help me prove reflexivity

Question

So given this problem, im trying to prove for part a) that $A\tau B$ is reflexive

so far this is what ive come up with:

for $A,B \in \mathcal P(X)$ if $A \tau B$ , then $A=B$.

since $A=B$, $\forall a \in A$ and $\forall b \in B$, $a=a$ and $b=b$

so $A\tau B$ is reflexive.

my answer is sort of an educated guess so i dont think its entirely correct. its always tempting to just ask for someone to correct me but I like discrete math and I want to be good at it so rather then giving me a quick fix, can someone guide me through this?

-Thanks

Answer 1

Asserting that $\tau$ is reflexive means that $\bigl(\forall A\in\mathcal{P}(X)\bigr):A\tau A$. This means that the sum of the elements of $A$ is equal to the number of elements of $A$, which is trivially true.

Answer 2

So $\tau$ is defined as $\{(A,B) \in \mathcal{P}(X) \times \mathcal{P}(X): \sum_{n \in A} n = \sum_{n \in B} n\}$.

Reflexive means that for any $A \in \mathcal{P}(X)$ we have $(A,A) \in \tau$. This holds iff $\sum_{n \in A} n = \sum_{n \in A} n$ which is trivially true.

Symmetry is also trivial: if $A$ has the same sum as $B$, then $B$ has the same sum as $A$. Transitivity is similarly trivial.

$\emptyset$ has sum $0$, as does $\{0\}$, so they form one class.

For sum $1$, we just have the elements $\{1\}, \{0,1\}$.

For sum $2$, we have the class $\{2\}, \{0,2\}$.

Similarly $\{3\}, \{0,3\}, \{1,2\}, \{0,1,2\}$ form one class.

etc.