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. $\blacksquare$
 A: This is my understanding of the hint you're interested in.
Look carefully at the definition of the identity relation, it says:

Given a set $A$, a binary relation $I_A$ is called an identity relation over $A$ if the following statement is true for every binary relation $R$ over $A$:
$I_A \circ R = R \land R \circ I_A = R$.

So, pay close attention to the phrase for every binary relation $R$. So like we can use any binary relation instead of $R$ and still get that $I_A \circ R = R \land R \circ I_A = R$ (of course if $I_A$ is identity relation over $A$.
So assuming that $I_{A1}$ and $I_{A2}$ are identity relations over $A$, we sure know that they are also binary relations over $A$. So, consider $I_{A1} \circ R = R \land R \circ I_{A1} = R$ and let $R$ = $I_{A2}$, then
$$I_{A1} \circ I_{A2} = I_{A2} \land I_{A2} \circ I_{A1} = I_{A2}$$
Next, consider $I_{A2} \circ R = R \land R \circ I_{A2} = R$ and let $R$ = $I_{A1}$, then
$$I_{A2} \circ I_{A1} = I_{A1} \land I_{A1} \circ I_{A2} = I_{A1}$$
Note, that $I_{A1} \circ I_{A2} = I_{A1}$ and $I_{A1} \circ I_{A2} = I_{A2}$. Thus, $I_{A1} = I_{A2}$.
You can argue why I assume that $R = I_{A1}$ and $R = I_{A2}$. In fact, I just picked any of all the binary relations over $A$. In the first case I picked $I_{A2}$ and in the second case I picked $I_{A1}$.
A: 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 }.
