# Limits, infinity, and cardinality (oh, and integrals too)

1. Mar 11, 2008

### pellman

These are some related questions in my mind, though I am rather confused about them.

1. What does $$\infty$$ at the "end" of the real number line have to do with $$\aleph_0$$, the cardinality of the integers, and C, the cardinality of the continuum? Is $$\infty$$ equal to one or the other (if such a thing is meaningful)? I'm pretty sure that this infinity is equal to $$\aleph_0$$ 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 $$\aleph_0$$. Ok. So how we does this pass over into something describing continuous functions? If we think of $$\int f(x)dx$$ as summation of infinitesimal areas $$f(x)dx$$, then since $$f$$ is continuous function there are C-many of these area elements, not merely $$\aleph_0$$.

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" $$\aleph_0$$-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

2. Mar 11, 2008

### pellman

Ok. Right after posting I figured out the answer to #1. $$\infty$$ is certainly equal to $$\aleph_0$$. If we just think of the number line as comprising only the integers, the infinity at the "end" is certainly $$\aleph_0$$. By definition. Well, why would filling in the spaces between the integers with reals make any difference?

If $$\aleph_{0}+4=\aleph_0$$, then wouldn't $$\aleph_{0}+\pi=\aleph_0$$ as well? If not, that would be too weird for me!

Last edited: Mar 11, 2008
3. Mar 11, 2008

### grmnsplx

Well, to begin with $$\aleph_{0}$$ is not equal to $$\infty$$ at all.
$$\aleph_{*}$$ is the cardinality of a set, not the number of elements in the set.

$$\aleph_{0}$$ describes a specific type of infinite set. that is to say that Some sets have an infinite number of elements in them, but a different kind of infinity. Namely, $$\aleph_{0}$$ means countably infinite.

When we say that a set S is countably infinite, we mean that there is a 1-to-1 mapping between the natural numbers an the elements in S.
So the set of even numbers is of cardinality $$\aleph_{0}$$ because we can map 1->2, 2->4, 3-6, ..., n->2n. Similarly the set of odd numbers is also countably infinite because we can map 1->1, 2->3, 3->5, ..., n->2n-1.
We can also count "larger" sets as well. The integers are easily counted by mapping the even natural numbers to the positive integers and the odd natural numbers to the negative integers.

I would say that $$\aleph_{0}+4=\aleph_0$$ makes no sense. At least not to me. I can only guess that you mean the $$Card(A \cup B) \leq Card(A) + Card(B)$$ with equality when $$Card(A) = \aleph_{0}$$ and $$Card(B) = n$$ where n is in N.

$$\aleph_{0} + \pi = \aleph_0$$ makes even less sense. The cardinality of a set is (the notion of) the number of elements in a set. How can a set have $$\pi$$ elements?

Make sense?

Last edited: Mar 11, 2008
4. Mar 11, 2008

### pellman

Ok. So cardinality is not the same thing as number, so we can't just go adding numbers to a ... (what is an instance of cardinality called. "A cardinality"? Like an instance of number is a number?)

The integers are subset of all cardinalities, right? We can have 3 members of a set or C members of a set and both are cardinalities. The only reason we can talk about 3 + pi is that there is other thing called "real numbers". But, as yet, there is no other thing to which $$\aleph_{0} + \pi$$ belongs. Do I have this right?

So is $$\infty$$ the same as $$\aleph_0$$ or not? There is a sense, number-wise not cardinality-wise, in which infinity for reals is the same as infinity for integers. If the definition of $$\infty$$ is, at least partly, a thing such that $$\infty>n$$ for every integer $$n$$, then we automatically get $$\infty>x$$ for every real number $$x$$.

And the cardinality $$\aleph_0$$ is certainly a thing such that $$\aleph_0<n$$ for every finite cardinality $$n$$, right? Of course, this is also true for C.

But really, never mind most of that. I'm just getting more confused, I think.

