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According to the definition of an Anti-Symmetric Relation if xRy and yRx then x = y

Which means, effectively, x is in relation with itself. Does this mean that anti-symmetry implies reflexive property as well?

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  • $\begingroup$ Oh yea my bad! Corrected it! $\endgroup$ – upapilot Jan 12 '15 at 23:02
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Your definition is wrong. The relation $R$ is antisymmetric if, whenever $x\mathrel{R}y$ and $y\mathrel R x$ it holds that $x=y$.

An example of a relation that is antisymmetric but not reflexive is $>$ on the set of integers.

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  • $\begingroup$ I don't see how > is an anti-symmetric relation. How can a number be greater than itself? $\endgroup$ – upapilot Jan 12 '15 at 23:03
  • $\begingroup$ That's why it's not reflexive. $\endgroup$ – Nishant Jan 12 '15 at 23:04
  • $\begingroup$ But it's not anti-symmetric either right? $\endgroup$ – upapilot Jan 12 '15 at 23:05
  • $\begingroup$ OH WAIT I see! It's vacuously anti-symmetric right? Since it's never the case that the antecedent is true the implication always holds. $\endgroup$ – upapilot Jan 12 '15 at 23:06
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    $\begingroup$ @Hi-Angel: I don't recall details four years later, but I suspect the OP may have edited the question after I posted the answer, but within the 5-minute grace period. $\endgroup$ – Henning Makholm Jan 9 at 13:18
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Not really. For example the empty relation is anti-symmetric, but is not reflexive unless the underlying set is empty as well.

I hope this helps $\ddot\smile$.

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    $\begingroup$ I don't understand the empty relation rationale, how does this work? $\endgroup$ – upapilot Jan 12 '15 at 23:12
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    $\begingroup$ @upapilot Consider an empty relation over an empty set, i.e. $R = \varnothing \subseteq \varnothing \times \varnothing$. Such relation is reflexive, because $\forall x \in \varnothing. x\ R\ x$ is true. $\endgroup$ – dtldarek Jan 12 '15 at 23:14
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    $\begingroup$ @upapilot In case you are asking about $R = \varnothing \subseteq U \times U$ where $U \neq \varnothing$, then $R$ is anti-symmetric because the premise is always false, i.e. there is no $x$ and $y$ such that $x\ R\ y$, so $\forall x,y\in U.\ x\ R\ y \land y\ R\ x \implies x = y$ is certainly true. $\endgroup$ – dtldarek Jan 12 '15 at 23:18
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A relation that is antisymmetric but not reflexive is said to be "strongly antisymmetric" or "asymmetric".

This implies : $$(xRy) \implies (\neg(yRx))$$

As if $(xRy)$ and $(yRx)$, then $x=y$ but $x\not R x$ because $R$ is not reflexive (which mean you actually can't apply antisymetry to deduce equality).

$>$ and $<$ are the most common examples.

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Anti-symmetry states that if xRy and yRx, then x=y.

Now, concerning a relation like >, it is clearly impossible to meet the conditions xRy and yRx (see why? no number can be strictly greater and strictly smaller than an other number at the same time).

Whenever, the conditions of some property cannot be met then we consider that it's true and that the property holds (because "A implies B" is always true if A is false). So > is anti-symmetric. Hope that helps !

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