6

I am reading about hyperreal numbers defined as (to my understanding) certain equivalence classes on all sequences of real numbers. $ω$ is defined as $(1, 2, 3, ...)$, and all functions are applied element-wise. This makes sense for sequences that have an infinite limit, like $e^ω$, which is simply a bigger infinity than $ω$, but it occurs to me that there must be certain "abominable" numbers that don't fit into the standard notion of reals, omegas, and epsilons.

By the transfer principle, values like $\sin{ω} = (\sin{1},\sin{2},\sin{3},...)$ and $\frac{\sin{ω}}{ω}=(\frac{\sin{1}}{1},\frac{\sin{2}}{2},\frac{\sin{3}}{3},...)$ should exist and be finite/limited. However, for the first value I am not able to show that it is equal to a real number, or at least infinitely close to a real number (which is a necessity according to the paper I am reading), and for the second number, although it is infinitely close to $0$, its sign is ambiguous, breaking trichotomy (in the same way as $\sin{ω}$) which should hold based on the transfer principle.

So, if $\sin{ω} = r$, what is the actual value of $r$?

IS4
  • 966
  • Even $$\lim_{n\rightarrow \infty} \sin(n)$$ does not exist, so I doubt that we can define $\ \sin{\omega}\ $ reasonably. – Peter Jun 18 '20 at 10:51
  • 5
    The "certain equivalence classes" business that you skip over is basically the key to answering all of your questions. The choice of nonprincipal ultrafilter determines whether the number you would call $(-1)^\omega$ is equal to $1$ or $-1$, for instance. If $\sin \omega$ has to be infinitely close to a real number then the choice of ultrafilter would determine which number in $[-1,1]$ that would be, etc. – Mark S. Jun 18 '20 at 11:12

1 Answers1

8

Your questions show that you have yet to internalize the following important fact: your source constructs the hyperreals $\:^\star\mathbb{R}$ in Section 1.3. by first fixing a non-principal ultrafilter $\mathcal{F}$ over $\mathbb{N}$.

You use $\omega$ to denote the fixed hyperreal arising from (the equivalence class of) the sequence $(0,1,2,\dots)$. But the properties of fixed hyperreals such as $\omega$ actually depend on the precise identity of the ultrafilter $\mathcal{F}$ you used to construct $\:^\star\mathbb{R}$.

Before we look at trigonometric functions, you should first think about something simpler, say the function $f: \mathbb{R} \rightarrow \mathbb{N}$ given as follows: $$f(x) = 1 \text{ if }x\text{ is an odd integer and }f(x)=0\text{ otherwise.}$$

By Section 1.6. of your source, we can get an extended function $^\star\!f: \:^\star\mathbb{R} \rightarrow \:^\star\mathbb{N}$. But what is the value of $^\star\!f(\omega)$? Since $\forall x \in \mathbb{R}. f(x) = 0 \vee f(x) = 1$, the Transfer principle says that either $^\star\!f(\omega) = 0$ or $^\star\!f(\omega) = 1$. But which?

It turns out that the answer depends on the ultrafilter $\mathcal{F}$ you used to construct $\:^\star\mathbb{R}$. You should check that if the ultrafilter contains the set of odd natural numbers, then $^\star\!f(\omega) = 1$, and if instead the ultrafilter contains the set of even natural numbers, then $^\star\!f(\omega) = 0$. Since for any set $A \subseteq \mathbb{N}$ $\mathcal{F}$ contains either $A$ or $\mathbb{N}\setminus A$, it must be one of these.

Once you understand these results, we can move on to your questions about the $\sin$ function.

However, for the first value I am not able to show that it is equal to a real number, or at least infinitely close to a real number (which is a necessity according to the paper I am reading

Of course $\sin(\omega)$ is infinitesimally close to some real number. This follows immediately from the general result that every limited hyperreal has a shadow (Theorem 3.3. in your source). And of course $\sin(\omega)$ is limited, since we have $\forall x. |\sin(x)| < 2$.

But which real number $r$ is $\sin(\omega)$ infinitesimally close to? Well, that depends strongly on the non-principal ultrafilter $\mathcal{F}$. In fact, since the image of $\sin(\mathbb{N})$ is dense in the interval $[-1,1]$, you can choose any number $x \in [-1,1]$ and find an ultrafilter $\mathcal{F}$ such that $\sin(\omega) \approx x$ in the hyperreal field constructed using $\mathcal{F}$.

and for the second number, although it is infinitely close to 0, its sign is ambiguous, breaking trichotomy

For similar reasons, the value of $\mathrm{sgn}\left(\frac{\sin(\omega)}{\omega}\right)$ is very far from ambiguous: it is either positive (so the $\mathrm{sgn}$ function takes the value $1$ or negative ($-1$), and not both. To deduce which possibility holds, you have to know more about the non-principal ultrafilter that was used to construct $\mathcal{F}$: if $\mathcal{F}$ contains the set $\left\{n \in \mathbb{N} \:|\: \frac{sin(n)}{n} > 0\right\} \subseteq \mathbb{N}$ then $\frac{\sin(\omega)}{\omega} > 0$ (exercise!) so its sign is positive. If $\mathcal{F}$ contains the complement of this set, then the sign is negative (exercise: why can't it be zero?). And $\mathcal{F}$ must contain one of these, by virtue of being an ultrafilter.

Z. A. K.
  • 11,359
  • Nicely explained! What I hadn't realized (and what should perhaps be more often iterated in popular explanations of hyperreals) is that the choice of the free ultrafilter matters and does resolve these things that I thought of as ambiguities. Does it mean that there is a class of mutually non-isomorphic hyperreal fields based on the choice of the ultrafilter? – IS4 Jun 18 '20 at 11:58
  • That depends on many factors (involving set theory, but also precisely what we mean by hyperreal fields). Here's a good starting point: https://math.stackexchange.com/questions/719131/uniqueness-of-hyperreals-contructed-via-ultrapowers/719180#719180 – Z. A. K. Jun 18 '20 at 12:04
  • Also just to be sure I understand the results, for every free ultrafilter, there exists a unique real $r$ such that $\sin{ω}-r$ is equivalent to an infinitesimal number simply by the virtue that the ultrafilter must contain a set of indices for which the sequence $\sin{ω}-r$ has an unambiguous sign and converges to $0$. And $\sin{ω}-r$ itself can never be $0$, as the sines of natural numbers are unique. Is that correct? – IS4 Jun 18 '20 at 12:15
  • 1
    Not sure about the meaning of "unambiguous sign". Even if $r$ is the shadow of $\sin(\omega)$, the object $\sin(\omega) - r$ is an equivalence class consisting of infinitely many real sequences, some of which start out positive, others negative or even zero. Of course they eventually all have the same sign. The rest looks correct to me upon one reading. – Z. A. K. Jun 18 '20 at 12:32