Do I have errata in my measure book: Measure, Integral and Probability II by Capinski

Question

The author goes to define what equivalence (from basic set theory) means on page 4. This book can be found online. But here is a short summary:

Given any set $E$, an equivalence relation on $E$ is a relation (i.e. a subset $R$ of $E \times E$, where we write $x \sim y$ to indicate that $(x,y) \in R$) with the following properties:

1. reflexive: $\forall x \in E$, $x \sim x$
2. symmetric: $x \sim y \implies y \sim x$
3. transitive: $x \sim y$ and $y \sim z$ implies $x \sim z$

An equivalence class for $x \in E$ is then defined as $[x] = \{z: z \sim x\}$. An equivalence relation partiions $E$ into disjoin equivalence classes. It is also obvious that $x \in [x]$ and thus $E = \cup_{x \in E}[x]$. But author proceeds to state that this union is disjoin if (all of the following I do not understand):

if $[x] \cap [y] \neq \emptyset$

(but by definition of disjoin union we actually want the intersections to be empty; so I though that this was a typo, and it should be $= \emptyset$; but I have not yet thought about this in perspective on the structure that we have defined above, and maybe that actually does not make sense; Now I have thought about it, if there is $E = \{1, 2\}$ and $R = \{(1,1), (2,2), (1,2), (2,1)\}$ then $[1] \cap [2]$ is not empty, they are the same.)

then there is $z \in E$ with $x \sim z$ and $z \sim y$, hence $x \sim y$ so that $[x] = [y]$.

So yeah... We have above that $[1] = \{1,2\} = [2]$ but the union $[1] \cup [2]$ is not disjoint... Or is my disjoint definition incorrect.

I had dejavu writing this, apparently I already asked this before but did not get an answer.

Answer 1

You misunderstand what the author is saying. They are not saying it's a disjoint union because $[x]\cap [y]\neq \emptyset$. They're saying it's a disjoint union because if $[x]\cap [y] \neq \emptyset$, then $[x]=[y]$. In other words, no two different equivalence classes can overlap.

For instance, let $E=\mathbb{N}$, and $R=\{(x,y) \mid x,y\in \mathbb{N} \text{ and } 3 \text{ divides } (x-y)\}$. We can write $$\mathbb{N}=\bigcup_{x \in \mathbb{N}} [x] = [0]\cup [1]\cup [2]\cup [3] \cup \cdots.$$ How is this union disjoint? Notice that $[0]=[3]=[6]=\cdots$, and similarly $[1]=[4]=[7]=\cdots$, and $[2]=[5]=[8]=\cdots$. So this union is really $$\mathbb{N}=[0]\cup [1]\cup [2]$$ which is disjoint.

Answer 2

If this is exactly what the author writes then there would be some clarification helpful.

Often one chooses a so called system $\Sigma \subset E$ of representatives from each equivalence class as follows:

• For each $x \in E$ there is a $\sigma \in \Sigma$ s.t. $[x] = [\sigma]$
• If $x,y \in \Sigma$ and $x \neq y \Rightarrow [x] \neq [y]$, hence $[x] \cap [y] = \emptyset$

Then we can write

$$E = \bigcup_{\sigma \in \Sigma}[\sigma]$$ where the union is disjoint.

Note, that one needs the axiom of choice to assure the existence of such a system of representatives in general.