Proof of $(\bigcup\limits_{i\in I} A_{i})\bigcap B =\bigcup\limits_{i\in I}(A_{i}\bigcap B)$ 
Let $A_{i}(i\in I$), $B$ are subsets of $X$  Prove that:
  $(\bigcup\limits_{i\in I} A_{i})\bigcap B =\bigcup\limits_{i\in
 I}(A_{i}\bigcap B)$

I was trying to prove it by using the method of mathematical induction.
For i=1 the equation looks like 
$(\bigcup\ A_{1})\bigcap B =\bigcup\ (A_{1}\bigcap B)= A_{1}\bigcap B$
Now need to prove equation for I=n+1, considering that it is true for I=n
$(\bigcup\limits_{i\in I=n+1} A_{i})\bigcap B =((\bigcup\limits_{i\in I=n} A_{i})\bigcap B) \bigcup(A_{n+1} \bigcap B) $  using distribution law
$((\bigcup\limits_{i\in I=n} A_{i})\bigcap B) \bigcup(A_{n+1} \bigcap B) = (\bigcup\limits_{i\in I=n}(A_{i}\bigcap B))\bigcup(A_{n+1} \bigcap B) = \bigcup\limits_{i\in I=n+1}(A_{i}\bigcap B)$
Is everything ok in this proof?
 A: Unfortunately, this proof doesn't quite work; you can't use induction on $I$, because $I$ may have infinite size (whereas induction would only work for $I$ having finite cardinality).
Rather than using induction, note that every element in the RHS is in $B$ and also in some $A_i$, and hence is an element of the LHS. Similarly, any element in the LHS must be in $B$ and in the union of all $A_i$, the latter of which means that it's in some $A_i$, and hence is an element of the RHS.
So both sides are subsets of each other, and so they're equal.
A: Your proof is not O.K, since $I$ is an arbitrary set (an index-set).


*

*Let $x \in (\bigcup\limits_{i\in I} A_{i})\cap B$, then there is $i_0 \in I$ such that $x \in A_{i_0} \cap B$ This gives $x \in \bigcup\limits_{i\in
 I}(A_{i}\cap B).$ Thus we have shown that 


$$(\bigcup\limits_{i\in I} A_{i})\cap B \subseteq \bigcup\limits_{i\in
 I}(A_{i}\cap B).$$ 


*It is now your turn to show that 


$$(\bigcup\limits_{i\in I} A_{i})\cap B \supseteq \bigcup\limits_{i\in
 I}(A_{i}\cap B).$$ 
