Relation problem Let $A = \{1,2,3,4\}$ and let $f: \mathcal{P}(A)\to\mathbb{N}$ be the function defined by saying that $f(X)$ is the sum of elements of $X$, for each $X\in \mathcal{P}(A)$. (If $X = \emptyset$, then by convention we say that $f(X)=0$). Define the relation $\sim$ on $\mathcal{P}(A)$ as follows:
$X\sim Y$ if and only if $f(X) > f(Y)$ or $X = Y$.
Write down whether $\sim$ is reflexive, symmetric, antisymmetric, transitive.

The part where I am struggling is to prove reflexivity.
Reflexive: $a\sim a$ for all $a$. So that means $f(X) > f(Y)$ is false, since $f(X) > f(X)$ is false.
But $X = Y$ part is true since, $X = X$ for all $X \in\mathcal P(A)$.
 A: I think you think about it correctly, but inefficient, what makes it confusing to you.
As you said, to prove reflexivity you have to show that $x\sim x$ for every $x$.
Now the set $X$ is in relation with itself, since $X=X$ always holds, and then you are already done. So there is no need to mention that $f(X)>f(X)$ does not hold (which is a correct observation).
A: Suppose you have a relation $W$ (for “whatever”) on a set $C$ and define, for $a,b\in C$,

$a\mathrel{R}b$ if and only if $a\mathrel{W}b$ or $a=b$

then the relation $R$ is necessarily reflexive: the predicate

$a\mathrel{W}a$ or $a=a$

is true for every $a\in C$; the truth value of $a\mathrel{W}a$ is completely irrelevant and should not even be mentioned in the proof.
Try and think in more abstract terms so the details of the exercise don't distract you from the main task.
In your case $C=\mathcal{P}(A)$ and the relation $W$ is defined by $X\mathrel{W}Y$ for $f(X)>f(Y)$.
Side note 1. A similar problem will show up with the proof of transitivity: be very careful.
Side note 2. An edit to the question masked off another mistake you made. The relation you had was called $S$, but in the attempted proof you called it $R$. This might just be a typo, but not using the correct symbol might earn you a lower grade.
