $R=\{(1,4),(1,5),(4,7)\}$.

My teacher said that this relation is anti-symmetric but I don't get how there isn't any $x; y$ that belongs to $A$ where $xRy$ and $yRx$ and $x=y$. Shouldn't a anti symmetric relation of $A$ be $R=\{(1,1),(4,4),(5,5),(7,7)\}$?

Help with numerical examples please.

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  • $\begingroup$ then how is the first one anti-symmetric? $\endgroup$ – Nareg Sep 26 '15 at 19:21
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A relation $R$ on a set $A$ is antisymmetric if for any $x,y\in A,$ we have $x=y$ when $x\:R\:y$ and $y\:R\:x.$ Your second relation satisfies $x=y$ when and only when $x\:R\:y$ and $y\:R\:x,$ meaning that the second relation is antisymmetric, and is also reflexive on $A.$ As a side note, the second relation is the only antisymmetric relation with domain $A$ that is also symmetric on $A$, as discussed here.

For the first relation, $x\:R\:y$ and $y\:R\:x$ is never satisfied, so it is vacuously antisymmetric.

Added: One fairly natural way to think about a (binary) relation $R$ on a set $A$ is as a subset of the "square" $A^2=\bigl\{\langle x,y\rangle: x,y\in A\bigl\}.$ We distinguish the diagonal of $A$ as the set of elements of $A^2$ whose entries are equal--more formally, $$\Delta_A:=\bigl\{\langle a,a\rangle: a\in A\bigl\}.$$ We then define the reflection across the diagonal of $A$ to be the function $\rho_A:A^2\to A^2$ given by $\langle x,y\rangle\mapsto\langle y,x\rangle.$

Then the reflexive relations on $A$ are precisely those that contain the diagonal of $A$--that is, those $R\subseteq A^2$ such that $\Delta_A\subseteq R.$ The symmetric relations on $A$ are those that are symmetric across the diagonal of $A$--that is, those $R\subseteq A^2$ such that $\rho_A[R]=R.$ The asymmetric relations on $A$ are those having no points in common with their reflections--that is, those $R\subseteq A^2$ such that $R\cap\rho_A[R]=\emptyset.$ Finally, the antisymmetric relations on $A$ are those that have only points of the diagonal in common with their reflection--that is, those $R\subseteq A^2$ such that $R\cap\rho_A[R]\subseteq\Delta_A.$

Hopefully, this aids in the intuition of why the first relation is antisymmetric.

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Let me tell u what antisymmetric means in simple language:

Anti symmetric - The word itself has a meaning against "Symmetric" {(a,b) and (b,a) exists}.

That is ,in the case of antisymmetric, it means that if (a,b) exists then obviously (b,a) cannot exist. [Because if (b,a) exists it becomes symmetric; and we already said that antisymmetric is against symmetric]. But under one condition if (a,b) exists, then (b,a) can also exists, i.e.

if one and only if a=b.

I'll explain with examples for better understanding:

A={1,2,3}

R1= {(1,1)(2,2)(3,3)} --> It is Anti symmetric because (1,1) exists and when it is reversed also we get (1,1) and it exists and 1=1. (Same with 2,2 and 3,3).

R2={(2,1)(2,3)(1,1)} -->It is Anti symmetric because (2,1) exists but (1,2) does not. (2,3) exists but (3,2) does not and (1,1) can exist because a=b.

R3={(2,3)(3,2)(2,2)(3,3)} --> It is NOT Anti Symmetric because (2,3) exists and (3,2) also exists, which is not prohibited.

And according to your question: A={1,4,5,7} R={(1,4)(1,5)(4,7)}

It is Antisymmetric because:

  • (1,4) exists but (4,1) does not.
  • (1,5) exists but (5,1) does not.
  • (4,7) exists but (7,4) does not.

I hope this helps...

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  • $\begingroup$ The question is almost 2 years old and has an accepted answer. Your answer doesn't really add anything, so spend your time more wisely, e.g. by learning LaTeX/MathJax to format maths here. $\endgroup$ – Henrik - stop hurting Monica Sep 10 '17 at 7:26
  • $\begingroup$ Thank you for your concern. I didn't notice the year since I'm a newbie here. I just wanted to help others to understand what I know. Next time, I would be more careful. $\endgroup$ – Sree Sep 10 '17 at 12:58

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