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.

  • $\begingroup$ 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 $g$? $\endgroup$ – Mark Kamsma May 5 at 10:53
  • $\begingroup$ I have added all the context I was given below the original question! $\endgroup$ – 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 $f(n)=h(n)$.

  • $\begingroup$ Thank you, understood! $\endgroup$ – Olly Reynolds May 5 at 11:07

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