I need to prove that the below relation is transitive.
$$\big\{(f,g)\in (\mathbb{R}\to\mathbb{R})\times(\mathbb{R}\to\mathbb{R})\ \ |\ \ \forall A\in \mathcal{P}(\mathbb{R}) \big(|A|=\aleph_0\Rightarrow (\exists x\in A[f(x)=g(x)])\big)\big\}$$
Thanks in advance for the help!
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Hint: Prove that if $(f,g)$ is in your relation, and $A\subseteq\mathbb{R}$ is countably infinite, then you can define an entire sequence $(x_n)_{n=0}^{\infty}$ of points in $A$ such that $f(x_n)=g(x_n)$ for every $n$. (You can define such a sequence recursively, by noting that removing one point from a countably infinite set yields a countably infinite set.)
Notice that (f,g) is in your relation if and only if $f(x)\neq g(x)$ only for finitely many points $x\in \mathbb{R}$: otherwise, the set $\{x\in\mathbb{R}:f(x)\neq g(x)\}$ would contain a countable set $A$, contradicting that $f$ is related with $g$.
After this, if $(f,g),(g,h)$ are in the relation, then the set of points $x\in\mathbb{R}$ where $f(x)\neq h(x)$ is at most $\{x\in \mathbb{R}:f(x)\neq g(x)\}\cup \{x\in \mathbb{R}:g(x)\neq h(x)\}$, which is finite as it is the union of two finite sets.