3 ms·
- Every real number can be represented as a limit of rational numbers, which are uncountable. - So for every countable subset you can find a (Cauchy) sequence o
by heinrichhartman 10y ago
- Every real number can be represented as a limit of rational numbers, which are uncountable.
- So for every countable subset you can find a (Cauchy) sequence of rationals that does not converge.
Hence, you loose one of the most important tools in Analyisis (Cauchy criterium for convergence).
You can still work with the remaining set, but formulating and proving theorems, is going to be much harder.
- If integrals over f and g exists, then the integral over f * g does not need to exists.
- E.g. Integrals over bounded regions will not always exists.
- Theorem of Montonic convergence fails
- Function spaces will not be complete (L2).
A nice theory of constructible "periods" has been developed by Kontsevich and Zagier (http://www.maths.ed.ac.uk/~aar/papers/kontzagi.pdf http://www.maths.ed.ac.uk/~aar/papers/kontzagi.pdf) but it relies heavily on the existing body of Analysis being available.
- waqf 10y agoI"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.