Antisymmetric and irreflexive relation which is not asymmetric

Question

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

score 2 · Accepted Answer · answered Dec 10 '12 at 12:08

2

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.

answered Dec 10 '12 at 12:08

Ragib Zaman

35,127

1

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
@ChristianIvicevic But it is asymmetric, since it contains (1,2) but not (2,1). – Ragib Zaman Dec 10 '12 at 13:46
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I see it now. Let's forget what I was talking about. – Christian Ivicevic Dec 10 '12 at 13:50

score 1 · Answer 2 · answered Oct 03 '16 at 05:24

Actually the answer is yes, since a relation is asymmetric if and only if it is antisymmetric and "irreflexive". There are three distinct properties of a relation, reflexive, irreflexive, and neither reflexive nor irreflexive. Plus, M = {1, 2} and R = {(1, 1), (1, 2)} is a correct counterexample. This R is antisymmetric and not reflexive but it is not asymmetric.

Antisymmetric and irreflexive relation which is not asymmetric

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