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From: Moebius <invalid@example.invalid>
Newsgroups: sci.logic,sci.math
Subject: Re: Replacement of Cardinality
Date: Tue, 13 Aug 2024 22:29:07 +0200
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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."