0
$\begingroup$

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.

$\endgroup$
5
  • 1
    $\begingroup$ Symmetry follows at once from that absolute value there: $\;|f(x)-g(x)|=|g(x)-f(x)|\;$ $\endgroup$ – DonAntonio Jun 16 '20 at 11:01
  • 1
    $\begingroup$ ...and who says it isn't an equivalence relation? Why do you thin it isn't? $\endgroup$ – DonAntonio Jun 16 '20 at 11:09
  • $\begingroup$ @DonAntonio It is not an equivalence relation if we consider also discontinuous function. For example $f(x) = 1, x \in \mathbb{Q}$ and $f(x) = -1, x \notin \mathbb{Q}$, is related to the null function. The null function is related to $g(x) = 1$. But $f$ is not related to $g$. $\endgroup$ – Enforce Jun 16 '20 at 11:21
  • $\begingroup$ @DonAntonio This relation is defined on the set of all functions $\mathbb{R}\to\mathbb{R}$, not only continuous ones. $\endgroup$ – mwt Jun 16 '20 at 11:22
  • $\begingroup$ @mwt I know. I only wanted the OP to answer why he/she believes the relation isn't an equivalence one... $\endgroup$ – DonAntonio Jun 16 '20 at 11:44
2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ You're right and thanks for pointing out my mistake in my (now deleted) answer. Lack of sleep does that haha. $\endgroup$ – Matheus Andrade Jun 16 '20 at 11:16
0
$\begingroup$

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 $x >= 0$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.