Proof equivalence relation with functions

Question

We define 2 function: $f : X \rightarrow X$ and $g : X\rightarrow X$ and we define $ V(f,g) = \{x \in X | f(x) \neq g(x) \} $. Next we define a relation $R$ on the set Fun($X,X$) of all functions from $X$ to $X$ by $(f,g) \in R$ if and only if V$(f,g)$ is a countable set. I have to proof that this is an equivalence relation. But I'm good with working with countable sets and kardinality. I got already stuck with the reflexivity. Suppose ($f,f$) $\in R$ then it doens't fullfi the prerequisite of V? Because $f(x)$ is always equal to $f(x)$. Thanks!

Answer 1

I guess that by countable you actually mean at most countable, which includes the cases wehre $V$ is finite or empty. Then the proof is quite easy:

• $(f,f) \in R$ because $V(f,f) = \emptyset$
• The symmetry is trivial
• Let $(f,g)$ and $(g,h)$ be in $R$. Then $V(f,g)$ and $V(g,h)$ are at most countable and $V(f,h) \subset V(f,g) \cup V(g,h)$ which is also countable.