0
$\begingroup$

I was asked on a test to find the number of relations on a set with 10 elements which is both symmetric and asymmetric.

Now since any element of the type

$$ (a,b) \quad where \quad b \neq a $$ cannot exist in such a relation (I think).

This problem essentially reduces to the statement whether a relation which only has elements of the type $ (a,a)$ (elements found on the main diagonal of the matrix form of R) is a relation which is both symmetric and asymmetric?

  • 1
    You seem to be confusing *antisymmetric* and *asymmetric*. What you have is true if the question had asked about relations that are both *antisymmetric* and symmetric. As you've stated the question (being about *asymmetric* relations), there is only one possible relation. (the empty relation) – 2017-02-07

1 Answers 1

3

No, for any such $R$, you can't have any $(a,a) \in R$ since that would go against asymmetry (If $(a,a) \in R$, then by asymmetry, $(a,a) \not \in R$ ... so we can't have $(a,a)$!)

The only relation that works is the empty relation, i.e. $R=\{ \}$, since if any $(a,b) \in R$, we must have (by symmetry) $(b,a) \in R$, but we also must have (by asymmetry) $(b,a) \in R$. Since we can't have both, we can't have any $(a,b) \in R$.

  • 0
    Antisymmetry means that $(a.b) \in R \land (b,a) \in R \Rightarrow b = a$. If antisymmetry was against reflexiveness then we couldn't have order relations. – 2017-02-07
  • 2
    @chelivery Correct ... but the OP asked about asymmetry, not anti-symmetry. – 2017-02-07
  • 1
    @chelivery The problem asked about *asymmetry*, not about *antisymmetry*. These are two different terms – 2017-02-07
  • 0
    Oh, damn: my bad. Of course. – 2017-02-07
  • 0
    No problem: easily confused! :) – 2017-02-07