The motivation to this question is: Is this an equivalence relation (reflexivity, symmetry, transitivity) :

Let $\theta(s):\mathbb{C}\to \mathbb{R}$ be a well defined function. I define the following relation in $\mathbb{C}$.

$\forall s,q \in \mathbb{C}: s\mathbin{R}q\iff\theta(s)\ne 0 \pmod {2\pi}$ (and)

$\theta(q)\ne 0 \pmod {2\pi}$

My question is: Can I consider the case with equality as the inverse relation of this one and it is also an equivalence relation.

  • 1
    $\begingroup$ I simply added the relation to which you are referring to this post, to help "round out" your question (so it's self-contained). I left your link. $\endgroup$ – Namaste Dec 21 '12 at 13:52

First: just to be sure you understand: your relation $R$ is not an equivalence relation if there exists $x$ such that it is not the case that $x R x$: when, e.g., $\theta(x) = 0 \pmod {2\pi}$, since in this situation, it is NOT the case that $\theta(x) \ne 0 \pmod {2\pi}\;$ AND $\;\theta(x) \ne \pmod {2\pi}$

That is, the relation IS reflexive if and only if FOR ALL $x \in \mathbb{C}$, it is true that $\theta(x) \ne 0 \pmod {2\pi}\;$ AND $\;\theta(x) \ne 0 \pmod {2\pi}$.

You need reflexivity (as well symmetry and transitivity), so if $x$ satisfies $\theta(x) = 0 \pmod {2\pi} $, then you do not have $x R x$, hence it is not an equivalence relation. Not knowing exactly how $\theta(x)$ is defined makes it difficult to ascertain whether or not it can be the case that reflexivity fails. But if, say, $\theta: \mathbb{C} \to \mathbb{R}$ is onto, then reflexivity fails.

Your relation defines what it means for $s$ to be related to $q$: $$sRq \iff (\theta(s) \ne 0 \pmod {2\pi} \land \theta(q) \ne 0\pmod {2\pi}.$$ So it defines the set of all ordered pairs $(s, q)$, $s, q \in \mathbb{C}$, which satisfy the given relation. To obtain the inverse relation, you need to determine what relation defines how $q$ is related to $s$: what relation $R^{-1}$ defines the set of all ordered pairs $(q, s)$, when $(s, q)\in R$?

In this case, it turns out that the inverse relation $R^{-1}$ defines exactly the same relation as does $R$: simply commute the conditions: $s \ R^{-1} q \iff \theta(q) \ne 0 \pmod {2\pi} \land \theta(s) \ne 0\pmod {2\pi}$. Then you have $q\ R \ s $.

  • $\begingroup$ Yes, you make things very clear. Thank you very much. $\endgroup$ – China Dec 21 '12 at 14:53

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