How to approach this set proof question (Iterative Union/Intersection)

Question

I'm writing today to ask for a heading/starting point in how I might approach this problem proposed to me involving Set Theory.

From my understanding, this proof is asking us to show that the set containing all the elements in A minus the set of all elements common to the sets B1,B2,...,Bn is equal to the set containing all the elements of A not in a particular set on Bn. I can see this making sense, as it's saying there's a set of values in B1, B2, Bn that are not in at least one of the iterations of the Union of (A-Bi), however I am very stuck on how to begin formally proving this using Set Theory. Can someone give me a direction/heading for this? I'm familiar with calculus and more number theory proofs but I'm completely lost on how to begin such a proof like this.

Answer 1

Let $x\in A-\bigcap_{i=1}^nB_i$.

Then $x\in A$ and not $x\in \bigcap_{i=1}^nB_i$ or equivalently some $i_0\in\{1,\dots,n\}$ exists with $x\notin B_{i_0}$.

Then we have $x\in A-B_{i_0}$ and this set is evidently a subset of $\bigcup_{i=1}^n(A-B_i)$ so that also $x\in\bigcup_{i=1}^n(A-B_i)$.

This proves: $$A-\bigcap_{i=1}^nB_i\subseteq\bigcup_{i=1}^n(A-B_i)$$

Give the other side a try yourself.

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edit (the other way).

Let $x\in\bigcup_{i=1}^n(A-B_i)$.

Then $x\in A-B_{i_0}$ for some $i_0\in\{1,\dots,n\}$.

Then $x\in A$ and $x\notin B_{i_0}$.

Then it cannot be true that $x\in\bigcap_{i=1}^nB_i$.

This because $\bigcap_{i=1}^nB_i\subseteq B_{i_0}$ so that $x\in\bigcap_{i=1}^nB_i$ leads to $x\in B_{i_0}$ which is not true.

We conclude that $$x\in A-\bigcap_{i=1}^nB_i$$

Answer 2

Use the fact that $A - B = A \cap B^{\complement}$, and then just typical algebra rules:

So:

$$A - \bigcap_{i=1}^n B_i = A \cap \big( \bigcap_{i=1}^n B_i \big)^{\complement} \overset{DeMorgan}=A \cap \big( \bigcup_{i=1}^n B_i^{\complement} \big)\overset{Distribution}=\bigcup_{i=1}^n \big( A \cap B_i^{\complement} \big)= \bigcup_{i=1}^n \big( A - B_i \big)$$