I'm needing help in determining if the relation $$R=\left\{(f,g)\mid \exists k,\forall x\in\Bbb Z, \ f(x) = g(k)\right\}$$
where f, g: $\Bbb Z \rightarrow \Bbb Z$ is an equivalence relation.
More specifically, i don't seem to understand the part where it says there exists a k for all x. I understand that there must be at least one k, but that's all, i can't seem to make sense of it in relations, like.. should there be a single k that should be used to prove the reflexivity, symmetry and transitivity properties of an equivalence relation? or will it suffice any k as long as i can find one k value as an example for each of the properties?
My method for qualifying a relation as an equivalent one is to step through the three conditions: reflexivity, symmetry, and transitivity.
Clearly $f\,R\,f$.
Next suppose $f\,R\,g$. Then $\exists\, k\, \forall x\in \mathbb{Z} : f(x) = g(k)$. Does it follow that $g\,R\,f$?
That is, $\exists\, j\, \forall x\in \mathbb{Z} : g(x) = f(\,j)$? My intuition tells me this need not follow.
Let's suppose $f(x) = 1$ and $g(x) = 1 + x^2$. We know that $f\,R\,g$, since when $k=0\, \forall x\in \mathbb{Z} : f(x) = g(0) \implies 1 = 1$. However, check out $g\,R\,f$. There does not exist an integer $j$ such that $\forall x\in\mathbb{Z}: g(x) = f(j) = 1$.
So unless I am misinterpreting the problem formulation, this is not an equivalence relation.