In Non-standard analysis, is the number of natural numbers a hyperreal number?

Question

In Non-standard analysis, is the number of natural numbers a hyperreal number? In other words, if $H$ is the hyperreal infinite unit, does the sum $\sum\limits_{n=1}^H 1$ yield the number of natural numbers? Or, put some other way, is there some hyperreal $H$ which would yield the number of natural numbers?

Answer 1

My anwser is a vague "no", because of the nature of the question. The vaguest expressions are written in italic.

The "number of natural numbers" is usually conceived as the cardinal number $\aleph_0$. The most common convention is that this corresponds as a set to the set $\mathbb{N}$ of positive integers.

This will not be naturally identifiable with a non-standard natural number $H$, of which there is no smallest. In particular the set of standard positive integers is not itself a non-standard positive integer since it is not an object of the universe in IST.

(By the way, we have $\forall h \in \mathbb{N}, h=\sum \limits_{k =1}^h 1$ so $\sum \limits_{n=1}^H 1$ is equal to $H$ for any non-standard positive integer, or any hyperatural number).

In generic systems of hyperreal numbers given by ultrapowers of $\mathbb{R}$ to the power $\mathbb{N}$, you can if you want decide that some hypernatural number such as the equivalence class of the sequence $(0,1,2,...)$ corresponds to $\aleph_0$, but this will most likely be arbitrary. Moreover, in such systems, there may not be anything that looks like any uncountable cardinal number. Two remarks to illustrate this:

1. I believe it is consistent with ZFC that every order embedding $(\omega_1,\in) \longrightarrow ({}^*\mathbb{N},{}^*<)$ is cofinal.
2. For any hypernatural number $H \in {}^*\mathbb{N}$, the set $\{N \in {}^*\mathbb{N} : N<H\}$ has cardinality continuum.

Nevertheless, there are systems of hyperreal numbers where one can identify such sums with ordinal numbers such as $\omega \equiv \aleph_0 \equiv \mathbb{N}$. Although this identification still suffers from the issue 2. above, it provides a more solid link between certain so-called euclidean numbers and certain ordinals.

Answer 2

No, because loosely, there is no hyper-natural that corresponds to countable infinity.

To be more technical, for any unlimited hyper-natural number h, the set { 1, 2, ..., h } has the cardinality of the continuum. There is no "in-between" hyper-natural between finite naturals, and unlimited hyper-naturals.

Answer 3

The answer is "yes and no," depending on what you want the question to mean.

The answer is no in the sense that there is no hypernatural number which is the supremum of all the naturals, and thus placed directly after all the naturals, and so on. nombre and Jonathan Hoyle have answered this part of the question pretty well. The notion of the size of a set is very well dealt with by using infinite cardinals, and there are many ways to embed infinite cardinals into the hypernaturals (at least up to some cardinal), but they aren't unique, so for whatever hypernatural you're calling $\aleph_0$, there will also be a $\aleph_0-1$ and so on. This may or may not be a dealbreaker, but the point is that which hypernatural you are identifying with this cardinality depends on your embedding, and it only represents the "size" of that set in a very, very abstract sense.

There is another way to interpret the question for which the answer could be a tentative "yes." For instance, one very different way to count the number of elements in some set $S$ of naturals is to sum over the indicator function for that set, which I'll call $I_S(n)$. We have that $I_S(n) = 1$ iff $n \in S$ and $I_S(n) = 0$ otherwise. Then if we sum over the indicator function, we get the following expression:

$$\sum_{n=1}^{\infty} I_S(n)$$

where I am excluding $0$ from the set of naturals because it just makes the notation a little bit simpler.

If $S$ is finite, this will simply be the number of elements in the set. If $S$ is an infinite set, on the other hand, this sum will diverge to infinity.

But, one thing we can do is to treat this sum as a purely formal object and see if we can come up with some kind of notion of how large it is. The way we can do this is to identify each sum with its sequence of partial sums, and then see if we can extend the usual ordering on reals to these sequences. This turns out to be equivalent to the hyperreal construction: we have basically just reinvented the idea of an ultrapower, but where we're interpreting the elements as partial sums rather than sequences - and since you can easily convert between one and the other, the difference is purely a matter of notation than anything. So for instance, we simply treat this as an element in our ultrapower:

$$\sum_{n=1}^{\infty} I_\Bbb N(n) = \sum_{n=1}^{\infty} 1 = 1+1+1+1+1+... \to (1, 2, 3, 4, 5, ...)$$

where the arrow $\to$ represents that we have simply changed the notation to write this infinite formal sum as a sequence of partial sums.

So in some sense, if you look at this view of things, there really is an answer to "how many naturals are there" within the hyperreals, which is the hyperreal element $(1, 2, 3, 4, ...)$.

