Could someone verify proof regarding equivalence relation and classes

Question

Exercise 11.3.4 from Book of Proof by Richard Hammack:

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Proof that R is an equivalence relation:

1. First we show $xRx\space\space \forall x \in A$. From the definition of $R$, $xRx$ if $x \in X$ for some $X \in P$. Now, since $x \in A$, it has to be in the partition of the set $A$, hence $x$ is in some $X \in P$. This shows $xRx$.

2. We show $xRy \implies yRx\space\space \forall x,y\in A$. Let $xRy$, then $x,y \in X$ for some $X \in P$. Now, since the order in a set does not matter, we may say $y,x \in X$ for some $X \in P$ [not sure about this argument]. Hence $yRx$ from the definition of $R$.

3. We show $((xRy) \wedge (yRz))\implies(xRz)$. Assume the antecedent. Then $x,y \in X$ for some $X \in P$ and $y,z \in X$ for some $X \in P$. Now, since $\bigcap_{X\in P}=\{\}$, $x,y,z$ must all be in some one $X \in P$. Hence $xRz$ from the definition of $R$. $\blacksquare$

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Proof that P is the set of equivalence classes of R:

Take any $X \in P$. Then for any two elements $x,y \in X$, $xRy$. Since $R$ is an equivalence relation and since $X$ consists of all elements that relate to $y$, that means $X=[y]$. $\blacksquare$

EDIT (in response to Bram28):

Define $E = \{ [x]| x \in A\}$. Now, let $X$ be an arbitrary set in $P$. That means for any $y,x \in X$, $yRx$. So $X=\{y\in A|yRx\}=[x]$ (by definition of equivalence class). So $X \in E$.

Conversely, let $X \in E$. Then $X=[x]=\{y\in A|yRx\}$. Then for any $y,x \in X$, $yRx$. Which means $X$ has to be in $P$. $\blacksquare$

Answer 1

Pretty good!

Your statement for 2) is indeed a bit funky. I would do this logically. that is, since $xRy$, we have that for some $X$: $x \in X$ and $y \in X$ ... And therefore for some $X$: $y \in X$ and $x \in X$. So $yRx$

3) can be sharpened a bit as well. We have that for some $X_1$: $x \in X_1$ and $y \in X_1$, and that for some $X_2$: $y \in X_2$ and $z \in X_2$ (you can't initally assume that $X_1$ and $X_2$ are the same set). but since $P$ is a partition of $A$, we have that for every $x \in a$ there is exacly one $X \in P$ such that $x \in X$. Since $y \in X_1$ and $y \in X_2$ we know $X_1 = X_2$, and thus there is some $X$ (Which is $X_1$ ) such that $x \in X $ and $z \in X$. Hence $xRz$.

Your last proof needs the most work. What you want to show here is that $P = E$ where $E = \{ [x]| x \in A\}$ where $[x] = \{ y | xRy \}$. And to do that, show that for every $X$: $X \in P$ iff $X \in E$. So try and do that one a little more systematically.