laminatedevildoll
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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)
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)
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