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

## 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?

• Your proof looks good to me. – Herr K. Sep 27 '18 at 21:20

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.