I have the set

$$ S = \{a,b,c,d\} $$

and the relation

$$G = \{(a,a)(b,b)(c,c)(d,d)(a,b)(b,a)(b,c)(c,b)\}$$

I want to define other various relations, such as symmetric, asymmetric, antisymmetric.

G is already symmetric because of the presence of $(a,b)(b,a)(b,c)(c,b)$, so for every element in the set S $ xRy => yRx $ holds. It's also reflexive.

If I add $(a,c)(c,a)$ to G it also becomes transitive. If I remove $(a,a)(b,b)(c,c)(d,d)$ it becomes irreflexive. If I were to leave $(a,a)$ in G it would be neither reflexive nor irreflexive.

If in G I only had $(a,b)(b,c)$ it would be an asymmetric relation, which is also automatically antisymmetric (if $aRb\;and\;bRa => a=b$) and irreflexive. Right?

If I had $(a,a)(b,b)(a,b)(b,c)$ it would only be antisymmetric, because there are some elements for which we have $xRy$, even if in this case it's always $x=y$ (it's not reflexive, but it could be if I added $(c,c)(d,d)$).

Is my reasoning right? I'm having some problems with asymmetric and antisymmetric relations (mostly because my professor inverted their names). Thank you for everyones help and time!


Yes, as a general theorem you have that a relation $R$ is asymmetric if and only if it is anti-symmetric and irreflexive, because:

$R$ is asymmetric if and only if there are not $x$ and $y$ such that $xRy$ and $yRx$.

So, assuming $R$ to be asymmetric:

if $R$ would not be irrelexive, then there would be some $x$ such that $xRx$, and therefore there are $x$ and $y$ (pick both to be that $x$) such that $xRy$ and $yRx$, and hence it would not be assymmetric. So: assymetric implies irreflexive

Also: $R$ is (trivially) anti-symmetric, since there simply is not a $x$ and $y$ such that $xRy$ and $yRx$, and hence the antecedent of $(xRy \land yRx) \to x = y$ is false, and hence the whole conditional true. So: assymetric implies anti-symmetric.

Indeed, both asymmetric and anti-symmetric agree on the claim that there is no $x \not = y$ such that $xRy$ and $y Rx$, meaning that the only difference between asymmetric and anti-symmetric is that for an anti-symmetric $R$ may contain elements $x$ such that $xRx$, while this is not true for an asymmetric $R$. Thus, if $R$ is anti-symmetric, but we then also rule out any $x$ such that $xRx$ (i.e. if $R$ is also irreflexive), then $R$ becomes asymmetric. In other words: if $R$ is anti-symmetric and irreflexive, then $R$ is asymmetric.

Which adds up to: $R$ is asymmetric if and only if $R$ is anti-symmetric and irreflexive.

Finally, some concrete examples may help to grasp the idea of asymmetric and anti-symmetric:

The 'smaller than' relation $<$ for numbers is asymmetric (and therefore also anti-symmetric): if $x<y$ we clearly can't have $y<x$. Note that $<$ is indeed irreflexive.

But the 'smaller or equal than' relation $\le$ is anti-symmetric: we can;t have two different $x$ and $y$ such that $x \le y$ and $y \le x$, but of course we do have that $x \le x$ fpor any $x$ (so it is reflexive ... and not asymmetric)

Similarly, the strict subset relation $\subset$ is asymmetric, but the 'normal' subset relation $\subseteq$ is anti-symmetric.


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