The map sending an index a of the initial segment [N] to N+1-a is a bijection of [N] onto itself, and a finite sum is unchanged when its summands are taken in reverse order.
Let be a field. Let be the set of natural numbers, with addition as in that definition and with the order ; let , and let be the initial segment determined by . Sums below are the finite sums of ; the field plays no role in claim 1. Then the following hold.
1. (The reversing map)¶ For every there is exactly one with , and this lies in . Writing for it, the map satisfies for every and is a bijection. Moreover, if is an ordered field, with multiplicative identity and with abbreviating , and is its canonical map, then
2. (Reversal)¶ Let be a map with values written . Then
Loading…
Prerequisites
No prerequisites tracked.
Dependents
No dependents yet.
Dependent proofs
No dependent proofs yet.
No relations recorded yet.