Is R an Equivalence Relation on Functions to [0,1]?

[eXorikos]

Homework Statement

Given is the set X. The set of functions from X to [0,1] we call Fun(X,[0,1]). On this set we consider the relation R. An ordered pair (f,g) belongs to R when $$f^{-1}(0)\setminus g^{-1}(0)$$ is a countable set.

a) Prove that R is transitive.

b) Is R an equivalence relation? Prove!

c) Prove that $$R \cap R^{-1}$$ is an equivalence relation.

Homework Equations

Transitive means that if (f,g) and (g,h) belong to the relation, that also (f,h) belongs to it.

Equivalence relation is a relation that is transitive, reflexive ($$(f,f) \in R$$ and symmetric ($$(f,g) \in R \Rightarrow (g,f) \in R$$.

The Attempt at a Solution

a) $$f^{-1}(0)\setminus g^{-1}(0)$$ is a countable set. So $$f{-1}(0)$$ is a countable set. This means that $$f^{-1}(0)\setminus h^{-1}(0)$$, because a subset of a countable set is also countable.

Is this correct?

b) Reflexivity is easy, because $$f^{-1}(0)\setminus f^{-1}(0)$$ is the empty set so that is obviously countable.
How do I prove that it is symmetric?

c) Is this just the subset of all the reflexive pairs?

 

[vela]

a) $$f^{-1}(0)\setminus g^{-1}(0)$$ is a countable set. So $$f^{-1}(0)$$ is a countable set.

You can't conclude $f^{-1}(0)$ is countable. For instance, suppose X is R, the set of real numbers, and f(x)=g(x)=0. Then $f^{-1}(0)\setminus g^{-1}(0)$ is the empty set, but $f^{-1}(0)=R$, which is not countable.