Equivalence relation and partition of set

Question

Suppose that $X$ is some set and $\sim$ is an equivalence relation on $X$ and $X\neq \varnothing$. For each $x\in X$ define $$R(x):=\{y\in X: y\sim x\}$$ an equivalence class of $x$. One can show that if $R(x_1)\cap R(x_2)\neq \varnothing$ then $R(x_1)=R(x_2).$

I would like to understand why these equivalence classes form the partition of $X$. So I need to show that there exist $\{P_i\}_{i\in I}$ such that $P_i\cap P_j=\varnothing$ and $X=\cup_{i\in I}P_i$, right?

Intuitively, I know that they form partition but cannot prove it rigorously. I was wondering how to specify these partitioning sets explicitly.

Answer 1

The set of partitions is just $V = \{R(x) \mid x \in X\}$.

More formally, let $P_v = v$ for all $v \in V$. Note that $P_v \subseteq X$ for all $v \in V$. I claim that $\{P_v\}_{v \in V}$ is a partition.

There are three conditions that must be checked. First, we must show that $\forall v \in V \exists x \in P_v$. To show this, consider some $v \in V$. Write $v = R(x)$ for some $x$. Then $x \in v = P_v$.

The second condition is that $\bigcup\limits_{v \in V} P_v = X$. To show this, suppose we have some $x \in X$. Then $x \in R(x)$. Let $R(x) = v$. Then $x \in v = P_v$. Therefore, $x \in \bigcup\limits_{v \in V} P_v$. Thus, we see that $\bigcup\limits_{v \in V} P_v = X$.

Finally, the third condition is that for all $v_1, v_2 \in V$, if $\exists x \in P_{v_1} \cap P_{v_2}$, then $v_1 = v_2$. Suppose there is some $x \in P_{v_1} \cap P_{v_2}$. Then we see that $R(x) = v_1$. And we see that $R(x) = v_2$. Therefore, $v_1 = v_2$.

Answer 2

I'll try to keep my explanation as simple as possible, bear with me.

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1. Each element of $X$ belongs to some equivalence class of $X$.

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This is pretty straightforward. Since $\sim$ is reflexive, every $x \in X$ satisfies $x \sim x$ and so $x \in R(x)$.

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2. Any two equivalence classes of $X$ are either disjoint or identical.

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Which means for any $x_1$ and $x_2$ in $X$, either $R(x_1)= R(x_2)$ or $ R(x_1) \cap R(x_2)= \varnothing$.

In simple words, the equivalence classes of any two elements are either the same or share no common elements at all. This is because if $x \in R(x_1)$ and $x \in R(x_2)$ then $x \sim x_1$ and $x \sim x_2$ and by symmetry and transitivity of $\sim$, we have $x_1 \sim x$ and $x \sim x_2 \Rightarrow x_1 \sim x_2$.

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3. Union of all equivalence classes is $X$.

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From 1, we know that $ x \in R(x), \ \forall x \in X$, so $\{x \} \subseteq R(x)$. Thus $\bigcup\limits_{x \in X} \{x \} = X \subseteq \bigcup\limits_{x \in X} R(x) \subseteq X$ since $R(x) \subseteq X, \ \forall x \in X$.

Hence,we have that $\bigcup\limits_{x \in X} R(x) = X$.

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