>>11103991>>11104015Gladly. Alright, we'll assume the rationals. Define a dedekind cut in the rationals to be a nonempty set which has an upper bound, and for any r in the set, if s < r, then s is in the set.
One can easily define an (order, field, topological, etc) isomorphism between the set of dedekind cuts with an obvious order, field structure, order topology, and the typical axiomatic definition of R as you might know it. find this in any introductory analysis text.
What interests us, rather, is the cut which contains 0.99999...99 for any number of 9's. Such a cut exists as the union of all cuts containing numbers less than 0.999...9 for some number of 9's. We aim to prove that this union cut is precisely the cut which corresponds to 1, i.e. {r in Q : r < 1}. This will prove the equivalence of 0.999... and 1 in R under this definition. By equivalence of this definition to any other (axiomatic, cauchy sequences in Q, decimal expansions) we'll have our result.
well we just have to prove 2 sets are equal! obviously the cut for 0.999..9 for some number of 9's is contained in the cut <1, so their union is contained. now let 0 < r < 1. we want to show that r < 0.99999...9 for some number of 9's. indeed, consider the number with floor(log_10(1 - r))-1 9's. what is this number? it is 1 - 10^(floor(log_10(1 - r))-1).
since floor(log_10(1 - r))-1 < log_10(1 - r),
1 - 10^(floor(log_10(1 - r))-1) > 1 - 10^(log_10(1 - r)) = 1 - (1 - r) = r.
thus r is contained in this cut.
this shows that the the cut <1 is contained in the union of the cuts <0.99...9. thus the cut <1 and the cut containing all 0.999...9s are equal, and the number 0.999.... is 1.
