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Defining:
"weak antisymmetric relation": $\forall a, b \left< a,b \right> \in R \land \left< b,a \right> \in R \Rightarrow a=b$

"Strong antisymmetric relation": $\forall a, b \left< a,b \right> \in R \Rightarrow \left< b,a \right> \notin R $

I've read that "strong" is in particular a "weak" relation, but as I understand:
"strong" isn't reflexive while "weak" is. So, how can "strong" be in particular "weak"?

I hope what I'm asking is understandable.
Thanks.

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  • $\begingroup$ Not every weak relation is reflexive. $\endgroup$ – mrkvon Oct 14 '19 at 10:49
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A strong antisymmetry is in particular weak because on the conditional statement that defines the weak antisymmetry, a strong relation will never satisfy the antecedent, thus making the conditional statement true. Therefore any strong antisymmetry is in particular weak.

It's useful to think of a weak antisymmetry as $\leq$, for whatever intuitive meaning of $\leq$ you're more comfortable with (for example the order relation on the real numbers). Similarly you can think of a strong antisymmetry as being $<$.

You can also observe that any strong antisymmetry can be extended to a weak antisymmetry by adding elements to the strong antisymmetry, but you can't extend a weak antisymmetry to a strong antisymmetry because a strong antisymmetry can't be reflexive - actually it doesn't even get to the point of a strong antisymmetry not being reflexive, it's stronger than that, it actually is antireflexive.

The fact that a strong antisymmetry is also weak, does not at all imply that is must have all the properties that a weak antisymmetry would have. I think the contrary is more intuitive, the broader concept is the one who will have all the properties of the smaller concept (this isn't true either, by the way).

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  • $\begingroup$ Superb. Thanks! $\endgroup$ – AnnieOK Jun 17 '14 at 15:57

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