The complement of an asymmetric relation is nonsymmetric?

381 Views Asked by At

Asymmetric relation: Given a set A and a relation R in A, R is asymmetric if it is never the case that for any ordered pair (x, y) in R, the pair (y, x) is in R.

Nonsymmetric relation: If for some (x, y) in R, the pair (y, x) is not in R then R is nonsymmetric.

Complement: The complement of a relation R contains all ordered pairs of the Cartesian product of A which are not members of R.

The book that I am reading says that the complement of an asymmetric relation is nonsymmetric.

Isn't this a counterexample: The set A is empty and relation R has no ordered pairs (i.e., R is empty).

R is asymmetric.

The complement of R is R, which is not nonsymmetric.

So, is the book incorrect?

1

There are 1 best solutions below

0
On

Assume $(x,y) \in R$; then by asymmetry: $(y,x) \in R^C$.

Obviously $(x,y) \notin R^C$.

Thus, we have found a pair $(a,b) \in R^C$ such that the pair $(b,a) \notin R^C$, that means that $R^C$ is not symmetric.