Elementary Set Theory: How is this relation symmetric and transitive?

Question

The question asks me to identify the properties of the following relation:

"is the mother of" (on the set of all men in this country)

The answer in the back of the book is symmetric and transitive, but I'm not sure I follow.

Here is my theory, in which I'm not at all confident: Since no man in this country is the mother of himself or any other man in this country, the relation is the null set. The null set is not reflexive because it does not contain $(a,a)$ for all $a$. The null set is (or cannot be shown not to be) symmetric because there is no case where $(a,b)$ and $a\neq b$. The null set is (or cannot be shown not to be) transitive because there is no case $(a,b)$ and $(b,c)$ but not $(a,c)$.

Thanks in advance for your help.

Answer 1

This is an example of a vacuous implication. The set is symmetric and transitive, simply because one can check these conditions only in the presence of a certain pre-condition (for symmetry, it's the membership of one pair, and for transitivity, the membership of two (not necessarily different) pairs). When that precondition is not satisfied anyway, this property follows vacuously.

For example, the statement "The sun rises in the west implies that it will rain today" is a true statement, simply because the sun will never rise in the west. This is a vacuous implication. A more concrete example is: An odd number in $\{2,4,6,8\} $ is a multiple of $3$. Of course this is true, simply because there is no odd number.

Answer 2

No man is the mother of anybody, so this is in fact true that it is the empty set.

However, the definition of symmetric is as follows: $$\forall a,b\in X(aRb\Leftrightarrow bRa)$$ In this case, $X$ is the set of all men. Because $aRb$ is always false and so is $bRa$, they are both always false and therefore both always equivalent. Thus, the empty set $is$ symmetric.

The definition of transitive is as follows: $$\forall a,b,c\in X((aRb\wedge bRa)\Rightarrow aRc)$$ Once again, since $aRb\wedge bRa$ is always false and so is $aRc$, false is implies itself and thus this holds once again.

Hope this helps!