Prove that there is exactly one identity relation $I_A$ for each set $A$ Originally, to solve this question. I found the 2 relations and checked if they were the subset of each other to prove that they were the same thing. But the question had a hint which stated, 
"As a hint, you can write a rigorous proof of this result without referencing what it means for one relation to be a subset of another. Look at the definition of an identity relation, which gives a series of equalities involving an identity relation. Can you make use of those equalities here?"
I came up with a proof but that used equalities but i'm not really sure if i've done it right. I was wondering if someone can check the proof and critique it. Also if you have another way of proving, that would be very helpful too. 
My proof:
Let's assume $A$ is an arbitrary set. We will prove that $A$ has exactly one identity relation.
Let $I_1, I_2$ be identity relations where for every $x$ and every $y$, $x=y$ and $(x,y) \in$ $A$x$A$. 
To show $I_1 = I_2$, we can say that if every $x$ and $y$ in $A$x$A$ such that $x = y$ then $(x,y) \in I_1 = (x,y) \in I_2$.
Therefore we can conclude that there is only one identity relation for each set A. $\blacksquare$

If I,J are identity relations for A, then
x in I iff exists a in A with x = (a,a) iff x in J.
Thus I = J since the identity relation for A is { (a,a) : a in A }.
