- #1
laminatedevildoll
- 211
- 0
Let X be a set. A partition of X is a subset [tex]\pi \subseteq P(X)[/tex] so
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]
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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