# Set Theory: General Intersection

How to properly prove the following:

For all integers positive integers n, if A1, A2,... and B are sets, then More is true. You do not have to intersect over countable index. For the proof, note that the following are equivalent.

• $x \in \cap_{\alpha \in I} (A_{\alpha} \setminus B)$;
• $x \in A_{\alpha} \setminus B$ for all $\alpha \in I$;
• $x \in A_{\alpha}$ for all $\alpha \in I$ and $x \notin B$;
• $x \in \cap_{\alpha \in I}A_{\alpha}$ and $x \notin B$;
• $x \in (\cap_{\alpha \in I}A_{\alpha}) \setminus B$.

Just chase elements: show that

$$\bigcap_{i=1}^n(A_i\setminus B)\subseteq\left(\bigcap_{i=1}^nA_i\right)\setminus B$$

by showing that if $x\in\bigcap_{i=1}^n(A_i\setminus B)$, then $x\in\left(\bigcap_{i=1}^nA_i\right)\setminus B$, and then show the opposite inclusion in the same way. I’ll get you started. If $x\in\bigcap_{i=1}^n(A_i\setminus B)$, then by definition $x\in A_i\setminus B$ for each $i\in\{1,2,\dots,n\}$. If $x\in A_i\setminus B$, then $x\in A_i$ and $x\notin B$. Thus, $x\in A_i$ for each $i\in\{1,2,\dots,n\}$. By definition this implies that $x\in\bigcup_{i=1}^nA_i$. You also know that $x\notin B$, so ... ?

A formal proof would require showing both sets are subsets of each other. Consider the set $\cap_{i=1}^{n} (A_i-B)$. This set is equal to the set:

$S=\{\alpha|\alpha \in [(A_1-B)\cap (A_2-B) \cap ... \cap (A_n-B)]\}$

Now consider any two sets $(A_i-B)$ and $(A_j-B)$ for $1 \le i,j \le n$ and $i \ne j$. Then I claim that $(A_i-B) \cap (A_j-B)=(A_i \cap A_j)-B$.

This easily follows since for any element x in $(A_i-B) \cap (A_j-B)$, this implies that $(x\in A_i \wedge x \not \in B) \wedge (x\in A_j \wedge x \not \in B)$. Thus $((x \in A_i \wedge x \in A_j) \wedge x \not \in B)$. This is exactly the set $(A_i \cap A_j)-B$ by definition. It is easy to see using a similar argument (very similar) that reverse implication holds.

We can use the claim inductively to show that $\cap_{i=1}^{n} (A_i-B) \subseteq (\cap_{i=1}^{n} A_i)-B$. Can you see how to prove that $(\cap_{i=1}^{n} A_i)-B \subseteq \cap_{i=1}^{n} (A_i-B)$ ?