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Math Help - Arbitrary Unions

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    Super Member Aryth's Avatar
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    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
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Aryth View Post
    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?
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    Quote Originally Posted by Aryth View Post
    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}<br />
{2}} \right)\,,\,A_2  = \left[ {\frac{1}<br />
{2},\frac{5}<br />
{4}} \right) \cdots A_k  = \left[ {\frac{1}<br />
{k},\frac{{2k + 1}}<br />
{{2k}}} \right), what is the union?
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    Super Member Aryth's Avatar
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    Since the sequence is continuously getting smaller, wouldn't the union just be A_1?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Aryth View Post
    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
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    Super Member Aryth's Avatar
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    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...
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Aryth View Post
    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
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    Super Member Aryth's Avatar
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    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.
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    Quote Originally Posted by Aryth View Post
    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}<br />
{2}} \right)
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    Super Member Aryth's Avatar
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    Ah, thanks. I get it now.
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