2 ms·
I"m not sure what you're saying. There are only countably many computable sequences of rational numbers, so why shouldn't they all converge?
by waqf 10y ago
I"m not sure what you're saying. There are only countably many computable sequences of rational numbers, so why shouldn't they all converge?
- heinrichhartman 10y agoMy argument aws: The constructible numbers are not "complete". You can always find cauchy sequences of constructible numbers that do not converge to a constructible number. If you add the requirement that the sequence itself must be constructible things _might_ change. It depends on how you define constructible sequence. I thought Cantors argument would show that there are uncountable constructible sequences: - Assume there are countable many C[1], C[2] - Construct a new one D = C[1][1] + 1, C[2][2] + 1, ... - D in constructible, hence it's on the list: C[j] = D - But D[j] =/= C[j][j] - Contradiction. This proof is however not correct. The sequence D does not need to be constructible.