# Show that the equivalence classes give a partition of $X$. [duplicate]

"... the following subset of $X$ denoted by $[x]$ $(x\in X)$ is called the equivalence class generated by $x$:

$$[x]:= \{y \in X; x\sim y \}$$

Show that the equivalence classes give a partition of $X$.

According to my book "Given a nonempty set $X$ we say that the family of subsets $\{X_\alpha\}_{\alpha \in I}$ is a partition of $X$ iff

$$\bigcup_{\alpha \in I} X_\alpha = X$$

$$X_\alpha \cap X_\beta = \emptyset \text{ if } \alpha \not = \beta$$

I missed my lecture on this topic and I dont have any clue where to start. As I understand it the equivalence classes is the set of all $x \in X$ s.t for $y \in X$ then $x\sim y$. But after here I am lost.

I guess I could denote the equivalence classes by

$$[x_i] := \{y \in X; x_i \sim y\}$$ And now try to prove that

$$\bigcup_{\alpha \in I} [x_{i}] = X$$ and

$$[x_i] \cap [x_j] = \emptyset \text{ if } j \not = i \text{ ?}$$

## marked as duplicate by Tim Raczkowski, Silvia Ghinassi, Leucippus, JMP, Claude LeiboviciJan 30 '16 at 5:52

• Your book actually omits an important requirement (or you accidentally left it out): all the $X_{\alpha}$ should also be nonempty. – BrianO Jan 30 '16 at 0:09
• $\text{“}$Given a nonempty set $X$ we say that the family of subsets $\{X_\alpha\}_{\alpha \in I}$ is a partition of $X$ iff $$\bigcup_{\alpha \in I} X_\alpha = X$$ and $$X_\alpha \cap X_\beta = \emptyset \text{ if } \alpha \not = \beta \text{ ''}$$ This should have said $\text{“}$the family of non-empty sets $\{X_\alpha\}_{\alpha\in I}.\text{''}$ $\qquad$ – Michael Hardy Jan 30 '16 at 2:21

Because $\sim$ is reflexive, for every $x\in X$ we have $x\in [x]$, thus $X\subseteq \bigcup_{[x] \in X/\sim} [x] \subseteq X$, which proves the first condition.

For the second condition, we prove the contrapositive. Suppose $[x] \cap [y] \ne \emptyset$; we show that $[x] = [y]$. Given $z\in [x] \cap [y]$, we have $z\sim x$ and $z\sim y$, so by symmetry and transitivity of $\sim$ we get $x\sim y$. It follows that $[x] = [y]$.