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Let $X,Y$ be infinite sets. Define $F$ as $F=\{f:X\rightarrow Y\}$ . We define a binary relation $R$ on $F$: $fRg$ if there is no countable $S\subseteq X$ such that $\forall x\in S \ f(x)\neq g(x)$. Is $R$ an equivalence relation?

I think it's not necessarily transitive, but I'm not sure. Thanks

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  • $\begingroup$ Of course it's not necessarily transitive because inequality is not transitive. Let X = Y = R, f(x) = 2^x, g(x) = 0 then fRg and gRf then f is not equivalence to f. $\endgroup$ – Zack Ni Jun 26 '16 at 9:42
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    $\begingroup$ @ZackNi f is not equivalence to g because $\forall x\in \Bbb N f(x)\neq g(x)$ $\endgroup$ – guest Jun 26 '16 at 9:50
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This is an equivalnce, but from a very trivial reason. Your relation is not defined well.

Even if you have one $x_0\in X$ such that $f(x_0)\ne g(x_0)$, then $S:=\{x\}$ is a countable set such that for every $x\in S$ (which is only $x_0$), $f(x)\ne g(x)$.

that means that if $fRg$ then for every $x\in X$, $f(x)=g(x)$. that is simply $f=g$! and this relation is an equivalnce.

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