How to prove $B \backslash (\bigcup\limits_{i∈I} A_{i}) = \bigcup\limits_{i∈I}(B \backslash A_{i})$

Question

DISCLAIMER: This was a typo in the third edition of the ebook, the correct statement to prove is: $B \backslash (\bigcap\limits_{i∈I} A_{i}) = \bigcup\limits_{i∈I}(B \backslash A_{i})$

I have been trying to solve Velleman's exercises in his book "how to prove it". However I have hit a snag with this specific proof. I'm used to fully writing it out in predicate form and then proving it. I have managed to solve $B \backslash (\bigcup\limits_{i∈I} A_{i}) ⊆ \bigcup\limits_{i∈I}(B \backslash A_{i}))$ but I'm struggeling to proof it the other way around: $\bigcup\limits_{i∈I}(B \backslash A_{i})) ⊆ B \backslash (\bigcup\limits_{i∈I} A_{i})$.

There is also being given that I is not an empty set.

I have analyzed the givens as follows: $\exists{i}(i \in I \wedge (a \in B \wedge a \notin A_{i}))$

And what I need to prove: $\forall{i}(i \in I \implies a \notin A_i)$ and $a \in B$

But I just don't see any way to show this is indeed true, if possible please provide an answer in predicate style and a direct proof (if that's possible) since I'm trying to learn and understand it. But any help is welcome really.

Note while this may seem simple I have done multiple attempts and can't figure it out. So I'm wondering if it's even correct, rules I have learned so far are Universal instantiation, generalization, same with existential, modus ponens, modus tollens, implication, bijunction, conjunction, negation. I have not seen other rules yet, if it's true I feel like it must be using some rule I'm not aware of.

Answer 1

The statement as written is wrong. The right statement is $\bigcup_{i\in I}(B\setminus A_i)=B\setminus\bigcap_{i\in I}A_i$. To see that the statement is wrong, consider $B=\{1,2,3,4\},\,A_1=\{1,2\},A_2=\{3,4\}$. Then: $$\bigcup_{i\in I}(B\setminus A_i)=\{1,2,3,4\}$$ $$B\setminus\bigcup_{i\in I}A_i=\emptyset$$

Answer 2

Proof that $\bigcup\limits_{i∈I}\Big(B \backslash A_{i}\Big) ⊆ B \backslash \Big(\bigcap\limits_{i∈I} A_{i}\Big)$ using set identites below:

$ \begin{array}{llll} \bigcup\limits_{i∈I} \Big( B \backslash A_{i} \Big) & = \bigcup\limits_{i \in I} \Big( B \cap A_i^C \Big) & \text{by definition of set difference} \\ & = B \cap \Big( \bigcup\limits_{i \in I} A_i^C \Big) & \text{by distributive law for sets} \\ & = B \cap \Big( \bigcap\limits_{i \in I} A_i \Big)^C & \text{by DeMorgan's Law for sets} \\ & = B \backslash \Big( \bigcap\limits_{i \in I} A_i \Big) & \text{by definition of set difference} \\ \end{array} $

By definition of equality between sets, we know $\bigcup\limits_{i∈I}\Big(B \backslash A_{i}\Big) \subseteq B \backslash \Big(\bigcap\limits_{i∈I} A_{i}\Big)$ and $B \backslash \Big(\bigcap\limits_{i∈I} A_{i}\Big) \subseteq \bigcup\limits_{i∈I}\Big(B \backslash A_{i}\Big)$. Therefore, $\bigcup\limits_{i∈I}\Big(B \backslash A_{i}\Big) \subseteq B \backslash \Big(\bigcap\limits_{i∈I} A_{i}\Big)$.