Dear Carlos,

for a mathematical part of the question - right, the number of intervals goes to alepho , but each of them is of cardinality continuum. It is similar in ℝn.

I agree with Peter, there is no paradox (as it was implicitely included in my answer).

However, I like your question, Carlos. I like this kind of doubts, they are pushing the science further.

Let me sketch a more direct explanation of the doubts.

1. When VERYFING integrability according to Riemann we have to check ALL possible finite partitions, their cardinality is at least continuum. Thus there is no paradox.

2. If the function IS Riemann integrable, than the definite integrals over [a,x], for all x \in [a,b] form a CONTINUOUS function on [a,b] which is determined uniquely (due to continuity, according to the Heine definition) by values on any dense (be it COUNTABLE) subset of [a,b].

3. Therefore also, the SUBSTANCIAL PROPERTIES of the integrated function are determined by the sequence of values of its integral. Let us not that it is in the core of the definition, that two functions may differ on a subset and even though their definite integral over [a,x] are equal for EVERY x. Namely this is the reason that Lebesque integrable functions form a Banach space only after identifying functions equal almost everywhere.

Regards, Joachim

O.K. And about the quantum physics paradox is concerned, I think the solution is analogous, for in making the transition from vectors with a finite basis to a function defined in a continuous interval, say [a, b], for each finite basis of size n we can take a one-to-one correspondence between the elements of said base and the elements of a partition of the interval in n subintervals, thus discretizing the continuous function to model a finite dimension space, and in the limit to infinity we proceed to model the continuum. And the case of all real numbers, it is the infinite limit of bounded intervals, and there is a one-to-one correspondence between the points of these intervals and the real numbers, and there is no problem in that case.

Thank you for helping me solve the problem.

But the paradox is so confusing that sometimes I am confused about what the paradox is. Let's explain details:

True, the intervals of any finite partition are continuous, but in the limit to infinity these intervals are reduced to points of cardinality aleph_0, the continuum being a set of points of cardinality aleph_1, and then we do not consider all our points in the path of the interval [a, b] when we make the limit to infinity. That is the problem that I pose.

This problem becomes even clearer in the quantum case, where, in the infinite limit, we end up with a numerable set of eigenfunctions, with their respective numerable eigenvalues, for a continuous interval.

Dear Carlos,

the intervals are shorter and shorter, but they remain intervals.

It is true that one cannot construct them physically, which is confusing. However, any interval - whatever small is its lenght - can be mapped onto ℝ by a bijection.

I usually say to my students: zoom it, zoom it again and again, ... what you´ll see? In the case of intervals, it is again an interval, never a point.

Okay, this first example is perhaps not very good, but in the second case of quantum physics I think the paradox is clearer, because we end up with a numerable base, and we must model our continuous interval with a space of dimension aleph_1.

Dear P.T. Followers of this question.

As it is already established by some of us, there is no paradox, since the family of equivalence classes of Lebesque integrable functions is a separable Banach space. Therefore there is a possibility to express the integrals as limits of sequences. HOWEVER, only if the function is integrable. But, checking the integrability, in particular the Riemann integrability it is not enough to calculate the limit of contably many sequences. As a matter of fact, this requires usage of uncountably many approximating sums.

Thus, since in the original question we read

" when we make a Riemann integral for a Riemann-integrable function of a variable in an interval [a, b] and compute the area under the curve, we perform a partition with a natural number n of subintervals and then we make n - -> Infinity "

which means that the assumption of R-integrability is taken into accoun. Conclusion: the usage of limit of one suitably chosen sequence related to finite partitions is correct.

Regards, Joachim

Right. Your explanations for the example of the integral are clear. But could anyone dare to comment on the paradox of the quantum example?, which I think is much clearer and more difficult to solve; proof of this is that there is no comment on it.

Dear Carlos, the answer to the case of quantum models is contained in my answer and in your correct explanation: The family of characteristic functions of suitably chosen intervals form again a countable basis in the space of measurable square integrable functions on the space R^n of admitted positions of the quantum particle. Passage to matrices is just a problem of equivalent representation of unitary operators. Obviously, there are many bases of L^2(R^n). Due to the separability of the space of integrable functions, rather strange seems existence of operators with continuum spectrum. But, this can also be expalined:)

Regards, Joachim

True. Although our base is discrete, it is a discrete base of orthogonal continuous functions, so we can represent any function in our space of integrable functions in [a, b] as a infinite serie of sums of our numerable base functions. As an example is the classic Fourier series in sines and cosines to represent any function of the space L²([a,b]).

Do you agree, Joachim?

My last question in the quantum example is how we can go from an accounting function (a vector in IR ^ n or in C ^ n) to a continuous function and not to a discrete numerable function in the limit to infinity.

This discretization can be seen in the eigenvalues and the basis of eigenvectors. And then in the limit we can not contain the continuous set of all the points of our intertvalue as I myself believed in my first answer.

We do not go to a continuous function as a sequence. We represent the continuous function by a series built of countably many continuous functions, roughly speeking. Same as in the case of Fourier series. Thus, the "counting function" is the beginning of the list of terms of the series, more precisely, the list of coefficients if the basis is already chosen. No mystery, no paradox.

Regards, Joachim

O.K. Thank you all for helping me understand these concepts.