In any case, here's my logic, take an expression 0.x1x2x3...xn... where xn is a digit from the decimal system, 0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9. Make a set with all the possible expressions. Two expressions 0.x1x2x3...xn... and 0.y1y2y3...yn... differ if there is at least one digit that differs, or if there exists an xn such that xn != yn. Find the cardinality of the set. Which is the continuum. Which means that there is a bijection between this set and the real numbers. Which would mean that there is no such thing as a real number having multiple decimal representations.
Only if the decimal expansion is finite.
By definition, the decimal expansion of some number
r is
Where {a1, a2, a3, ...,} is some sequence. Notice that the set {a1, a2, a3, ...} is necessarily infinite. (Hence the number 1 can also be expanded as 1.000000...).
The problem is that infinite sums are
not actual sums.
They are the limit of the sequence of partial sums -- in this case {a1/10, a1/10 + a2/100, a1/10 + a2/100 + a3/1000, ...}.
So you can't just compare each of your xn's and yn's.
It's totally legitimate that the sequences {a1, a2, a3, ...} and {b1, b2, b3, ...} might not be equal, but the
limits of the sequence of their partial sums are the same. In this case the set of an's and bn's are, {1, 0.0, 0.00, 0.000, ...} and {0, .9, .09, .009, ...}.
If you must provide a proof via set theory, I'd probably go after some argument using the suprema and infima of the respective sets of partial sums. That's really just an analysis argument, though.
Also, I'm not exactly sure what you mean by "the cardinality of the set." Which set are you talking about? R, or your sequences {x1/10, x2/100, ... xn/10^n, ...} and {y1/10, ... yn/10^n, ...}?
