- #1
pellman
- 684
- 5
These are some related questions in my mind, though I am rather confused about them.
1. What does [tex]\infty[/tex] at the "end" of the real number line have to do with [tex]\aleph_0[/tex], the cardinality of the integers, and C, the cardinality of the continuum? Is [tex]\infty[/tex] equal to one or the other (if such a thing is meaningful)? I'm pretty sure that this infinity is equal to [tex]\aleph_0[/tex] but I'm interested in anyone else's take.
2. When we find the area under an integrable curve (whether we use Lebesgue or Reimann sums, or whatever) we take the limit of sums of an ever-increasing number of diminishingly-small areas. The number of area elements in this sum, being an integer at any point in the limiting process, is certainly approaching [tex]\aleph_0[/tex]. Ok. So how we does this pass over into something describing continuous functions? If we think of [tex]\int f(x)dx[/tex] as summation of infinitesimal areas [tex]f(x)dx[/tex], then since [tex]f[/tex] is continuous function there are C-many of these area elements, not merely [tex]\aleph_0[/tex].
My guess is that this is something like Fourier series. A function which can be respresented as a Fourier series, even though it is continuous and has C-many points, can be fully described in terms of its Fourier coefficients, of which there are "only" [tex]\aleph_0[/tex]-many. This is not true of every function you can imagine; only those satisfying the conditions which allow them to be represented by Fourier series.
Similarly, if my hunch is correct, it is precisely those functions which are integrable whose areas-under-the-curve can be calculated from a countably-infinite number of area-elements, in spite of the fact that continuous functions have an uncountable number of points. Maybe finite, continuous, but non-integrable functions might have something like an "area" under their graphs, but it would take a sum of C-many area elements to calculate it. I'm totally speculating here.
Heck, for all know, "able to be calculated from a countably-infinite number of area elements" is precisely what integrable means--no more, no less.
Any thoughts from anyone are most welcome.
Todd
1. What does [tex]\infty[/tex] at the "end" of the real number line have to do with [tex]\aleph_0[/tex], the cardinality of the integers, and C, the cardinality of the continuum? Is [tex]\infty[/tex] equal to one or the other (if such a thing is meaningful)? I'm pretty sure that this infinity is equal to [tex]\aleph_0[/tex] but I'm interested in anyone else's take.
2. When we find the area under an integrable curve (whether we use Lebesgue or Reimann sums, or whatever) we take the limit of sums of an ever-increasing number of diminishingly-small areas. The number of area elements in this sum, being an integer at any point in the limiting process, is certainly approaching [tex]\aleph_0[/tex]. Ok. So how we does this pass over into something describing continuous functions? If we think of [tex]\int f(x)dx[/tex] as summation of infinitesimal areas [tex]f(x)dx[/tex], then since [tex]f[/tex] is continuous function there are C-many of these area elements, not merely [tex]\aleph_0[/tex].
My guess is that this is something like Fourier series. A function which can be respresented as a Fourier series, even though it is continuous and has C-many points, can be fully described in terms of its Fourier coefficients, of which there are "only" [tex]\aleph_0[/tex]-many. This is not true of every function you can imagine; only those satisfying the conditions which allow them to be represented by Fourier series.
Similarly, if my hunch is correct, it is precisely those functions which are integrable whose areas-under-the-curve can be calculated from a countably-infinite number of area-elements, in spite of the fact that continuous functions have an uncountable number of points. Maybe finite, continuous, but non-integrable functions might have something like an "area" under their graphs, but it would take a sum of C-many area elements to calculate it. I'm totally speculating here.
Heck, for all know, "able to be calculated from a countably-infinite number of area elements" is precisely what integrable means--no more, no less.
Any thoughts from anyone are most welcome.
Todd