# The union is not disjoint in the example from the book

I am reading 'Measure, Integral and Probability' book. There is this following para in the book:

An equivalence relation $$\sim$$ on $$E$$ partitions $$E$$ into disjoint equivalence classes: given $$x \in E$$, write $$[x] = \{z : z \sim x\}$$ for the equivalence class of $$x$$, i.e. the set of all elements of $$E$$ that are equivalent to $$x$$. Thus $$x \in [x]$$, hence $$E = \cup_{x \in E} [x]$$. This is a disjoint union: if $$[x] \cap [y] \neq \emptyset$$, then there is $$z \in E$$ with $$x \sim z$$ and $$z \sim y$$, hence $$x \sim y$$, so that $$[x] = [y]$$. We shall denote the set of all equivalence classes so obtained by $$E/\sim$$.

First of all, I do not see how $$\cup_{x \in E} [x]$$ is a disjoint union. Take, for example, a set

$$S = \{1, 2, 3, 4, 5 \}$$ with the following equivalence on it:

$$\sim = \{(1,1),(2,2),(3,3),(4,4),(5,5),(1,2),(2,1),(2,3),(3,2),(1,3),(3,1) \}$$

then

$$[1] = \{1,2,3\}$$, $$[2]=\{1,2,3\}$$, $$[3]=\{1,2,3\}$$, $$[4]=\{4\}$$, $$[5]=\{5\}$$. Already, you have that sert $$[1]$$ is the same as $$[2]$$. So that the union $$\cup_{x \in E} [x] = \{1,2,3\} \cup \{1,2,3\}$$ and that's not disjoint. Unless I do not understand what disjoint union means.

Also, what is $$E/\sim$$?

• It should probably say take distinct representatives for each equivalence class, then that would be a disjoint union. – Rellek Dec 13 '18 at 22:18
• what about $E/\sim$ is that just $\cup_{x \in E} [x]$? – i squared - Keep it Real Dec 13 '18 at 22:24
• No it would just consist of the representatives for each equivalence class. That is, $E / \sim = \{ [x] \ | \ x \in E \}$. In your example, $E/ \sim = \{ [1] , \ [4], \ [5] \}$. Of course you could use $[2]$ and $[3]$ for representatives of the class of $[1]$ instead, but they are the same under the relation $\sim$. – Rellek Dec 13 '18 at 22:28