Proof Check in Algebra I was resolving a question about partition, quotient set and equivalence relation, but I don't sure if my proof is correct. Anyone can help me? I'm grateful right now for all help. Follows the question and my attempt, respectively.
Question:
Let be $A$ an set non-empty and $P(A)$ power set of $A$. We say a non-empty set $\mathbb{P} \subset P(A)$ is a partition of a set $A$ if:
(i) $\forall B_1$ and $B_2 \in \mathbb{P}$, $B_1$ $\neq$ $B_2$ $\Longrightarrow$ $B_1$ $\bigcap$ $B_2$ $=$ $\emptyset$.
(ii) $\bigcup_{B \in \mathbb{P}}$ $B$ $=$ $A$.
Proof that if $x,y \in A$ and $x \sim y$ $\iff$ $\exists B \in \mathbb{P}$ such that $x,y \in \mathbb{P}$, so $\sim$ define an equivalence relation on set $A$. Furthermore, $A/_{\sim}$ $=$ $\mathbb{P}$.
My attempt:
We want show that $\sim$ is an equivalence relation on set $A$.
sets of the form $B$ $=$ { x }\, $\forall x \in A$ satisfy conditions (i) and (ii), so $\exists B \in \mathbb{P}$ such that $x,y \in \mathbb{P}$ $\iff$ $x,y \in A$ and $x \sim x$ (reflexive property checked).
$x \sim y$ $\iff$ $\exists B \in \mathbb{P}$ such that $x,y \in \mathbb{P}$ $\iff$ $y \sim x$ (symmetric property checked).
$\exists B_1 \in \mathbb{P}$ such that $x,y \in \mathbb{P}$ $\iff$ $x \sim y$
$\exists B_2 \in \mathbb{P}$ such that $y,z \in \mathbb{P}$ $\iff$ $y \sim z$
$y \in B_1 \bigcap B_2$ $\Longrightarrow$ $B_1 \bigcap B_2$ $\neq$ $\emptyset$  $\Longrightarrow$ $B_1 = B_2$ (negation of the codition (i) ).
Therefore, $\exists B_1 \in \mathbb{P}$ such that $x,y,z \in \mathbb{P}$ $\iff$ $x \sim y$, $y \sim z$ and $x \sim z$.
We want show now that $A/_{\sim}$ $=$ $\mathbb{P}$.
Enough to show that $B_x = \overline{x}$ is a set of $\mathbb{P}$ that satisfy (i) and (ii).
(i) is satisfied, because $\overline{x} \neq \overline{y} \Longrightarrow \overline{x} \bigcap \overline{y} = \emptyset$
(ii) is satisfied, because $\bigcup_{B \in \mathbb{P}}$ $B$ $=$ $\bigcup_{x \in A}$ $\overline{x}$ $=$ $A$.
P.S.: sorry for some mistakes, but I don't know english very well.
 A: Your proof that you have an equivalence relation seems OK, except for the reflexive property. You are given $\mathbb{P}$, so you don't get to know things like "sets of the form $B=\{x\}$" belong to $\mathbb{P}$. Instead, you need to show that there exists $B\in \mathbb{P}$ such that $x\in B$. This follows because $\bigcup_{B\in\mathbb{P}}B=A$.
Now, let's look at your proof that $A/\sim\, =\mathbb{P}$.
First, you want to show that if $x\in A$, then $\overline{x}=B$ for some $B\in \mathbb{P}$. Why is that? Well, as in your proof of the transitive property, you have that $x$ belongs to a unique $B\in \mathbb{P}$. Now you deduce 
$$y\in B\;\;\;\Leftrightarrow\;\;\;x\sim y\;\;\;\Leftrightarrow\;\;\;y\in\overline{x}.$$
Therefore, $B=\overline{x}$
Finally, we want to show that if $B\in \mathbb{P}$, then there is some $x\in A$ such that $B=\overline{x}$. Here we run into a problem if $\emptyset\in \mathbb{P}$, so we'll want to adjust our definition of a partition by including 
(iii) $\emptyset\notin \mathbb{P}$.
With this addition, finishing the proof is easy. Since $B\neq\emptyset$, there exists $x\in B$ and we have already established that $\overline{x}=B$.
A: Here's a mental trick to understand a part:
If you begin by taking a partition $\Bbb P$ for $A$, then 
your relation can be paraphrased as saying that $x,y\in A$ are related if
they are in the same subset $B$ in $\Bbb P$.
Now reflexivity follows from the fact the $x\in B$ for some $B$ in $\Bbb P$, since $\Bbb P$ is partition. 
Symmetry reads "$x,y\in B$ implies $y,x\in B$".
For transitivity: if $x,y\in B$ and $y,z\in C$ for some $B,C$ in $\Bbb P$, the presence of $y\in B\cap C$ and being that $B\cap C=\varnothing$ or $B=C$, hence $x,z\in B$ also.
