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

[Esran]

Homework Statement

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."

Homework Equations

Not Applicable.

The Attempt at a Solution

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

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

My current approach is the following:

\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)

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

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

This is where I am having problems.

 

[clamtrox]

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

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