Look at question #2 in the OP. Can we meaningfully speak of the cardinality of the set of all area elements in the limit of the Reimann sum? As we take the limit, the cardinality (of the set of area elements) for every Reimann sum is an integer. So can we speak of the cardinality of the limit? And if so, is that cardinality $$\aleph_0$$?

5. Mar 11, 2008

### HallsofIvy

Staff Emeritus
"$\infty$ itself is NOT a number and is NOT "at the end of the number line". Saying "x goes to infinity" means that x get arbitrarily large, not that it "goes to" anything called infinity.

No, infinity is not the same as $$\aleph_0$$. "Infinity" is, as I said before, a "symbol" used, more or less sloppily, to indicate something getting large without bound. $$\aleph_0$$ is the cardinality of the set of all natural numbers.

As for your question #2, we only have "area elements" (I take it you mean the rectangles used in defining the Reimann sum) for Reimann sums with a finite number of intervals. Once you take the limit, there is no longer a Reimann sum nor any "area elements".

6. Mar 11, 2008

### ice109

which is of course exactly that: the number of elements in the set.
1-1 and onto.

of course infinite cardinal plus finite cardinal makes sense. so
$$\aleph_{0}+4=\aleph_0$$ is true because cardinals addition is the max of the added cardinals. though i don't know about finite cardinals that aren't from the naturals.

Last edited: Mar 11, 2008
7. Mar 11, 2008

### Hurkyl

Staff Emeritus
They have very little to do with each other. Essentially, the only thing they have in common is that $+\infty$ and $\aleph_0$ happen to be the least upper bound of the set of natural numbers in their respective ordered structures. (The extended real numbers and the cardinal numbers, respectively)

Depending on your epistemological views, either:
(1) This is simply a mental fiction used to guide intuition
(2) The notion of an integral is the correct transcription of what our intuition is saying

This makes sense because there is a (canonical) way to interpret a natural number as a cardinal number, and we can interpret "+" to be the addition operation on the class of cardinal numbers.

This, however, is nonsense. There is not a (canonical) way to interpret a real number as a cardinal number, so "+" cannot denote addition of cardinals. None of the usual meanings of those symbols permit that arrangement of symbols to be grammatically correct.

Last edited: Mar 11, 2008
8. Mar 11, 2008

### pellman

But there is a thing called the extended real number line which is the reals with two elements denoted by $$+\infty$$ and $$-\infty$$, which by definition are greater than and less than all finite real numbers, respectively. See http://en.wikipedia.org/wiki/Extended_real_number_line#Arithmetic_operations for more.

Yes, I know. It just bugged me today how--although we generally have discrete things associated with the countably infinite and the uncountably infinite with the continuous--yet, integrals are on operation on continuous functions but are gotten by an apparently countable sum of element areas.

9. Mar 12, 2008

### grmnsplx

whether or not it makes sense, it doesn't seem the slightest bit interesting.
it doesn't seem worth the bother to define such an operation on such a small set of elements [i.e. aleph_n's and n's]. Why not simply apply the definition I mentioned which applies much more generally.

Furthermore, we should avoid using such an operation as it can lead to confusion (i.e. aleph_0 + pi). Instead of thinking about adding cardinalities (yuck!), we should consider the cardinality of the union of sets. And I think anyone can see that there can never be a set with cardinality pi, or root two, or .5...

10. Mar 12, 2008

### ice109

your definition of cardinal addition is wrong. two cardinalities are equal iff there exists a bijection between them. your definition, just like your first mistake, implies there is only an injection.

anyway your quibling over nothing. adding cardinal numbers and considering the cardinalities of the union of sets is the same thing, they exist for exactly that reason. no one that knows what cardinal numbers are will addirrational cardinals because they don't exist.

11. Mar 12, 2008

### Hurkyl

Staff Emeritus
Disjoint union.

12. Mar 12, 2008

yes. forgive