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Definition: Let $A$ be a set and let $G$ be an equivalence relation in A. If $x ∈ A$, then the equivalence class of $x$ modulo $G$ is the set $G_x$ defined as follows: $$G_x=\{y\in A\mid (y,x)\in G\}=\{y\in A\mid y\sim _G x\}.$$

  1. Let $G$ and $H$ be equivalence relations in $A$. Prove that each equivalence class modulo $G ∩ H$ is the intersection of an equivalence class modulo $G$ with an equivalence class modulo $H$. More exactly, $$(G\cap H)_x\subseteq G_x\cap H_x,\quad \forall x\in A.$$
  2. Let $G$ and $H$ be equivalence relations in $A$, and assume that $G ∪ H$ is an equivalence relation in $A$. Prove that each equivalence class modulo $G ∪ H$ is the union of an equivalence class modulo $G$ with an equivalence class modulo $H$. More exactly, $$(G\cup H)_x=G_x\cup H_x,\quad \forall x\in A.$$

My attempt:

  1. Let $x\in A$. $$\begin{align*}y\in (G\cap H)_x&\Rightarrow (y,x)\in (G\cap H)\ \wedge\ y\in A\\ &\Rightarrow (y,x)\in G\ \wedge \ (y,x)\in H\ \wedge\ y\in A\\ &\Rightarrow y\in G_x\ \wedge\ y\in H_x\\ &\Rightarrow y\in G_x\cap H_x. \end{align*}$$
  2. Let $x\in A$. $$\begin{align*}y\in (G\cup H)_x&\Leftrightarrow (y,x)\in (G\cup H)\ \wedge\ y\in A\\ &\Leftrightarrow [(y,x)\in G\ \vee\ (y,x)\in H]\ \wedge\ y\in A\\ &\Leftrightarrow ((y,x)\in G\ \wedge\ y\in A) \vee\ ((y,x)\in H\ \wedge\ y\in A) \\ &\Leftrightarrow y\in G_x\ \vee \ y\in H_x\\&\Leftrightarrow y\in G_x\cup H_x \end{align*}$$

My proof is right?

asd asd
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  • Looks good to me. – Dylan C. Beck Mar 22 '21 at 21:49
  • You don’t need to carry the “$y\in A$” around in the implications: Just say “let $y\in A$” after “Let $x\in A$” to make it a contextual assumption valid throughout the argument. – Arturo Magidin Mar 22 '21 at 21:58
  • There’s either a problem with your interpretation of $G\cup H$, or there is a problem with the premise of the problem. If $G$ and $H$ are equivalence relations on $A$, in general $G\cup H$, the set-theoretic union, need not be an equivalence relation on $A$. For example, take $A={x,y,z}$, take $G={(x,x),(y,y),(z,z),(x,y),(y,x)}$, and let $H={(x,x),(y,y),(z,z),(y,z),(z,y)}$. Then both $G$ and $H$ are equivalence relations, but $G\cup H$ is not transitive: $(x,y),(y,z)\in G\cup H$, but $(x,z)\notin G\cup H$. But the notation assumes $G\cup H$ is an equivalence relation. (cont) – Arturo Magidin Mar 22 '21 at 22:01
  • First of all, thank you for taking the time to review the exercise for me. And responding to the last comment, the exercise already says that $ G \cup H $ is assumed to be an equivalence relation. – asd asd Mar 22 '21 at 22:04
  • Ah; I missed that assumption. Good. Because it’s false without it. – Arturo Magidin Mar 22 '21 at 22:05

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