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If $\succsim$ is transitive but irreflexive, then it is asymmetric.

this is my proof:

Suppose $\succsim$ is not asymmetric, which means that for any $x,y \in X$ $x\succsim y \rightarrow y \succsim x$. By definition $\succsim$ is transitive, i.e. for any $x, y, z \in X$ we have $x \succsim y$ & $y\succsim z$ $\rightarrow$ $x \succsim z$. So since we are supposing $\succsim$ is symmetric: $y\succsim x \rightarrow x \succsim$ y. Now since $\succsim$ is transitive $y \succsim x$ & $x\succsim y \rightarrow y \succsim y$ but $\succsim$ is irreflexive so it's a contradition and $\succsim$ is asymmetric.

What do you think of my proof, is it wrong or not clear enough?

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    $\begingroup$ Your proof looks good to me. $\endgroup$ – Herr K. Sep 27 '18 at 21:20
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A relation that is not asymmetric need not be symmetric. But if $\succsim$ is not asymmetric, there must be elements $y$ and $x$ such that both $x\succsim y$ and $y\succsim x$ hold and these elements are all you need to arrive at a contradiction.

Also, $x\succsim y\implies y\succsim x$ can hold for an asymmetric relation, namely when $x\succsim y$ does not hold.

I'd like also to point out that "any" is ambiguous here and might be read as "all" instead of the intended "some."

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  • $\begingroup$ Ok I undestand that I need to improve the form, but is saying "for any" wrong? I intended it as ∀, why should I say "some"? I just want to be as precise as possible. $\endgroup$ – Zhang_anlan Sep 28 '18 at 8:52
  • $\begingroup$ I don't think it is wrong, but "some" is clearer in this context. $\endgroup$ – Michael Greinecker Sep 28 '18 at 9:43

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