Proving an Equivalence Relation: Tips & Examples

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To prove that a relation R(z, d) is an equivalence relation, it must satisfy reflexivity, symmetry, and transitivity. The discussion emphasizes that to demonstrate these properties, one must clearly define how R relates to a function f from set A. Specifically, if R(z, d) is defined such that d equals f(z), then one must show that for all members of A, f(z) equals z for reflexivity, if d equals f(z) then z equals f(d) for symmetry, and if x equals f(z) and y equals f(x), then y equals f(z) for transitivity. The conversation highlights the importance of ensuring that the function f is invertible for the equivalence relation to hold. Understanding these properties in the context of the specific relation is crucial for proving equivalence.
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Iam not clear on how to prove a equivalence relation? I know that is has to have three properties reflexivity, symmetry, and transtivity, but I am unsure how to check.

For example Iam given f is a function from z to d. R(z,d) = binary relation

How do I prove that R(z,d) is a equivalence relation?? Iam unclear on how to approach this and work with the properties?
 
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A relation ~ on a set A, is an equivalence relation if

Let x,y,z belong to A

i) x~x (reflexive)
ii)If x~y then y~x (symmetric)
iii)if x~y and y~z then x~z (transitive)

To show that a given relation is an equivalence relation you would need to show that these 3 properties hold true for the relation.
 
As both you and d_leet said, you would prove R is an equivalence relation by showing that each of the conditions holds. How you would do that depends strongly on what R is!

"For example Iam given f is a function from z to d. R(z,d) = binary relation"
How is R(z, d) related to f? Do you mean R(z,d) if and only if d= f(z)?

Rather than saying "f is a function from z to d" it would be better to say "f is a function on set A" (with z and d members of set A).

If, indeed R(z,d) if and only if d= f(z), then you must prove:
1. Reflexive. That f(z)= z so R(z,z) for every member of set A.
2. Symmetric. If d= f(z), then z= f(d).
3. Reflexive. If x= f(z) and y= f(x), then y= f(z).

From 2, it looks like we are saying that f must be invertible. If that is the case, then for any d in A, there is only one x, such that d= f(x). Combining that with 1, we clearly must have f(x)= x for any member of A. That would mean that R(d,x) is equality: d= x.
 
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The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense. Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with ## (M^{T})^{T}=M##, and just wanted to see if it made sense to talk...

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