Re: Replacement of Cardinality

Am 13.08.2024 um 21:25 schrieb Chris M. Thomasson:
 

It depends on context. Are we going from 1/1 all the way down, or from 0 all the way up to 1/1.

 Again, 1/1 is the first term in the sequence (1/1, 1/2, 1/3, ...), but it's the largest/last unit fraction in respect to < (as defined on IR, and hence on the unit fractions).
 ... < 1/3 < 1/2 < 1/1.
 

0 is not a unit fraction, which means there is no smallest one... Fair enough?

 Right. There is no smallest unit fraction (in respect to < as defined on the IR).
 Proof: If u is a unit fraction, 1/(1/u + 1) is a smaller one.
 Using symbols: Au e {1/n : n e IN}: 1/(1/u + 1) e {1/n : n e IN} & 1/(1/u + 1) < u.
 Or with the definition:
            1/IN := {1/n : n e IN}
just:
           Au e 1/IN: 1/(1/u + 1) e 1/IN & 1/(1/u + 1) < u.
 This implies:
            Au e 1/IN: Eu' e 1/IN: u' < u.
 "For each and every unit fraction, there is a smaller one."