# Arbitrary Unions

• Oct 9th 2008, 02:56 PM
Aryth
Arbitrary Unions
Let $\Omega = \{A_n:n= 1,2,3,...\}$ where $A_n = \left[\frac{1}{n}, 1 + \frac{1}{2n}\right)$. Find $\bigcup \Omega$
• Oct 9th 2008, 02:59 PM
Jhevon
Quote:

Originally Posted by Aryth
Let $\Omega = \{A_n:n= 1,2,3,...\}$ where $A_n = \left[\frac{1}{n}, 1 + \frac{1}{2n}\right)$. Find $\bigcup \Omega$

hint: graph the first element on a number line, then the second, then the third. you should of course be seeing a pattern by then.

doing it without graphing is not much harder. what happens to 1/n for various values of n? what about 1 + 1/(2n)? how do these change when n gets large?
• Oct 9th 2008, 03:29 PM
Plato
Quote:

Originally Posted by Aryth
Let $\Omega = \{A_n:n= 1,2,3,...\}$ where $A_n = \left[\frac{1}{n}, 1 + \frac{1}{2n}\right)$. Find $\bigcup \Omega$

Looking at these $A_1 = \left[ {1,\frac{3}
{2}} \right)\,,\,A_2 = \left[ {\frac{1}
{2},\frac{5}
{4}} \right) \cdots A_k = \left[ {\frac{1}
{k},\frac{{2k + 1}}
{{2k}}} \right)$
, what is the union?
• Oct 9th 2008, 06:47 PM
Aryth
Since the sequence is continuously getting smaller, wouldn't the union just be $A_1$?
• Oct 9th 2008, 07:24 PM
Jhevon
Quote:

Originally Posted by Aryth
Since the sequence is continuously getting smaller, wouldn't the union just be $A_1$?

no

look at where the left endpoint is going

remember, we're looking at intervals here
• Oct 9th 2008, 07:37 PM
Aryth
Isn't the left endpoint getting smaller as k gets bigger?

1, 1/2, 1/3, 1/4, ... , 1/k

All intervals after $A_1$ fit inside of $A_1$... So, the arbitrary union of all the intervals would then be the one interval that contains them all... Or, that's how my professor made it seem anyway...
• Oct 9th 2008, 07:39 PM
Jhevon
Quote:

Originally Posted by Aryth
Isn't the left endpoint getting smaller as k gets bigger?

1, 1/2, 1/3, 1/4, ... , 1/k

All intervals after $A_1$ fit inside of $A_1$... So, the arbitrary union of all the intervals would then be the one interval that contains them all... Or, that's how my professor made it seem anyway...

as i said, you seem to be having trouble visualizing this without drawing it. draw a picture. A_1 is the interval [1, 3/2), A_2 is [1/2, 5/4), notice that the second interval is outside of the scope of the first. A_2 contains the points on [1/2, 1) which are NOT in A_1, and so, A_1 cannot represent the union, because we can see that A_2 has points that are not in A_1

getting smaller means the intervals are "expanding" to the left. the left endpoint keeps getting farther from 1
• Oct 15th 2008, 11:59 AM
Aryth
I have tried for awhile to figure this out. But what my professor said isn't enough to solve this problem. So if you could solve this one for me so I can see the method, that would be a good help.

Thanks.
• Oct 15th 2008, 12:35 PM
Plato
Quote:

Originally Posted by Aryth
Let $\Omega = \{A_n:n= 1,2,3,...\}$ where $A_n = \left[\frac{1}{n}, 1 + \frac{1}{2n}\right)$. Find $\bigcup \Omega$

$A_n = \left[\frac{1}{n}, 1 + \frac{1}{2n}\right)$.
$\bigcup\limits_{n } {A_n } = \left( {0,\frac{3}
{2}} \right)$
• Oct 15th 2008, 12:57 PM
Aryth
Ah, thanks. I get it now.