# Proving $\bigcup A - \bigcup B \subset \bigcup(A-B)$

I need to prove:

1. $\bigcup\limits_{i} A - \bigcup\limits_j B \subset \bigcup\limits_j (A-B)$
2. $\bigcap\limits_{i} A - \bigcap\limits_j B \subset \bigcup\limits_j (A-B)$

So when can the equality hold?

• You can use $\TeX$ on this site by enclosing formulas in dollar signs; single dollar signs for inline formulas and double dollar signs for displayed equations. You can see the source code for any math formatting you see on this site by right-clicking on it and selecting "Show Math As:TeX Commands". Here's a basic tutorial and quick reference. In the present case, the commands you need are \cup for $\cup$, \cap for $\cap$ and \subset for $\subset$. There's an "edit" link under the question. – joriki Sep 15 '12 at 5:28

1. Consider an $x\in\cup A_i-\cup B_j$. So $x\in\cup A_i$, but $x\notin \cup B_j$. Since $x\in\cup A_i$, $x$ is a member of some $A_i$, say $x\in A_5$. Since $x\notin\cup B_j$, $x$ is not a member of any $B_j$. In particular, $x\notin B_5$. So $x\in A_5-B_5$, but $A_5-B_5\subseteq\cup(A_i-B_i)$.

2. A similar argument would work here.

It’s a little hard to tell from your notation, but I’m guessing that you have families $\{A_i:i\in I\}$ and $\{B_i:i\in I\}$ of sets and want to show that $$\bigcup_{i\in I}A_i\setminus\bigcup_{i\in I}B_i\subseteq\bigcup_{i\in I}(A_i\setminus B_i)\tag{1}$$ and $$\bigcap_{i\in I}A_i\setminus\bigcap_{i\in I}B_i\subseteq\bigcup_{i\in I}(A_i\setminus B_i)\;.\tag{2}$$

Both $(1)$ and $(2)$ can be proved by ‘element-chasing’: pick an arbitrary member of the lefthand side, and show that it must belong to the righthand side.

For $(1)$, for instance, suppose that $x\in\bigcup_{i\in I}A_i\setminus\bigcup_{i\in I}B_i$. Then $x\in\bigcup_{i\in I}A_i$, so there is an $i_0\in I$ such that $x\in A_{i_0}$. You also know that $x\notin\bigcup_{i\in I}B_i$, so $x\notin B_{i_0}$. What does this tell you about the relationship between $x$ and $A_{i_0}\setminus B_{i_0}$? Can you finish the proof of $(1)$ from here?

Try a similar argument for $(2)$. If $x\in\bigcap_{i\in I}A_i\setminus\bigcap_{i\in I}B_i$, then $x$ is in every $A_i$, but $x$ is not in every $B_i$. (Why?) Use this to show that there must be at least one $i\in I$ such that $x\in A_i\setminus B_i$.

When agreement on writing union as sum, intersection as product and complement-with-respect-to-total-set as overline, the total set as $1$ and the empty set as $0$, then $$\nonumber \big(\bigcup_i a_i - \bigcup_i b_i \big)= \big(\sum_i a_i \big)\overline{\big(\sum_i b_i \big)} = \sum_i a_i \big(\prod_j \bar b_j\big)$$ As $s s = s$ (and $s+s=s$) this equals $$\nonumber \sum_i a_i \bar b_i \big( \prod_j \bar b_j \big)= \big( \sum_i a_i \bar b_i\big)\big( \prod_j \bar b_j\big).$$ Adding the disjoint expression $$\nonumber \big( \sum_i a_i \bar b_i\big) \big(\overline { \prod_j \bar b_j }\big)$$ and using the rule $xy+x\bar y = x(y+\bar y)=x1=x$) the result is $$\nonumber \sum_i a_i \bar b_i.$$ In caps and cups it says $$\nonumber \bigcup_i a_i - \bigcup_i b_i \subseteq \bigcup_i (a_i - b_i).$$ The condition for equality is that what is added is zero,
$$\nonumber \big( \sum_i a_i \bar b_i\big) \big(\overline { \prod_j \bar b_j }\big)=0.$$ Using that $x \bar y=0$ gives $x \bar y + x y = xy \leq y$ gives $$\nonumber \sum_i a_i \bar b_i \leq \prod_j \bar b_j= \overline{\sum b_j}.$$ Written with caps and cups it says $$\nonumber \bigcup_i( a_i- b_i) \subseteq \big(\bigcup b_i\big)^c.$$ The other equation: $$\nonumber \big(\bigcap_i a_i - \bigcap_j a_j\big) = \big( \prod_i a_i\big)\big( \overline{\prod_j b_j}\big)= \big(\prod_i a_i\big)\big(\sum_j \bar b_j\big)=\big(\sum_j \bar b_j \big) \big( \prod_i a_i \big)= \big(\sum_i \bar b_i a_i\big) \big( \prod_j a_j \big)\qquad .$$ If one adds $$\nonumber \big(\sum_i \bar b_i a_i \big)\big(\overline{\prod_j a_j}\big),$$ one gets the right hand side $$\nonumber \sum_i a_i \bar b_i.$$ The added part is disjoint, and the result is bigger, whith equality only if added part is zero, $$\nonumber \big(\sum_i \bar b_i a_i \big)\big(\overline{\prod_j a_j}\big)=0$$ With x and y the ab-sum and a-product, and from $x = x(y+\bar y)=xy + x\bar y = xy \leq y$ the result is $$\nonumber \sum_i a_i \bar b_i \leq \big(\prod_j a_j\big),$$ or in cap-cup, $$\nonumber \bigcup (a_i - b_i) \subseteq \bigcap_i a_i.$$