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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)

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

The function $\pmod {2\pi}$ is the addition $\pmod {2\pi}$

My question: Is this an equivalence relation (reflexivity, symmetry, transitivity)?

The formula of $\theta(s)$ is not important for this question.

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Your relation is $$sRq\iff \theta(s)\not \equiv 0\text{ and }\theta(q)\not \equiv 0 \mod 2\pi$$ for $s,q\in \mathbb{C}$.

For symmetry: $$sRq\iff \theta(s)\not \equiv 0\text{ and }\theta(q)\not \equiv 0 \mod 2\pi \iff qRs$$

For transitivity: $$sRq\text{ and }qRp\iff \theta(s)\not \equiv 0\text{ and }\theta(q)\not \equiv 0\text{ and }\theta(p)\not \equiv 0 \mod 2\pi\implies sRp$$ Reflexitivity is: $$sRs\iff \theta(s)\not \equiv 0\mod 2\pi$$ That clearly depends on your choice of $\theta$. Therefore, $R$ is an equivalence relation iff $$\theta(\mathbb{C})\cap \left\{2k\pi:k\in \mathbb{Z}\right\}=\emptyset$$

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  • $\begingroup$ Yes, I must define the relation in the set $Ω={s∈ℂ:θ(s)≠0}$ $\endgroup$ – China Dec 20 '12 at 18:38
  • $\begingroup$ @user53124 $\theta (s)\not\equiv 0\mod 2\pi$ $\endgroup$ – Nameless Dec 20 '12 at 18:41
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    $\begingroup$ @copper.hat So you other have a meaningless equivalence relation or no equivalence relation at all $\endgroup$ – Nameless Dec 20 '12 at 18:52
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    $\begingroup$ Yes; that is a meaningful statement! $\endgroup$ – copper.hat Dec 20 '12 at 18:59
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    $\begingroup$ It is not an equivalence relation if there exists a point in $\mathbb{C}$ for which any property fails. There exists infinitely many points q in $\mathbb{C}$ for which $\theta(q) \equiv 0 \pmod {2\pi}$, and for each of these, reflexivity fails. Hence, as defined (an equivalence relation holding $\forall q \in \mathbb{C})$, it is not an equivalence relation, as copper.hat makes clear. One cannot arbitrarily exclude points from the set on which the relation is defined, unless that was the point of the exercise: "Define a set such that the relation is an equivalence relation." $\endgroup$ – Namaste Dec 20 '12 at 20:04
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It is not an equivalence relation if there exists $x$ such that $\theta(x) = 0 \mod 2\pi $.

You need reflexivity, so if $x$ satisfies $\theta(x) = 0 \mod 2\pi $, then you do not have $x R x$, hence it is not an equivalence relation.

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