On Fri, Nov 16, 2012 at 11:57 PM, David Wilson <davidwwilson@comcast.net> wrote:
It occurred to me that the standard proof of 0.9999... = 1 depends on the continuity of the reals numbers, which I believe the surreals lack.
I don't know what you mean by "the continuity of the reals"; do you mean completeness? I know what it means for a function to be continuous, but not a set. But I don't understand why you think the proof that .9999... = 1 requires continuity. It seems to me that if we write this as a two-way-infinite series, ... + 0 + 0 + 0 + .9 + .09 + .009 + ..., with a_1 = .9 Then all we need to assume about such sums to prove that if it has a sum, that sum must be 1 is exactly the three properties you describe below:
If m in R and b_i = m a_i for all i in Z, then sum(b) = m sum(a). If c_i = a_i + b_i for all i in Z, then sum(c) = sum(a) + sum(b).
We would want sums to be preserved by shifting the sequence left or right, e.g:
If k in Z and b_i = a_{i+k} for all i in Z, then sum(b) = sum(a).
If the sum is S, the first property tells us that the sum of ... + 0 + 0 + 0 + 9 + .9 + .09 +... with a_1 = 9, is 10 * S The third property says that the sum of this is still 10 * S when we shift it over by 1, so that a_0 = 9 and a_1 = .9. Now the second property (subtracting rather than adding, but we can use the first property to multiply the first sequence by -1) says that the sum of the sequence ...0 + 0 + 0 + 9 + 0 + 0 + 0 + ... So any notion of summability that assigns the sum 9 to this series, and satisfies your three properties, must assign the sum 0 to the sequence ... 0 + 0 + 0 + .9 + .09 + .009 + ....., with no assumption of "continuity" needed. Andy