Interactive Real Analysis
- part of Examples 7.1.2(c):
Show that if P' is a refinement of P
then | P' | 
| P |
Back
To prove this fact is more confusing than enlightening. It seems
clear that if one or more points are inserted into the partition
P to form the refinement partition P',
the largest distance between the points of P' must
now be less than (or equal to) that of the points of P.
| P |
BackBut alas, even things that "seem clear" still need formal proof, so ...
Since | P | is a maximum, there must be at least one integer j such that | P | = xj+1 - xj. Take all such points from the partition P, i.e. all points such that | P | = xj+1 - xj. Now consider the refinement P'.
- Suppose none of the additional points are inside the intervals
[xj, xj+1]. Then the original
maximum has not changed so that | P | = | P' |
- Suppose at least one of the additional points is inside at
least one of the subintervals
[xj, xj+1]. Then this subinterval
can no longer contribute to the maximum of P' so that
| P' | | P |