Let X be a set. A partition of X is a subset

[laminatedevildoll]

Let X be a set. A partition of X is a subset $$\pi \subseteq P(X)$$ so

that for every $$x \in X$$ there is precisely one $$A \in \pi$$ so

that $$x\in A$$. If R is an equivalence relation on X, then

$$\pi_R = {R(x): x \in X}$$is a partition.

If $$\pi$$ is partition of X then

$$R_\pi= \bigcup$$ $$A$$ x $$A$$ is an equivalence

relation, where

$$A\in \pi$$ Furthermore,

$$\pi_(R_\pi)=\pi$$ where $$\pi_(R_\pi)$$ (pi sub R sub pi)

$$R_(\pi_R)= R$$

To prove this theorem, I have started out by proving that $$R_\pi$$ is

an equivalence relation, for reflexitivity, symmetry, and transitivity. In order

to complete the proof, do I need to prove $$\pi_(R_\pi)=\pi$$

$$R_(\pi_R)= R$$

Do I do that by using the following conditions?

1. $$x \in R(X)$$ for each $$x \in X$$

2. If $$y \in R(x)$$, then $$x \in R(y)$$

3. If $$R(x) \cap R(y) \noteq \empty$$, then $$R(x)= R(y)$$

 

[TenaliRaman]

Your two statements (which are two be proved) can be interpreted as,
1. Given that R is an equivalence relation, show that the collection of relative sets R(x) form a partition
2. Given that P(pi) is a partition, show that the above described relation is an equivalence relation.

So yes, i guess you need to show these statements as true.
The conditions you want to use will be useful but just remember the three properties of equivalence relation and use whatever you can accordingly.

-- AI