MHB Show that ~ is an equivalence relation

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The discussion focuses on proving that the relation defined on the set M, where x is equivalent to y if they belong to the same subset in the partition P, is an equivalence relation. The participants confirm that the relation is reflexive, as each element is related to itself. They also establish symmetry, noting that if x is related to y, then y is related to x. Transitivity is demonstrated by showing that if x is related to y and y to z, then x is related to z. The conversation concludes with a suggestion to clarify the definitions of the equivalence classes for better understanding.
mathmari
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Hey! :o

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)
 
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Well, how obvious to you consider each of your claims?

It might be useful to specify that [1]= [3]= [4]= {1, 3, 4}, that [2]= [8]= {2, 8}, that [7]= {7}, and that [5]= [6]= [9]= [10]= {5, 6, 9, 10}.

 
More generally, for all sets $A$ and $B$ and functions $f:A\to B$ the relation $x\sim y\iff f(x)=f(y)$ is an equivalence relation on $A$.
 

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