Let X be a set. A partition of X is a subset [tex]\pi \subseteq P(X)[/tex] so(adsbygoogle = window.adsbygoogle || []).push({});

that for every [tex]x \in X[/tex] there is precisely one [tex]A \in \pi[/tex] so

that [tex]x\in A[/tex]. If R is an equivalence relation on X, then

[tex]\pi_R = {R(x): x \in X} [/tex]is a partition.

If [tex]\pi[/tex] is partition of X then

[tex]R_\pi= \bigcup[/tex] [tex]A[/tex] x [tex]A[/tex] is an equivalence

relation, where

[tex]A\in \pi[/tex] Furthermore,

[tex]\pi_(R_\pi)=\pi[/tex] where [tex]\pi_(R_\pi)[/tex] (pi sub R sub pi)

[tex]R_(\pi_R)= R[/tex]

To prove this theorem, I have started out by proving that [tex]R_\pi[/tex] is

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

to complete the proof, do I need to prove [tex]\pi_(R_\pi)=\pi[/tex]

[tex]R_(\pi_R)= R[/tex]

Do I do that by using the following conditions?

1. [tex]x \in R(X)[/tex] for each [tex]x \in X[/tex]

2. If [tex]y \in R(x)[/tex], then [tex]x \in R(y)[/tex]

3. If [tex]R(x) \cap R(y) \noteq \empty[/tex], then [tex]R(x)= R(y)[/tex]

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

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