Equivalence relation example. How is this even reflexive?

Question

Is the below question a mistake? How is this an equivalence relation? For example, how would it even be reflexive? E.g if you pick any A $\subseteq$ $U$, say A = {a, b}, then A ~ A is not true, because both sets are even. Am I just missing something obvious here? In fact it wouldn't be reflexive for any odd # of element set either, e.g. A = {a} since A ~ A would mean neither has an even number of elements.

We have a universal set of lowercase alphabet letters, $U$ = {a, b, ...., z} . For sets A,B $\subseteq$ $U$ we can define a relation, A ~ B as long as the number of elements that are in either A or B, but not both is even. Prove that ~ is an equivalence relation.

Answer 1

You are misinterpreting the phrase "the number of elements that are in either $A$ or $B$, but not both is even."

It does not mean, "one and only one of the sets $A,B$ has an even number of elements".

Correctly understood, it means "the set of all elements belonging to one but not both of the sets $A,B$ has an even number of elements"; i.e., "$(A\cup B)\setminus(A\cap B)$ has an even number of elements".

If $A=B$ then $(A\cup B)\setminus(A\cap B)$ is the empty set, which has an even number of elements.

Answer 2

You need to consider the number of elements in the set $(A \cup B) \setminus (A \cap B)$.

If $A = B$ the number of elements is zero, which is even.

Answer 3

The set of elements that are in either $A$ or in $B$ but not both is the symmetric difference $A\setminus B \cup B\setminus A$. The symmetric difference of $A$ and $A$ is empty, and 0 is even!