proving a relation has the transitive property

Question

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!

Answer 1

Let's assume $f$ and $g$ are almost identical, and also that $g$ and $w$ are almost identical.

Let $$A=\{i\in\mathbb N\mid f(i)\ne g(i)\}$$ (the "set on which $f$ and $g$ differ") and $$B=\{i\in\mathbb N\mid g(i)\ne w(i)\}$$ (the "set on which $g$ and $w$ differ"). Because of our assumption, the sets $A$ and $B$ are finite. Take $C=A\cup B$ and it is also finite as a union of two finite sets.

Now, let $$D=\{i\in\mathbb N\mid f(i)\ne w(i)\}$$ (the "set on which $f$ and $w$ differ"). If we can prove that $D\subset C$, then $D$ will be finite, thus proving that $f$ and $w$ are almost identical. Take $i\in D$. This implies $f(i)\ne w(i)$. At least one of $f(i)=g(i)$ and $g(i)=w(i)$ must then be false, so either $i\in A$ or $i\in B$. In any case, $i\in C$.

Answer 2

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 $|X| < |\mathbb N |$.