# An equivalence relation is a partition of A?

• I
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## Summary:

equivalent class is a Cartesian product

## Main Question or Discussion Point

Hi equivalent class is a Cartesian product of A*A. Then shouldn't it's union be a partition of A*A, instead being a partition of A

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Math_QED
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This question doesn't really make sense. Can you try to rephrase it?

Given a (non-empty) set $X$, an equivalence relation $R$ is a subset $R \subseteq X \times X$ such that

(1) $(x,x) \in R$ for all $x \in X$
(2) For all $x,y \in X: (x,y) \in R \implies (y,x) \in R$
(3) For all $x,y,z \in X: (x,y) \in R, (y,z) \in R \implies (x,z) \in R$

An equivalence class of an element $x\in X$ is then the set of all elements that are in relation with $x$, i.e. the set $[x]:=\{y \in X: (x,y) \in R\}$.

It is true though that $X = \bigcup_{x \in X} [x]$ and that every element of $X$ is in precisely one equivalence class. Thus the equivalence classes partition the set $X$.

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Shouldn't their union be X*X where * is the Cartesian product. Think about it this way you have sets of element of (a1,a2). How can you union them and then they give you elements of (a), R is a subsets of X*X, take the union of all the partitions you should get X*X not the set X

Math_QED
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Shouldn't their union be X*X where * is the Cartesian product. Think about it this way you have sets of element of (a1,a2). How can you union them and then they give you elements of (a)
No, the equivalence classes are subsets of $X$. Recall that the equivalence class of $x$ is the subset $\{y \in X: (x,y) \in R\} \subseteq X$. This is a subset of $X$ (by definition!), not of $X \times X$. Thus any union of such sets remains in $X$.

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From my university textbook

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Math_QED
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From my university textbook
Yes, this is true. But I say that the union of the equivalence classes partition $X$. Just look carefully at the definitions I wrote down.

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No, the equivalence classes are subsets of $X$.
But they aren't they are subset of X*X

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