# Can this sequence be a hyperreal number? What would be its real part?

Consider the sequence $\{a_n\} = \{\sin(n) \mid n\in \mathbb N \}$. Can this sequence be viewed as a hyperreal number? What could be its real part? Any intuition would be highly appreciated :)

Recently I have been wondering about some special cases of hyperreal numbers. For example, I know that the interpretation of the sequence $\{b_n\} = \{1,0,1,0, ... \}$ depends on how we have decided to construct the ultrafilter of quasi-big sets. Following the construction as detailed in the (wonderful) book "Infinitesimal Calculus" by Henle and Kleinberg, we would first add all cofinite sets of naturals to the filter. Then, one by one, we would add other sets of naturals. If we decide to add the set of even naturals, then the set of uneven naturals would not make out part of the filter. Then the set $\{n \mid b_n = 0\}$, which is equal to the set of even naturals, would be considered quasi big, so $\{b_n\}$ would be considered equal to $0$. If however, we had added the set of uneven naturals to the filter instead, then $\{b_n\}$ would equal $1$.

I feel like a similar reasoning can be applied to $\{\sin(n) \mid n\in \mathbb N \}$, but that sequence is tricky in that no two of its elements are ever equal. I am having a hard time finding a real $r$ and an infinitesimal $h$ such that $\{n \mid \sin(n) = r + h_n\}$ is quasi-big.

## 1 Answer

First of all, a hyperreal number, in the ultrapower construction is an equivalence class of sequences, not just one particular sequence. So formally speaking it isn't a hyperreal number.

But letting this slide, the answer really depends on the choice of ultrafilter. Really that the sequence is in fact dense in $[-1,1]$. If you happen to have a set $I$ in the ultrafilter such that $\{\sin(n)\mid n\in I\}$ monotonously converges to $0$ from above, the real number you look for is $0$. If the same is true for $\frac12$, it is that. If other sets happen to be in the ultrafilter you will get other results.

And since we cannot explicitly define a free ultrafilter, nor we usually want to, we have much freedom in assuming things like above happen or not happen.

• Ah! The phrase If you happen to have a set I in the ultrafilter such that {sin(n)∣n∈I} monotonously converges to 0 from above was really what I needed for understanding this problem. Also, that was an amazingly quick answer :) Thanks a lot! – Michael Langbein Aug 11 '15 at 9:10
• (Also it would have been faster, I'm typing on my phone...) – Asaf Karagila Aug 11 '15 at 9:13
• Where is monotonous convergence needed? – Primo Petri Aug 22 '15 at 9:39
• @Primo: Well, it's not really needed. But if the sequence jumps between positive and negative, then the sign of the resulting hyperreal would also depend on the choice of sets in the ultrafilter. It's not really important for the end result, sure, but it simplifies understanding the situation. – Asaf Karagila Aug 22 '15 at 9:51