The relation on the set of functions

Question

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!

Answer 1

We can make $\Phi$ be the empty relation, by setting $\phi ( x , y ) = 1+ \frac{1}{ | x - y | }$ is $x\neq y$ and $\phi(x, x ) = 1$, in this way you will not have $\lim_{x\to 0} \phi(f(x) , g(x) ) = 0$ in any way.

Answer 2

Put $\phi(x,y)=0$ if $x,y>1$ and either $x=y^{2}$ or $y=x^{2}$; otherwise $\phi(x,y)=1$. Then you have $f\Phi g$ if and only if $\lim\limits_{x\rightarrow +\infty} f(x)=\lim\limits_{x\rightarrow +\infty} g(x) =1^{+}$ and $|\ln(f(x))/\ln(g(x))|=2$ for $x$ big enough. The latter condition ensures that $\Phi$ is never-transitive, and $(x\mapsto 1+1/x, x\mapsto(1+1/x)^{2})\in \Phi$, for example.