how to prove original intervals and canonical form of intervals have the same interval graph

Question

According to this paper,a family of intervals is said to be canonical if the coordinates of the endpoints of the intervals are distinct integers between 1 and 2n where n is the number of intervals.

Here is the example Original intervals = {[2,5], [3,11], [6,9], [7,13], [8,18], [12,16], [15,19]}

Canonical form of intervals = {[1,3], [2,8], [4,7], [5,10], [6,13], [9,12], [11,14]}

These original intervals and canonical form of the original intervals give the same interval graph, but how can I prove that it is always the case mathematically ?

Edit: This is the method I use to create canonical intervals: I sort the intervals according to their starting points in ascending order Then, I assign the first interval's [2,5] starting point as 1. Then, I compare it with the following intervals to see the overlappings. Then, I arrange the intervals accordingly.

I repeat this process for all intervals.

Examples:

[2,5] contains only the starting point of interval 2, so

[2,5] becomes [1, (start of interval 2), 3] = [1,3]

[3,11] contains, start of interval 3, start of interval 4, end of interval 3, and start of interval 5, so

[3,11] becomes [2, 3, (start of interval 3), (start of interval 4), (end of interval 3),(start of interval 5),8] = [2,8]

note: interval 2 can not start here because at point 3 because it is where interval 1 ends. All start and end points should be distinct.

Answer 1

You haven't described how a system of intervals are converted to canonical form, but here is one way to do it. We find an order-preserving mapping (satisfies $f(x) < f(y)$ iff $x < y$) from the coordinates in the original system to the integers $\{1,\ldots,2n\}$ (we do this by sorting all of the original coordinates). We get the same interval graph since the mapping is order-preserving – that's a nice exercise.