proving a relation has the transitive property I am trying to solve the following question:
For all $ f,g∈ \Bbb N^ \Bbb N $ we say that f and g are almost identical if there does not exist $ X⊆ \Bbb N,where|X|=∞$ ,such that $ ∀i∈X:f(i)≠g(i) $ .Given a relation R,where $ R⊆ \Bbb N^\Bbb  N×\Bbb N^\Bbb N $ ,and defined∶  R≡{ $(f,g) ∈\Bbb N^\Bbb N×\Bbb N^\Bbb N $ | f and g are almost identical }. Prove that R is a transitive relation.
I had the idea to suppose that R is not transitive ,and to arrive to a contradiction by somehow find a natural number t such that $f(t)=g(t)$  and $g(t)=w(t)$  and conclude that $f(t)=w(t)$(which is a contradiction), but it seems  not easy to do so.
I would be happy for some help or hints!

Hint
If $f$ and $g$ are "almost identical", this means that there is only a finite number of arguments for which they differ. But the arguments are numbers; thus there is an $n_{(f,g)} = \max \{ x \mid f(x) \ne g(x) \}$.
The same for $g$ and $w$: $n_{(g,w)} = \max \{ x \mid g(x) \ne w(x) \}$.
This two facts imply that, with $n= \max \{ n_{(f,g)}, n_{(g,w)} \}$, we have:

$f(x)=g(x)$ and $g(x)=w(x)$, for every $x > n$.

Thus, if $X = \{ x \mid f(x) \ne w(x) \}$, we have that $|X| < |\mathbb N |$.
