Let R be a reflexive and transitive relation on X. Show that R intersection R^-1 is an equivalence relation on X.
My proof:
Let R be reflexive and transitive, therefore R (x,x) (y,y) (z,z) (x,y) (y,z)
and R^-1 (x,x) (y,y) (z,z) (y,x) (z,y).
By the intersection rule we have R intersection R^-1 (x,x) (y,y) (z,z).
Is this correct?
How can I show that R intersection R^-1 is an equivalence relation on X?
Thank you.
relation is transitive R then (x,z) (x,y) (y,z)
R^-1 is also transitive (z,x) (y,x) (z,y)
relation is reflexive R then (x,x) (y,y) (z,z)
relation is reflexive R^-1 then (x,x) (y,y) (z,z)
Then R (x,z) (x,y) (y,z) (x,x) (y,y) (z,z)
and R^-1 (z,x) (y,x) (z,y) (x,x) (y,y) (z,z)
How is R intersection R^-1 is an equivalence relation?
relation is transitive R then (x,z) (x,y) (y,z)
R^-1 is also transitive (z,x) (y,x) (z,y)
relation is reflexive R then (z,x) (y,x) (y,z)
relation is reflexive R^-1 then (x,z) (y,x) (y,z)
Then R intersection R^-1 is an equivalence relation
and include {(x,z) (x,y) (y,z) (z,x) (y,x) (y,z)}
You can't use a name without first saying what it refers to. If I said that Jim likes Jane, is it true or not? This depends on who I mean by Jim and Jane.
Every variable you use must be introduced, for example, by "for all x," "there exists x," "let x be such that," etc.
Is this good ?
relation is transitive R then (x,z) (x,y) (y,z)
R^-1 is also transitive (z,x) (y,x) (z,y)
relation is reflexive R then (z,x) (y,x) (z,y)
relation is reflexive R^-1 then (x,z) (y,x) (y,z)
Then R intersection R^-1 is an equivalence relation
and include {(x,z) (x,y) (y,z) (z,x) (y,x) (y,z)}
No, you used x, y and z without introducing them.
Note, for example, the definition of reflexivity: R is reflexive if for all x we have R(x,x).
Also, "(x,z) (x,y) (y,z)" does not make sense. The fact that some x and y are related by a relation R is usually denoted by R(x,y), xRy, or (x, y) ∈ R.
Let R be transitive (x,y) (y,z) then (x, z).
R^-1 transitive will be opposite as (y,x) (z,y) then (z,x)
and then R is reflexive and transitive: (x,y) (y,z) (x, z) (y,x) (z,y) (z,x).
and R^-1 is reflexive and transitive: (y,x) (z,y) (z,x) (x,y) (y,z) (x,z).
Therefore R intersection R^-1 is also symmetric and R intersection R^-1 is an equivalence relation.
No, since R is transitive, we have that for all x y and z, if R(x, y) and R(y, z), then R(x, z). Just renaming the variables, we get that for all x, y, z, if R(z, y) and R(y, x), then R(z, x). Now, for all x, y and z, iff , iff and iff . Therefore, the previous statement means that for all x, y, z, if and , then , i.e., is transitive, as Plato wrote above. This is a lemma that is useful in proving that is transitive.
This is not clear.
what about reflexive?
Let R be reflexive, then we have a relation R on a set X that is reflexive if (x,x) elements of R for every x element of X. So we get (x,x), (y,y) (z,z).
What is reflexive of R^-1? the same as R?
The blue part is not needed.
This is not clear. Why not continue it to (t,t), (u,u), etc? What does (x,x) mean? It's nether true or false, but R(x,x) is true or false. So, R is reflexive means by definition that for every x, R(x,x), or (x,x) ∈ R.
I am not sure what you mean by "reflexive of R^-1." The fact is, if R is reflexive, then so it R^-1, which has to be proved, but which is easy. Remember only that the definition of reflexivity starts with "for all x ∈ X, ...", so a proof of such statement must start with "Consider an arbitrary x ∈ X."