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

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- Oct 9th 2008, 02:56 PMArythArbitrary Unions
Let $\displaystyle \Omega = \{A_n:n= 1,2,3,...\}$ where $\displaystyle A_n = \left[\frac{1}{n}, 1 + \frac{1}{2n}\right)$. Find $\displaystyle \bigcup \Omega$

- Oct 9th 2008, 02:59 PMJhevon
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 PMPlato
- Oct 9th 2008, 06:47 PMAryth
Since the sequence is continuously getting smaller, wouldn't the union just be $\displaystyle A_1$?

- Oct 9th 2008, 07:24 PMJhevon
- Oct 9th 2008, 07:37 PMAryth
Isn't the left endpoint getting smaller as k gets bigger?

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

All intervals after $\displaystyle A_1$ fit inside of $\displaystyle A_1$... 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 PMJhevon
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 AMAryth
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 PMPlato
- Oct 15th 2008, 12:57 PMAryth
Ah, thanks. I get it now.