Zero divisors in the hyperreal numbers

Question

I am currently reading this introduction to hyperreal numbers. On the first page, to illustrate the problem with just taking hyperreal numbers to be sequences of reals, the following example is used: $$ (0,1,0,1,...) \cdot (1,0,1,0,...) = (0,0,0,...),$$ demonstrating that this fails to be a field. Thus, the construction using ultrafilters and taking sequences mod a certain equivalence relation is motivated.

I see how this construction works for convergent series. However, in the example above, I fail to see that this actually solves our problem. It still seems that we have zero divisors. If $(0,1,0,1...) \sim (0,0,0,0...)$ and $(1,0,1,0,...) \sim (0,0,0,0,...)$, then the sets $2\mathbb{N}, 2\mathbb{N}-1$ are contained in the ultrafilter, and since ultrafilters are closed under intersection, their intersection $\emptyset$ is as well. Thus, these two elements must be nonzero and we have zero divisors.

There must be an error in my reasoning, but unfortunately I don’t see where. The introduction cited uses the transfer principle to show that the hyperreal numbers do indeed form a field, but this feels kind of unsatisfactory given this example.

Answer 1

As you have noted, $2\mathbb{N}$ and $2\mathbb{N}-1$ are the complement of each other in $\mathbb{N}$. By definition of an ultrafilter, this means that precisely one of these lies in the ultrafilter; say $2\mathbb{N}$ does, for example. You are correct that this means that the equivalence class of $(0,1,0,1,\dots)$ is non-zero. But this also means that the equivalence class of $(1,0,1,0,\dots)$ is equal to the equivalence class of zero (why?). So, even though the product of these two equivalence classes is zero, this does not give an example of non-trivial zero divisors, since one of the elements in question is equal to zero. In short, the fact that $2\mathbb{N}$ and $2\mathbb{N}-1$ are complementary means that precisely one of the elements you describe is non-zero, not that both of them are.

Answer 2

If $a$ and $b$ are complementary binary sequences, then $$\begin{aligned}[] [a] = 0 &\iff \{n \mid a_n = 0\} \in\mathcal F \iff \{n \mid b_n = 1\} \in\mathcal F \iff [b]=1 \end{aligned}$$

So by design it is guaranteed that either $[a]=0$ and $[b]=1$ or $[a]=1$ and $[b]=0$.