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by i2go
563 days ago
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did not read the whole document. just the proof of theorem 1 and it has a minor error. it is not obvious that C(n) != x. that is because it is not true. to see that, consider a map C(n) such that C(n) = 0.10000000... that is: c_1(n) = 1 and c_i(n) = 0 for all other i. If you let \overline{c}_1(n) = 0 and c_i(n) = 9 for all other i,
then we have that c_i(n) != \overline{c}(c)_i(n) for all i but C(n) = x. This is a common mistake people make. To fix it, one needs to be more careful when defining \overline{c}_i. |
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I think the usual way to fix it is actually to be more careful when defining C(n) - that is, to specify that the value of C(n) is a sequence of digits, with the sequences in 1:1 correspondence with [0,1], not itself a value that belongs to the interval [0,1]. Then you can just say that trailing zeroes, or trailing nines, aren't allowed, giving you a unique representation for every value in [0,1]. At that point c-bar will work as defined.