Can an Ultrafilter be defined in terms of the Convergence of a Sequence?

Question

We know that the set $\{\sin(0), \sin(1), \sin(2), ... \}$ is dense in the interval $(-1,1)$.

So now consider the sequence $S$ = $\langle \sin(0), \sin(1), \sin(2), \ldots \rangle$. This sequence does not converge, but since it is dense and bounded in $(-1, 1)$, there are subsequences of $S$ that converge to any element in $[-1,1]$.

Let $U$ be any subset of $N$, and let $S(U)$ be the subsequence of $S$ over the indices of $U$.

Now, let $x$ be an arbitrary element of $[-1,1]$, and let $F$ be the collection of all subsets $U$ of $N$ such that $S(U)$ converges to $x$.

Can $F$ be used to generate a non-principal ultrafilter on $N$?

It seems to me it might, since any other choice of $U$ outside $F$, would cause $S(U)$ to converge to some value other than $x$ (or not converge at all). And when viewed as a hyperreal, $\langle \sin(0), \sin(1), \sin(2), \ldots \rangle$ must be in the halo of some specific real number inside $[-1,1]$, determined exclusively by the choice of ultrafilter.

Or is $F$ somehow "too large" to be useful as the basis for an ultrafilter?

Answer 1

This can be done in the following sense. For every number $r\in[-1,1]$, and every subsequence of the sequence $\sin n$ converging to $r$, let $S_r$ be the complement in $\mathbb N$ of its index set. For $r\not=r'$, a union of type $S_r\cup S_{r'}$ are a cofinite set by construction, which is fine if we are trying to construct a nonprincipal ultrafilter. Now consider the intersection $S_r\cap S_{r'}$. Let us show that this intersectoin is necessarily infinite. Indeed, it contains every index of a term in the sequence which is sufficiently close to points of $[-1,1]$ other than $r$ and $r'$. Thus $S_r\cap S_{r'}$ is necessarily an infinite set. A similar argument applies to finite intersections of the sets $S_r$. Thus we can complete the collection to a filter. Then we use Zorn's lemma to find an ultrafilter extending such a filter, which will be necessarily nonprincipal. Note that we can't hope to find an ultrafilter without using Zorn's lemma or something analogous, since there are models of ZF with no nonprincipal ultrafilters at all.

Answer 2

I also see, from much simpler reasoning that I missed, why such an approach is not unique. For with being the cardinality of the continuum, there are 2 possible non-principal ultrafilters on ℕ, but only possible choices in [-1,1] for subsequences of $\langle sin(n) \rangle$ to converge to (when they converge at all). Thus, there must still be 2 distinct ultrafilters for each possible choice.

Therefore, specifying an ultrafilter that has $\langle sin(n) \rangle$ → r, for some r ∈ [-1,1], does little more than would specifying an ultrafilter that contains the set of even numbers. For all intents and purposes, I would essentially have the same number of ultrafilters to choose from, had I made no such specification.

Thanks again.