Equivalence relation - Proof question Prove that the relation, two finite sets are equivalent if there is a one-to-one correspondence between them, is an equivalence relation on the collection $S$ of all finite sets.
I'm sure I know the gist of how to do it, but I'm a beginner in proofs, and I'm not sure if I've written it down correctly. I absolutely encourage nitpicking in the following proof, as I wish to learn how proofs are correctly written. Thanks!
Proof
Let $S$ be the class of all finite sets.
Let $A,B$ and $C$ be three finite sets.
Reflexive property
Now, $n(A)=n(A)$, and hence there exists a one-to-one correspondence between $A$ and $A$
Therefore, $A≈A$ ------------------$(1)$
Symmetric property
Let $A≈B$
$⇒n(A)=n(B)$
$⇒n(B)=n(A)$, and hence there exists a one-to-one correspondence between $B$ and $A$
$⇒B≈A$
Therefore, $A≈B⇒B≈A$----------------$(2)$
Transitive property
Let $A≈B$ 
$⇒n(A)=n(B)$---------------------$(3)$
Also, let $B≈C$
$⇒n(B)=n(C)$---------------------$(4)$
From $(3)$ and $(4)$, $n(A)=n(C)$
$⇒A≈C$
Therefore, $A≈B$ and $B≈C⇒A≈C$--------(5)
From $(1), (2)$ and $(5)$, it is clear that the relation, two finite sets are equivalent if there is a one-to-one correspondence between them, is an equivalence relation.
Q.E.D
 A: Your argument for reflexivity is circular, I’m afraid. When you write that $n(A)=n(A)$, you’re already assuming that $A\approx A$: both statements mean that there is a one-to-one correspondence between $A$ and $A$. To prove them, you must demonstrate that there really is such a correspondence. Fortunately, this isn’t at all difficult: we just use the identity map
$$\varphi:A\to A:a\mapsto a\;.$$
You make a similar mistake in your argument for symmetry: when you say that $n(A)=n(B)$ implies that $n(B)=n(A)$, you’re assuming the conclusion that you’re trying to reach. All that you’re actually given is that $A\approx B$. This means that there is a bijection $\varphi:A\to B$. You want to show from this that there is also a bijection from $B$ to $A$. Since $\varphi$ is a bijection (one-to-one correspondence), it has an inverse, $\varphi^{-1}$, that is also a bijection, and it goes from $B$ to $A$. That is, $\varphi^{-1}:B\to A$ is a bijection, and therefore $B\approx A$, as desired.
And you’ve done it again in the argument for transitivity, but this time I’ll just get you started on the right track. Your hypothesis is that $A\approx B$ and $B\approx C$. This means that you have bijections $\varphi:A\to B$ and $\psi:B\to C$. How can you combine $\varphi$ and $\psi$ to show that there is a bijection from $A$ to $C$, thereby showing that $A\approx C$ and hence that $\approx$ is transitive?
