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I'm currently learning proofs and elementary set theory. I would like to have feedback on my proof since I'm self-studying. Are some part superfluous or unclear? My proof goes as follows:

I will prove that $(\bigcap_i X_i)-A = \bigcap_i(X_i - A)$, where "$-$" is the set difference:

  1. Suppose that $x \in (\bigcap X_i) - A$, then $x \in \bigcap X_i$ but $x \notin A$. Therefore $x \in X_i$ for all $i$, and $x \notin A$, which means that for all $i$, $x \in X_i$ and $x \notin A$. Thus for all $i$ then $x \in X_i - A$, and as such $x \in \bigcap_i (X_i - A)$.

  2. I then proceed in the same way to show that $x \in \bigcap_i (X_i - A) \Longrightarrow x \in (\bigcap X_i) - A$.

  3. Since all elements in $(\bigcap_i X_i)-A$ are also in $\bigcap_i(X_i - A)$ and vice versa, then $(\bigcap_i X_i)-A = \bigcap_i(X_i - A)$.

In (1.), I say that $x \notin A$ and $(\forall i) [x \in X_i]$ is equivalent to $(\forall i) [x \notin A$ and $x \in X_i]$. Is it ok to make such a step or should it be proved too? And if so, how could it be proved?

Thank you for your help!

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up vote 1 down vote accepted

Your proof is correct and well-written, so good job. As for your second question, it is certainly okay to use this fact without proof here. The fact that you can move the $x\in A$ inside the scope of the universal quantifier without changing the truth value is a special case of the fact from first-order logic that if $\phi$ and $\psi$ are formulas such that $x$ is not free in $\phi$, then $\forall x (\phi \wedge \psi)$ is equivalent to $\phi \wedge \forall x(\psi)$. Proving that this holds in general is not too hard, but requires a bit of formal-logic machinery to do precisely. Don't worry too much about proving facts like that (as long as they are true, of course!).

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