Hacker News new | past | comments | ask | show | jobs | submit login

Why not, can you give an example.

Most of the "common" derivatives and integrals I can think of would work just fine, unless I'm missing something obvious?




- 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) but it relies heavily on the existing body of Analysis being available.


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?


My 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.


Are the computable numbers continuous? If not, is that a problem for analysis? If so, can you construct some computable analogue of continuity which is sufficient?


No, for example the integral sqrt(1 - x^2) from x = 0 to x = 1. Or any other integral over a circle/sphere for that matter.


Nothing wrong with that, it's still a computable number -- I'm not suggesting doing away with all real numbers, just the non-computable ones.

I don't claim to have thought through every detail, but having now gone and done some reading, this seems fine in the world of constructivism, which only requires computable numbers (and therefore only countable infinities of numbers)




Consider applying for YC's Fall 2025 batch! Applications are open till Aug 4

Guidelines | FAQ | Lists | API | Security | Legal | Apply to YC | Contact

Search: