I'm attempting a problem:
Find $\bigcap_{i=1}^\infty Ai$ and $\bigcup_{i=1}^\infty Ai$, if for every positive integer $i$ $Ai=(0,i)$, that is, the set of real numbers $x$ with $0<x<i$.
I first start with $\bigcap_{i=1}^\infty Ai$.
I know that when you expand it out, you'll get ${(0,1), (0,2), (0,3),…} $ which will keep continuing all the way to infinity. Since it is asking for the intersection of all the sets, it is asking for the set which has elements that are members of all the sets in the collection. Would the correct answer be $A_n = (0,Z^+) $, where $Z^+$ is the set of positive integers?
Now for $\bigcup_{i=1}^\infty Ai$.
This is asking for the set which has elements that are members of at least one set in the collection. Since $i$ goes all the way up to infinity, isn't there an infinite possible amount of answers here? For example, one answer could be $(0,1)=A_{n_1}$. Or it could even be $A_{n_2} = (0,Z^+) $?
I think I have this right and am going about correctly. Or am I?
Best Answer
Definition: $A\cap B =\{x: x\in A \text{ and } x\in B\}$. That means any elements that belongs to $A\cap B$ must be in both $A$ and $B$. So
$\cap_{i=1}^{n}(0,i) =(0,1)$. As $n\to \infty$, we have $\cap_{i=1}^{\infty}(0,i) =(0,1)$.
On the other hand, $A\cup B =\{x: x\in A \text{ or } x\in B\}$. That means any elements that belongs to $A\cup B$ must be in either $A$ or $B$. So
$\cap_{i=1}^{n}(0,i) =(0,n)$. As $n\to \infty$, we have $\cup_{i=1}^{\infty}(0,i) =(0,\infty)$.