4 ms·

Good luck developing analysis with only countable infinities. Limits will take you out of the realm of countable spaces. Most of your derivatives and integrals

Good luck developing analysis with only countable infinities. Limits will take you out of the realm of countable spaces. Most of your derivatives and integrals won't exists, if you force them to take values in countable sets.

CJefferson 10y ago
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?
- kutkloon7 10y ago
  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.
  - CJefferson 10y ago
    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)
- twic 10y ago
  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?
- 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 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 ago
      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.
pacala 10y ago
One of the observations in programming language semantic proofs is that one doesn't need coinduction. Statements can be proven by good old fashion induction, by indexing them with the number of steps in the computation, then proving they hold for any finite number of steps. I suspect the same technique applies to analysis. We don't need to prove a statement holds for reals with "infininte" number of digits, but merely for all rationals with finite k digits, where k is a parameter of the proof, thus can be arbitrarily large.
- macawfish 10y ago
  Reminds me of computing points in a Mandelbrot or Julia set

CJefferson 10y ago

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?

kutkloon7 10y ago
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.
- CJefferson 10y ago
  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)
twic 10y ago
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?
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 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 ago
    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.

pacala 10y ago

One of the observations in programming language semantic proofs is that one doesn't need coinduction. Statements can be proven by good old fashion induction, by indexing them with the number of steps in the computation, then proving they hold for any finite number of steps. I suspect the same technique applies to analysis. We don't need to prove a statement holds for reals with "infininte" number of digits, but merely for all rationals with finite k digits, where k is a parameter of the proof, thus can be arbitrarily large.

macawfish 10y ago
Reminds me of computing points in a Mandelbrot or Julia set