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I'm approaching the study of pure Algebra, starting from equivalence relations, sets and quotient sets. I have a background in Algebra, but it's only a bit of linear algebra and I never worked with equivalence relations, so I am gonna ask you if the two examples I am writing are correct or not. They are taken from two exercises, but there is no solution.

First Be $X = \mathbb{Z}$ and define an equivalence relation between elements of $X$ such that: $x \sim y \iff x-y$ is even.

I proved that that is indeed an equivalence relation, and then I wrote the "class elements" (are they called that way?):

$$[A] = \{ 0; \pm 2; \pm 4; \ldots \}$$ $$[B] = \{ \pm 1; \pm 3; \pm 5; \ldots \}$$

There are no other "classes", hence I can say $X/\sim = \{ [A], [B] \}$. Is this the correct way to write? I thought that I could also write that $X/\sim = A \cup B$. Would this be correct?

Second This one is a bit strange to me (or maybe it's just obvious). $X$ as before, and $x \sim y \iff |x-y| = 2$.

For this, I started to write the class elements:

$$[A] = \{ 0; 2 \}$$ $$[B] = \{ 2; 4 \}$$ $$[C] = \{ 4; 6 \}$$

And so on, but also I have $[a] = \{ -2 ; 0 \}$, $[b] = \{-1; 1\}$ and so on...

Am I right so far or is this wrong?

I would (if correct) then write

$$X/\sim = A\cup B \cup C \cup \ldots \cup a \cup b ... $$

Tell me if I'm right please.

Also I ask you if you know some GREAT reference about those topics, quotient sets and so on.

Thank you so much!

EDIT

I feel dumb, but I realised, thank to the comments, that the second one is not an equivalence relation. It's not transitive nor reflexive...

Bill Dubuque
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Heidegger
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    Well, the second one isn't an equivalence relation (why not?). – lulu Jun 12 '24 at 18:19
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    @lulu Ah right! Because if $|x-y| = 2$ and $|y-z| = 2$ then it's not true that $|x-z| =2$... Hence it's not transitive – Heidegger Jun 12 '24 at 18:21
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    @Heidegger That's right, but there's an even easier reason... – J.D Jun 12 '24 at 18:22
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    @J.D Because perhaps $x$ is not in relation with itself? I really didn't think before starting the exercise, I feel dumb now :D – Heidegger Jun 12 '24 at 18:23
  • @J.D Thank you again for having made me thing! Anyway, Can you please tell me if the way of writing $X$ as the union of the "class equivalence" is right? Also, do you have some good notes about? So far I just found brief and short papers... – Heidegger Jun 12 '24 at 18:26
  • The notion of "equivalence class" really only pertains to equivalence relations. It's not clear what (if any) meaning they have for a general relation. – lulu Jun 12 '24 at 18:28
  • @lulu How would I call the $[A]$, $[B]$ and so on..? Just sets? – Heidegger Jun 12 '24 at 18:29
  • I have no idea what you mean by those sets. Again,. for an equivalence relation $[A]$ denotes the set of elements equivalent to $A$. Those then have the property that any two are either disjoint or coincident. But none of that is true in this case. – lulu Jun 12 '24 at 18:35
  • @lulu In the first example, how do you call $[A]$ and $[B]$? Just sets? They are set such that $X = A \cup B$. – Heidegger Jun 12 '24 at 18:51
  • @lulu "for an equivalence relation [] denotes the set of elements equivalent to " I don't get this part: what is $A$? – Heidegger Jun 12 '24 at 18:52
  • $A$ is any element in the set on which the equivalence relation is defined. For your first relation, which is an equivalence relation, $\cdots = [0]=[2]=[4]=\cdots$ and all just equal the set of even numbers. For that relation there are only two equivalence classes, the Evens and the Odds. Again, there is no analog for the second relation. – lulu Jun 12 '24 at 18:57

1 Answers1

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For the first relation, you are correct that there are two equivalence classes. They are the even integers and the odd integers.

What students find weird about the way we denote equivalence classes is that $[a] = [b]$ iff $a \sim b$. In your first example relation, the even integers are all equivalent to $0$, so you can call this equivalence class $[0]$. The odd integers are all equivalent to $1$, so you can call this equivalence class $[1]$. Now $2 \sim 0$, so $[2]$ is just another name for $[0]$, the even integers. and $[-1] = [1]$, the odd integers. Et cetera.

Since there are no other equivalence classes, you can say that $\mathbb{Z} = [0] \cup [1]$, or $\mathbb{Z}/\mathord{\sim} = \{[0],[1]\}$. But $\mathbb{Z}/\mathord{\sim} = 0 \cup 1$ would be wrong; $0$ and $1$ aren't sets so their union is undefined.

As discussed in the comments to your question, the second relation is not an equivalence relation. It's symmetric but neither reflexive nor transitive. We don't consider “classes” for relations that aren't equivalence relations. The best you could say is that the smallest equivalence relation which contains this relation is the first example relation.

There are lots of questions on this site about algebra textbooks. Here's one.

Matthew Leingang
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