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Can anyone give me a counterexample for a relation $R\subset M\times M$ for the statement $$R\text{ antisymmetric} \wedge R\text{ not reflexive}\implies R\text{ asymmetric}$$

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up vote 3 down vote accepted

No, because a relation is asymmetric if and only if it is antisymmetric and not reflexive.

To see that your implication is always true, we could check the contrapositive statement: If R is symmetric then R is not antisymmetric or R is reflexive. This is easily seen to be true since if R is symmetric and anti-symmetric, it is a sub-relation of the equality relation, in which case it is obviously reflexive.

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Thank you for the second part - i was totally confused here. – Christian Ivicevic Dec 10 '12 at 12:14
I just got a question right now. What aboout $M=\{1,2\}$ and $R=\{(1,1),(1,2)\}$? – Christian Ivicevic Dec 10 '12 at 13:42
@ChristianIvicevic What about it? – Ragib Zaman Dec 10 '12 at 13:43
Isn't this example antisymmetric, not reflexive and not asymmetric? If not, why? – Christian Ivicevic Dec 10 '12 at 13:45
I see it now. Let's forget what I was talking about. – Christian Ivicevic Dec 10 '12 at 13:50

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