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:

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


$[1] = \{1,2,3\}$, $[2]=\{1,2,3\}$, $[3]=\{1,2,3\}$, $[4]=\{4\}$, $[5]=\{5\}$. 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 $E/\sim$?

  • $\begingroup$ It should probably say take distinct representatives for each equivalence class, then that would be a disjoint union. $\endgroup$ – Rellek Dec 13 '18 at 22:18
  • $\begingroup$ what about $E/\sim$ is that just $\cup_{x \in E} [x]$? $\endgroup$ – i squared - Keep it Real Dec 13 '18 at 22:24
  • 1
    $\begingroup$ 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 $\sim$. $\endgroup$ – Rellek Dec 13 '18 at 22:28

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