Why is this not an equivalence relation on real functions? $(\exists c\in\mathbb{R})(\forall x\in\mathbb{R})|f(x)-g(x)|=c$

Question

Say we have the following relation on the set of all functions $\mathbb{R} \to \mathbb{R}$

$$(\exists c \in \mathbb{R})(\forall x \in \mathbb{R})|f(x) - g(x)| = c$$

I'm having trouble understanding why this relation isn't an equivalence relation.

I know that the relation is reflexive, as $f(x) - f(x) = 0$, $0 \in \mathbb{R}$.

But I'm having trouble when it comes to symmetry and transitivity.

Answer 1

Say $f(x)= -1$ if $x <0$ and $f(x) = 1$ if $x \geqslant 0$. Then $|f(x) - 0| = 1$ and $f$ is related to the null function.

The constant function $g(x) = 1$ is also related to the null function.

But $f$ and $g$ are not related as $|f(x)-g(x)|$ is non-constant!

I think if you restrict your relation on continuous functions, you avoid this kind of behavior and maybe it would define an equivalence relation on this set.

Answer 2

To show that the relation is not transitive let's take three functions $f,g,h : \mathbb{R} \to \mathbb{R}$ defined as follows: \begin{align} f(x) &= 0 \\ g(x) &= \left\lbrace\begin{array} --1 &\text{ for $x < 0$} \\ 1 &\text{ for $x >= 0$} \\ \end{array}\right. \\ h(x) &= g(x) + 1 \end{align}

For all $x \in \mathbb{R}$ we have $|f(x) - g(x)| = 1$ and $|g(x) - h(x)| = 1$. However $|f(x) - h(x)|$ equals either $0$ (for $x < 0$) or $2$ (for $x >= 0$).