Prove that $B \setminus (\bigcup_{i \in I} A_i) = \bigcap_{i \in I} B \setminus A_i$. This is an exercise from Velleman's "How To Prove It". I saw a similar question asked on here, but I am still confused. Also, those questions were closed for being off-topic for some reason.


  
*Suppose $B$ is a set, $\{A_i | i \in I\}$ is an indexed family of sets, and $I \neq \emptyset $. 

  
*b. Prove that $B \setminus (\bigcup_{i \in I} A_i) = \bigcap_{i \in I} B \setminus A_i$.
  
  

Here is my attempt at some of the proof. I have indicated the part where I get stuck:
Proof: Let $x$ be arbitrary. Suppose $x \in B \setminus (\bigcup_{i \in I} A_i)$. Then $x \in B$, and for all $i \in I$, $x \notin A_i$. Let $j \in I$ be arbitrary. It follows that $x \notin A_j$. Thus, $x \in B \setminus A_j$. Since $j$ was arbitrary, $x \in \bigcap_{j \in I} B \setminus A_j$.
Now suppose  $x \in \bigcap_{i \in I} B \setminus A_i$. So $\forall i \in I (x \in B)$ and $\forall i \in I(x \notin A_i)$. Suppose $x \in \bigcup_{i \in I} A_i $. Then we can choose a $j \in I$ such that $x \in A_j$. But since $j \in I$, it follows that $x \notin A_j$, which is a contradiction. Thus, $x \notin \bigcup_{i \in I} A_i$. [How do we show that $x \in B$ to complete the proof?] $\square$
Proving the statement with a string of equivalences kind of makes sense:
\begin{align}
x \in B \setminus (\bigcup_{i \in I} A_i) &\leftrightarrow x \in B \wedge x \notin \bigcup_{i \in I} A_i \\
& \leftrightarrow x \in B \wedge \neg (\exists i \in I (x \in A_i)) \\
& \leftrightarrow x \in B \wedge \forall i \in I(x \notin A_i) \\
& \leftrightarrow \forall i \in I(x \notin A_i \wedge x \in B ) & \text{since $x\in B$, of course $x \in B$ for all $i \in I$. But why the converse?}\\
& \leftrightarrow \forall i \in I(x \in B \setminus A_i)\\
& x\in \bigcap_{i \in I}B \setminus A_i \\
\end{align}
I do not understand how we can go from $\forall i \in I(x \in B)$ to $x \in B$, since if $x \in B$ is true, then that means $x \in B$ even for some $j \notin I$. I think I am misunderstanding some fundamental rules. Thanks in advance!
 A: As far as I see, your problem is to understand why you can infer $x \in B$ from 
\begin{align}\tag{1}
\forall i \in I \,(x \in B)
\end{align}
knowing that $I \neq \emptyset$.
Your question is legitimate because in $(1)$, $x \in B$ under the hypothesis $i \in I$ (while in the conclusion $x \in B$ there is no further hypothesis).
Indeed, a formally proper way to write $(1)$ is the following:
\begin{align}\tag{2}
\forall i \, (i \in I \to x \in B)
\end{align}
Intuitively, from $(2)$, or equivalently $(1)$, you can infer $x \in B$ (without any further hypothesis) because the statement $x \in B$ does not depend on $i$, since $i$ does not occur in $x$ or in the definition of $B$.
Hence, the hypothesis $i \in I$ does not play any role to conclude $x \in B$ and you can discard it. But you can do it provided that your hypothesis $i \in I$ is true, i.e. $I$ must be non-empty.
More formally, since $I$ is non-empty, there exists $i \in I$. According to $(2)$,  for such a $i$ we have $i \in I \to x \in B$. By modus ponens (since $i\in I$ and $i \in I \to x \in B$) you can conclude that $x \in B$.

Note that the hypothesis that $I$ is non-empty is crucial. If $I = \emptyset$ then $(2)$, or equivalently $(1)$, is vacuously true: since the hypothesis $i \in I$ is false, then the implication $i \in I \to x \in B$ is true regardless of $x \in B$ or $x \notin B$ (for every $i$ in the universe).
So, for $I = \emptyset$ you cannot conclude whether $x \in B$ or not.
As a consequence, when $I = \emptyset$, we have that $B \setminus (\bigcup_{i \in I} A_i) \neq \bigcap_{i \in I} B \setminus A_i$ (unless $B$ is the whole universe), because it can be easily shown that, for $I = \emptyset$, we have $B \setminus (\bigcup_{i \in I} A_i) = B$ while $\bigcap_{i \in I} B \setminus A_i$ is the whole universe.
