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I'm currently trying to solve the following tasks:

1) Prove that the relation $\mathcal{R} = \{(x, y) \in \mathbb{C} \mid x - y \in \mathbb{R}\}$ is an equivalence relation.

2) Find all equivalence classes of $\mathcal{R}$.

I already solved the first one, but I'm having "trouble" with the second one. If I'm not mistaken, there is an infinite amount of equivalence classes, right?

  • $x \sim 0 = \{0\}$, because for "$x - 0 \in \mathbb{R}$" to be true, $x$ has to be $0$.

And you can continue doing this, for an infinite amount of equivalence classes:

  • $x \sim 1 = \{1\}$

  • $x \sim 2 = \{2\}$

  • $x \sim 3 = \{3\}$

  • $\cdots$

If this is actually correct, how do I correctly write it down?

Can I just write something like "$x \sim i = \{i\}$ for $i \in [0, \infty[$"?

Blue
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Gereon99
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  • I assume that you mean $\mathcal{R}$ is a relation ${(x,y)\in\Bbb C\mid x-y\in\Bbb R}$ where the $\Bbb C$ is in reference to the complex numbers and $\Bbb R$ is in reference to the real numbers... yes? Let us look at the equivalence class of $0$ in more detail... Can you describe the subset of complex numbers who when subtracted by zero result in a real number? Sure... $0-0$ is a real number. Is $5-0$ a real number or a nonreal number? Is $5$ then related to zero? – JMoravitz Jan 31 '20 at 16:48
  • I don't agree with your assertion of $x=0$ being a necessariy condition in order for $x-0$ to be in $\Bbb R$. –  Jan 31 '20 at 17:07
  • @Gae. S Yes, you're absolutely right, I think I messed that up, because the imaginary part has to be 0. – Gereon99 Jan 31 '20 at 17:18

1 Answers1

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Hint:

Try to show that given a complex number $x\in\Bbb C$ where we write $x$ in the form $x=a+bi$ with $a,b\in\Bbb R$ and another complex number $y$ written as $y=c+di$ with $c,d\in\Bbb R$...

...that $x$ is related to $y$ if and only if $b=d$.

So... $2+i$ is related to $5+i$ since $(2+i)-(5+i)=-3$ is real for example and $5$ is related to $0$ since $(5)-(0)=5$ is real and so on...

As to the question of the equivalence classes... if we were to picture the complex numbers as appearing on the complex plane, what does it mean for $b=d$?

So, the set of all numbers related to some complex number $a+bi$ will be the set of all complex numbers whose imaginary part is equal to $b$... how might you write that? How might you write the collection of all such equivalence classes?

JMoravitz
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  • Lets talk about x ~ 0 again: You can write this as 'a + bi - (c + di)' and this has to be part of the real numbers. This is equivalent to '(a - c) + i (b - d)'. For this to be a real number the imaginary part has to be zero, 'b - d = 0' -> 'b = d' So the imaginary part of x has to be the same as the imaginary part of 0, which is 0. This means that x ~ 0 = {a - c} and since c equals 0: x ~ 0 = {a} with a being any real number. x ~ 1 can be disassembled the same way and is already included in x ~ 0, since x ~ 1 = {a - 1}, correct? So theres only one equivalence class? – Gereon99 Jan 31 '20 at 17:16
  • @Gereon99 is $0+2i$ included in the class of $0$? Is $5+82i$ included in the class of $0$? How about $\sqrt{2}+\pi i$? – JMoravitz Jan 31 '20 at 17:18
  • @Gereon99 the punchline will be that there are uncountably infinitely many different equivalence classes. They can be pictured as the set of horizontal lines (parallel to the real axis) of the complex plane. – JMoravitz Jan 31 '20 at 17:22
  • Its not, because 0 + 2i - (0 + 0i) = 2i which isn't a real number. Any number x with an imaginary component thats not 0 will end up being a complex number, since y has no imaginary component to subtract it with to end up with an z = (a-c) + 0i. Isnt the condition for (x - y) to be part of the real numbers that x has to have the same imaginary part as y (which is always 0), thus x being any real number? – Gereon99 Jan 31 '20 at 17:25
  • The equivalence class of $0$ is the set of all real numbers. The equivalence class of $i$ is the set of all numbers which have imaginary part $i$. The equivalence class of $a+bi$ with $a$ and $b$ fixed will be $[a+bi] = {a'+bi~:~a'\in\Bbb R}$. The set of all such equivalence classes is ${[a+bi]~:~b\in\Bbb R}$, or if you want to combine those last two sentences together will be ${{a+bi~:~a\in\Bbb R}~:~b\in\Bbb R}$ – JMoravitz Jan 31 '20 at 17:28
  • Im sorry, Im definitely missunderstanding something... but how can lines parallel to the real axis be part of an equivalence class? Like, they are complex numbers then, arent they? So they arent included in the relation in the first place. Can I define equivalence classes x ~ z with z being some complex number? (like z = 2i) Then it would make sense to me, but I always thought z has to be some natural number. – Gereon99 Jan 31 '20 at 17:31
  • "I always thought that $z$ has to be some natural number." Absolutely not. All that an equivalence class is is a maximal subset of the domain of the relation such that everything in the subset is related to everything else in the subset, leaving nothing out. There is no rule that states that the only equivalence classes that need be considered are of the form $[n]$ with $n$ a natural number. – JMoravitz Jan 31 '20 at 17:34
  • Okay, I apologize for not knowing simple stuff as this. To my defense: This was my first "complex" equivalence relation. Thank you for explaining this to me. Man, I feel stupid now... – Gereon99 Jan 31 '20 at 17:34