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Sum of lengths of a finite number of overlapping segments > length of their union.

  1. Nov 20, 2009 #1
    1. The problem statement, all variables and given/known data

    I know this is probably fairly trivial, but for the life of me I cannot remember or reconstruct the proof for the proposition, "The sum of the lengths of a finite number of overlapping open intervals is greater than the length of their union."

    2. Relevant equations

    Not Applicable.

    3. The attempt at a solution

    Suppose [tex]G=\left\{\left(a_{i},b_{i}\right)\right\}^{n}_{i=1}[/tex] is a finite collection of overlapping open intervals (any two intervals in [tex]G[/tex] have nonempty intersection). We wish to show:

    [tex]b_{n}-a_{1}\leq\sum^{n}_{i=1}\left(b_{i}-a_{i}\right)[/tex].

    My current approach is the following:

    [tex]\sum^{n}_{i=1}\left(b_{i}-a_{i}\right)=b_{n}-a_{1}+\left(b_{n-1}+b_{n-2}+\ldots+b_{1}\right)-\left(a_{n}+a_{n-1}+\ldots+a_{2}\right)[/tex]

    It remains to show that [tex]\left(b_{n-1}+b_{n-2}+\ldots+b_{1}\right)-\left(a_{n}+a_{n-1}+\ldots+a_{2}\right)[/tex] is a positive number.

    Well, we know [tex]\left(b_{n-1}+b_{n-2}+\ldots+b_{2}\right)-\left(a_{n-1}+a_{n-2}+\ldots+a_{2}\right)[/tex] is a positive number, so we just need to establish that [tex]b_{1}-a_{n}<\left(b_{n-1}+b_{n-2}+\ldots+b_{2}\right)-\left(a_{n-1}+a_{n-2}+\ldots+a_{2}\right)[/tex].

    This is where I am having problems.
     
  2. jcsd
  3. Nov 20, 2009 #2
    Re: Sum of lengths of a finite number of overlapping segments > length of their union

    [tex] b_n - a_n + b_{n-1} - a_{n-1} + ... = b_n +(b_{n-1}-a_n) + (b_{n-2} - a_{n-1}) +... -a_1 [/tex]

    and since the intervals overlap, you can sort them in such a way that [tex]b_{k-1} > a_k [/tex] for all k.
     
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