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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?

Bob Shannon
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    You're going in the opposite direction for intersections, Bob; intersections become smaller; notice that , e.g., $2$ is not in $(0,1)$, same for $3$, etc. – DBFdalwayse Oct 31 '13 at 03:01
  • I see. So would the intersection be $A_n = (0,n),(0,n+1),(0,n+2),...$, where n starts at 1 and goes to $\infty$? If so I'm not sure how to write it as a single set. – Bob Shannon Oct 31 '13 at 03:03
  • Can you see that $(0,1)$ is in every set, and it is the only interval that is in every other set? Usually you need to give some argument to this effect. – DBFdalwayse Oct 31 '13 at 03:05
  • Ah. Yes, I recognize that $(0,1)$ would be in every set. But so would $(0,2)$, and $(0,3)$...? – Bob Shannon Oct 31 '13 at 03:07
  • I'm interpreting it as a single set with members ${(0,1),(0,2),(0,3),...}$ – Bob Shannon Oct 31 '13 at 03:07
  • No; how is $(0,2)$ contained in $(0,1)$? You're looking for the collection of elements that are in every set. No point in $(1,2)$ as a subset of $(0,2)$ is contained in $(0,1)$. – DBFdalwayse Oct 31 '13 at 03:08
  • I see. So wouldn't the set $(0,Z^+)$ be the intersection? – Bob Shannon Oct 31 '13 at 03:10
  • Or if that notation doesn't work where you use $Z^+$, it seems to me that it would be $0$ instead. – Bob Shannon Oct 31 '13 at 03:11
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    Let me post as an answer, to avoid too many comments, and then let me know if you have a followup. – DBFdalwayse Oct 31 '13 at 03:11

2 Answers2

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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 $\cup_{i=1}^{\infty}(0,i) =(0,\infty)$.

Worawit Tepsan
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The intersection of a collection of sets consists of the elements that are in every one of the sets. One thing you may see is that the intersection of sets is not larger than the smallest of the sets ( when smallness applies, like in this case).

In this case, you want to find all Reals that are in each of :

$(0,1), (0,2),...., (0,n), (0, n+1),.....$

Now, can you see that, e.g., $2$, or $3$ will not be in every one of the sets?

Notice too, that these sets are nested, i.e., $(0,1)\subset (0,2) \subset (0,3) \subset....$

Does that help you see what the intersection set is ?

The union of sets , will contain all the elements that are in at least one of the sets, basically the elements that are in either $(0,1)$ or in $(0,2)$, or.....

DBFdalwayse
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