- #1

mathmari

Gold Member

MHB

- 5,049

- 7

Hey!

Let $M:=\{1, 2, \ldots, 10\}$ and $\mathcal{P}:=\{\{1,3,4\}, \{2,8\}, \{7\}, \{5, 6, 9, 10\}\}$.

For $x \in M$ let $[x]$ be the unique set of $\mathcal{P}$ that contains $x$.

We define the relation on $M$ as $x\sim y:\iff [x]=[y]$.

Show that $\sim$ is an equivalence relation.

For that we have to show that the relation is reflexive, symmetric and transitive.

Is everything correct and complete? Or do we have to justify each property with more details, i.e. using the definition of $[x]$ ? (Wondering)

Let $M:=\{1, 2, \ldots, 10\}$ and $\mathcal{P}:=\{\{1,3,4\}, \{2,8\}, \{7\}, \{5, 6, 9, 10\}\}$.

For $x \in M$ let $[x]$ be the unique set of $\mathcal{P}$ that contains $x$.

We define the relation on $M$ as $x\sim y:\iff [x]=[y]$.

Show that $\sim$ is an equivalence relation.

For that we have to show that the relation is reflexive, symmetric and transitive.

- Reflexivity:

Let $x \in M$. Then it holds, trivially, that $[x]=[x]$. Therefore $x\sim x$. So $\sim$ is reflexive. - Symmetry:

Let $x,y \in M$ and $x\sim y$. Then $[x]=[y]$. Equivalently it holds that $[y]=[x]$ and therefore $y \sim x$. So $\sim$ is symmetric. - Transitivity:

Let $x,y,z\in M$ and $x\sim y$ and $y\sim z$. Then it holds that $[x]=[y]$ and $[y]=[z]$. So we have that $[x]=[y]=[z]$, so $[x]=[z]$ and therefore $x\sim z$. So $\sim$ is transitive.

Is everything correct and complete? Or do we have to justify each property with more details, i.e. using the definition of $[x]$ ? (Wondering)

Last edited by a moderator: