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I need to prove that $(X \cap Y) $\ $E $ $\subset (X$ \ $E) \cap (Y$ \ $E)$, where $E$ is an equivalence relation over $A$ and $X,Y \subset A$.

I don't know where to begin. I know that $X$ \ $ E$ denotes the set of all the equivalence classes {$ [a]_E : a \in X$}. To prove inclusion, I say that $y \in (X \cap Y) $\ $E$. I don't know what to do next. How can I proceed?

Thanks.

Lstoi
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1 Answers1

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The statement $(X \cap Y) \setminus E \subset (X \setminus E) \cap (Y \setminus E)$ is false. Counterexample: $X=\{1,2\}, Y=\{2,3\}, [1]=X$. Then $(X \cap Y) \setminus E=\{\{2\}\}\not\subset X\setminus E=\{\{1,2\}\}$.

If you want to show the converse inclusion, here is a hint: $A\in X\setminus E$ is a set, $A\subset X$, and $\exists x\in A : (z\in A \Leftrightarrow z\in X \wedge xEz)$. Write down equivalent statements for $Y\setminus E$ and $(X\cap Y)\setminus E$. Proving the inclusion should be straightforward from there.

Ansgar Esztermann
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