Let where . Find

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- October 9th 2008, 02:56 PMArythArbitrary Unions
Let where . Find

- October 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? - October 9th 2008, 03:29 PMPlato
- October 9th 2008, 06:47 PMAryth
Since the sequence is continuously getting smaller, wouldn't the union just be ?

- October 9th 2008, 07:24 PMJhevon
- October 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 fit inside of ... 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... - October 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 - October 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. - October 15th 2008, 12:35 PMPlato
- October 15th 2008, 12:57 PMAryth
Ah, thanks. I get it now.