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Let $\varphi: \mathbb{R}^{2} \to \mathbb{R}$ be a symmetric (not necessarily continuous) function (so, $\varphi(x,y)=\varphi(y,x)$ $\forall (x,y)\in \mathbb{R}^{2}$),

let $\mathcal{F}$ be the set of all functions $f: \mathbb{R} \to \mathbb{R}$ (not necessarily continuous),

and let $\Phi$ be the nonempty symmetric relation on $\mathcal{F}$ defined as follows in association with the function $\varphi$: $$ (f,g) \in \Phi \subset \mathcal{F} \times \mathcal{F} \Longleftrightarrow \Big( \lim\limits_{x\to +\infty} |f(x)-g(x)|=0 \, \& \, \lim\limits_{x\to +\infty} \varphi(f(x),g(x))=0 \Big). $$

Does there exist such a function $\varphi: \mathbb{R}^{2} \to \mathbb{R}$ that the corresponding relation $\Phi$ satisfies the following condition: $$ \forall f,g,h \in \mathcal{F} \quad \Bigg( \Big( (f,g) \in \Phi \, \& \, (g,h) \in \Phi \Big) \Rightarrow (f,h) \notin \Phi \Bigg) \, ? $$ So, $\Phi$ should be the 'never transitive' relation.

Thank you very much in advance!

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    $\begingroup$ Yes. If $\varphi(x,y)=17$ is constant, then $\Phi$ is empty and your condition holds. Have you left out a condition? $\endgroup$ – Joel David Hamkins Aug 31 '14 at 13:53
  • $\begingroup$ But what if 17 is not constant? (An actually serious comment: what is the motivation for this? Without knowing that, it's not clear how to alter the problem to avoid Joel's examples and still yield something interesting.) $\endgroup$ – Noah Schweber Sep 1 '14 at 0:45

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