I am aware that relations can be both symmetric and antisymmetric, or either one of the two. However, I am still a little bit confused as to why they can not be both (i.e. not symmetric and not antisymmetric)?
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Suggest delete first "not" in title. Also maybe replace "why they cannot be both..." by "whether they can be both not symmetric and not antisymmetric" in body. [both for clarity of grammar] – coffeemath Feb 04 '19 at 07:10
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1But they can be both. – Did Feb 04 '19 at 07:17
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@Did Yes can be both. But that's what I thought OP was asking about. – coffeemath Feb 04 '19 at 09:00
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@coffemath I edited the title since you suggested to remove the not from the title – Bandolero Feb 04 '19 at 12:41
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On the set $\{1,2,3\}$, take the relation $$ \{(1,2),(2,1),(1,3)\} $$ It has $(1,3)$ but not $(3,1)$ so it isn't symmetric. Also, it has both $(1,2)$ and $(2,1)$, but $1\neq 2$ so it isn't antisymmetric.
Arthur
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it's just a bad terminology. If $Sym$ is a symmetric relation on the set $X$ then the set difference $X \times X \setminus Sym$ not necessarily to be an antisymmetric relation
Leox
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Sure, they can be both.
The most common non-symmetric relations in mathematics come from partial orders. Most of the time, that's going to give us an antisymmetric relation, but it doesn't have to. We just need some larger equivalence classes of elements. For example, consider the divisibility relation on $\mathbb{Z}$, or on the polynomial ring $F[x]$. It's clearly not symmetric, but it's also not antisymmetric; $n$ and $-n$ divide each other in $\mathbb{Z}$, and any two constant multiples of the same polynomial divide each other in $F[x]$.
jmerry
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