Re: Cauchy completeness of Cauchy reals

[Dusko Pavlovic <dusko@kestrel.edu>]

Andrej Bauer wrote:

>There seems to be an open question in regard to this, advertised by
>Alex Simpson and Martin Escardo: find a topos in which Cauchy reals
>are not Cauchy complete (i.e., not every Cauchy sequence of reals has
>a limit). For extra credit, make it so that the Cauchy completion of
>Cauchy reals is strictly smaller than the Dedekind reals.
>
this will give me negative credits one way or another, but here it goes:

let a = (a_i) be a cauchy sequence of rationals between 0 and 1. (cauchy
means |a_m - a_n| < 1/m + 1/n, as in andrej's message.)

let b = (b_i) be the subsequence b_i = a_{2^i+3}. b and a are equivalent
in the sense from the message, because

     |a_m - b_n| < 1/m + 1/(2^n+3) < 2/m + 2/n

note that

     |b_i - b_{i+k}| < 1/(2^i+3) + 1/(2^(i+k)+3) < 1/(2^i+2)

now define x_i to be the simplest dyadic that falls in the interval
between b_i - 1/2^(i+2) and b_i + 1/2^(i+2). in other words, to get x_i,
begin adding 1/2 + 1/4 + 1/8... until you overshoot b_i - 1/2^(1+2). if
you also overshoot b_i + 1/2^(1+2), and the last summand was 1/2^k, skip
it, and try adding 1/2^(k+1), etc. one of them must fall in between. a
less childish way to say this is that x_i is the shortest irredundant
binary (no infinite sequences of 1) such that

     |b_i - x_i| < 1/2^(i+2)

so x =  (x_i) is a cauchy sequence equivalent to b, with

     |x_i - x_{i+k}| <= |x_i-b_i| + | b_i - b_{i+k}| + |b_{i+k} - x_{i+k}| <
                     <  1/2^(i+2) + 1/2^(i+2)        + 1/2^(i+k+2) <
                     <  1/2^i

this means that the first i digits of x_i and x_{i+k} coincide.

now let X be the binary number such that its first i digits are the same
as in x_i, for every i. (if it ends on an infinte sequence of 1s,
replace it by the corresponding irredundant representative.) this is
clearly a constant cauchy sequence equivalent to x, so it is its limit.
since x is equivalent to b

    |b_m - x_n| <= |b_m - b_n| + |b_n - x_n| < 1/2^(m+3) + 1/2^(n+3) +
1/2^(n+2) < 2/m + 2/n

and b is equivalent to a, X is also the limit of a.

is there an error in the above reasoning? i can't find it. on the other
hand, i printed out the paper by alex and martin, and the conjecture is
stated rather strongly, so i guess i must be missing something.

-- dusko