Uncountably many functions coinciding only finitely many values

Solution 1:

In fact, you can't even have 3 such functions!

Suppose $f_0, f_1, f_2 : \mathbb{N}\mapsto \{0;1\}$ satisfy $A_i := \{n \in\mathbb{N}\ \mid \ f_0(n) = f_i(n)\}$ is finite for $i=1,2$.

Then, for every $n\notin A_1\cup A_2$, we must have $f_1(n) = f_2(n)$. Hence $f_1$ and $f_2$ coincide on an infinite set.

Solution 2:

No, let $f,g,h:\mathbb{N}\rightarrow\{0,1\}$ and suppose that $f$ and $g$ coincide pairwise on only finitely many values. Let $$X=\{n\in\mathbb{N}:f(n)\neq g(n)\}.$$ Note that $X$ is cofinite. Now suppose $f$ and $h$ coincide pairwise on only finitely many values. Note that $$Y=\{n\in\mathbb{N}:f(n)\neq h(n)\}$$ is cofinite. Then, since for all $n$ we have that if $f(n)\neq g(n)$ and $f(n)\neq h(n)$, then $g(n)=h(n)$, we find $$X\cap Y=\{n\in\mathbb{N}:f(n)\neq h(n),\ f(n)\neq g(n)\}=\{n\in\mathbb{N}:g(n)=h(n)\}$$ is infinitely large, as the intersection of two cofinite sets if cofinite.