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I would like someone to review my understand of why the relation "is less than" ($<$ ) is anti-symmetric on $\mathbb{Z}$. My reasoning is as follows:

Anti-Symmetry says that, if $(a,b) \in R$ and $(b, a) \in R$, then $a = b$.

In this case, we cannot have $a < b$ and $b < a$, so we cannot even get to the point where a violation of the condition of anti-symmetry is possible. And so, by default, the relation is anti-symmetric, despite the fact that it contains no elements $(a,b) \in R$ and $(b, a) \in R$ such that $a = b$.

I would appreciate it if people could please take the time to review my reasoning.

The Pointer
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3 Answers3

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You have the defnition of "antisymmetric" wrong. It's not speaking of whether $(a,b)$ and $(b,a)$ are in $\mathbb Z\times\mathbb Z$, but whether they are in the relation that you're asking whether is antisymmetric. In this case that is the less-than relation.

That being said, you're right that $(<)$ is vacuously antisymmetric, for the reason you give in: There's nothing to contradict it, and a "for all" property is true whenever there's no counterexample.

hmakholm left over Monica
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  • Ahh, you're right. I knew that, so I'm not sure why I wrote $\mathbb Z\times\mathbb Z$ instead of $R$! I will change it now for the sake of correctness. – The Pointer Mar 31 '18 at 01:10
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Your argument applies to any strict inequality, and yes they're vacuously antisymmetric. You can prove we can never have $a<b$ and $b<a$ using transitivity and irreflexivity.

bitesizebo
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Because if you have $p\to q$ and $p$ is wrong then the first proposition is correct. That's exactly your situation since there's no $x$ and $y$ which give $x<y \land y<x$, exactly as you said.

Parsa Noori
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