# Equivalence relation $f\sim g \iff f(n)=g(n)$

I have this equivalence relation defined on $$\mathbb{N}$$ and $$f\sim g \iff f(n)=g(n)$$ for some $$n$$.

I was testing this relation for reflexivity, symmetry and transitivity:

1) It is reflective since $$f(n)=f(n)$$

2) If $$f(n)=g(n)$$ then $$g(n)=f(n)$$, so symmetric

3) But for transitivity I think it fails, however I cannot find a counter example for this to work.

Can anyone point me in the right direction?

Context: Let S denote the set of real-valued functions on N. For each relation given below, determine whether it is an equivalence relation. If it is not an equivalence relation, then indicate an axiom – reflexivity, symmetry, or transitivity – which fails.

• You actually define an equivalence relation on some space of functions, not on $\mathbb N$ itself. Not that it really matters for the answer, but what is the intended codomain of $f$ and $g$? – Mark Kamsma May 5 at 10:53
• I have added all the context I was given below the original question! – Olly Reynolds May 5 at 11:00

Choose $$f(n)=1,$$ $$g(n)=1\text{ for }n=1,\qquad g(n)=0\text{ otherwise}$$ and $$h(n)=0.$$
This is a counterexample for transitivity as $$f\sim g$$ (choose $$n=1$$) and $$g\sim h$$ (choose e.g. $$n=2$$), but there exists no $$n\in\mathbb N$$ so that $$f(n)=h(n)$$.