0
$\begingroup$

How to properly prove the following:

For all integers positive integers n, if A1, A2,... and B are sets, then enter image description here

$\endgroup$
3
$\begingroup$

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 $x \notin B$;
  • $x \in (\cap_{\alpha \in I}A_{\alpha}) \setminus B$.
$\endgroup$
1
$\begingroup$

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 ... ?

$\endgroup$
1
$\begingroup$

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 $(\cap_{i=1}^{n} A_i)-B \subseteq \cap_{i=1}^{n} (A_i-B)$ ?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.