Now, one objection to this is that without specifying what ultrafilter we're using, we don't really have many ideas about what properties this number has. We don't know if it's odd or even, for instance, or if it's prime or composite, and so on. In fact, it all depends on which indices are in the ultrafilter, so in some sense, you could say that we haven't completely specified which hypernatural this number is at all.

But it turns out that we can get quite far with this approach anyway, even if we're leaving that open. For instance, we can immediately see that for any ultrafilter, we will have that the above number:

1. will always be a hypernatural,
2. will always be infinite and greater than any standard natural,
3. will always be the largest possible hypernatural you can obtain by summing over indicator functions in this way.

We can also play around with some interesting related infinite quantities that we can likewise get from summing over different indicator functions. In general, for any subset $S$ of the naturals, we will say that $N_S$ is the hyperreal element that you get by looking at the sequence of partial sums of the indicator function in this way. So we have the following

$$ \sum_{n=1}^{\infty} I_\Bbb N(n) = 1+1+1+1+1+... \to (1, 2, 3, 4, 5, ...) = N_{\Bbb N}\\ \sum_{n=1}^{\infty} I_{\Bbb N - {1}}(n) = 0+1+1+1+1+... \to (0, 1, 2, 3, 4, ...) = N_{\Bbb N}-1\\ \sum_{n=1}^{\infty} I_{A \cup B}(n) = N_A + N_B\\ \sum_{n=1}^{\infty} I_{A - B\subset A}(n) = N_A - N_B\\ $$

Thus in general, these hypernatural entities obey basic arithmetic properties that regular naturals do, and in a way that even cardinal arithmetic doesn't. We can also look at the sizes of odd and even numbers:

$$ \sum_{n=1}^{\infty} I_{2\Bbb N}(n) = 0+1+0+1+0+... \to (0, 1, 1, 2, 2, ...) = N_{2\Bbb N}\\ \sum_{n=1}^{\infty} I_{2\Bbb N+1}(n) = 1+0+1+0+1+... \to (1, 1, 2, 2, 3, ...) = N_{2\Bbb N+1}\\ $$

Now, depending on the ultrafilter, these quantities can either be equal (if the even indices are in the ultrafilter), or the number of odd numbers is one greater than the number of even numbers (if the even indices are in the ultrafilter). Which one this is, is equivalent to determining whether $N_\Bbb N$ is even or odd. (Note that this would be flipped if we'd counted zero as well, rather than starting at one, as then it's the set of even naturals which may be larger instead.)

But we do know that for whatever ultrafilter we choose, we have that $N_{2\Bbb N} + N_{2\Bbb N+1} = N_{\Bbb N}$. And we also have

$$ (N_\Bbb N)/2 = (0.5, 1, 1.5, 2, 2.5, ...) $$

If the even indices are in the ultrafilter, meaning that $N_\Bbb N$ is an even number, then this is also a hypernatural and it's equal to the sizes of both the even and odd naturals above. If it's not, then we have that this is 1/2 less than the size of the odd numbers and 1/2 greater than the size of the even numbers (and is thus equal to the average of both). You can also play around with extending this to integers, where you get the result that $N_\Bbb Z = 2 N_\Bbb N + 1$ -meaning the number of integers must be an "odd" hypernatural, for instance.

You can play around with this and compute some sizes of various infinite sets of naturals - you can easily see there are more composites than primes and all that. In some sense this is very similar to the idea of "natural density" -- and if you like, you can divide the result by $N_\Bbb N$ to get a value in between $0$ and $1$, like the standard natural density. The result will always be a hyperrational whose denominator is at most $N_\Bbb N$ (since it was originally a hypernatural and we divided by $N_\Bbb N$, which is the largest quantity that can be formed in this way). If your set has a standard natural density, this resulting hypernatural quantity will always be infinitesimally close to the standard density - in fact, if you take the standard part of the result, you get the standard density - so we may as well call this construction a nonstandard density. But these nonstandard densities give you quite a bit more information than the standard density, in that there are also infinitesimal densities for finite sets, and even infinite sets with standard density 0 (such as the set of perfect squares) will have positive but infinitesimal nonstandard density in this way. Sets that have the same natural density can also differ infinitesimally in this metric based on how quickly they converge to the standard density in question. We also have that every set has a nonstandard density, whereas only some sets have a standard density, but if your set doesn't have a standard density, then the size of the resulting nonstandard density will be entirely left up to the whims of the ultrafilter. The nonstandard density will always be a hyperrational whose denominator is at most $N_\Bbb N$ (since it was originally a hypernatural and we divided by $N_\Bbb N$, which is the largest quantity that can be formed in this way).

Terry Tao has some really good ideas on this kind of "partially constructive" nonstandard analysis that I've done here: https://terrytao.wordpress.com/2012/04/02/a-cheap-version-of-nonstandard-analysis/