structure of the hypernaturals

Question

I want to understand the structure of the hypernaturals a little better. Let me recall the ultraproduct construction of the hypernaturals. On the set of all sequences of $\mathbb{N}$, we define an equivalence relation $(x_n)_n\sim (y_n)_n$ by $\{n~\vert~x_n = y_n\}\in \mathcal{U}$, where $\mathcal{U}$ is some free ultrafilter on $\mathbb{N}$. Anyway, my questions are these:

1) What's the cardinality of $\mathbb{N}^*$?

2) What does the lattice structure look like? More specifically, if I take the equivalence relation $x\sim y$ as $x$ and $y$ are in the same galaxy, then what does the ordered set $\mathbb{N}^*/\sim$ look like? Is it a dense total ordering? Is it isomorphic to something known?

3) It is possible to order-embed some ordinals in $\mathbb{N}^*$. For example, we can put $\omega$ as $(1,2,3,...)$. And $\omega^2$ as $(1,4,9,...)$. How much ordinals can we order-embed in the hypernaturals? Can we order-embed all countable ordinals?

Any good references to answer these questions are also appreciated!

Answer 1

This is a partial answer.

This answer shows that $|\Bbb N^*|=2^\omega$.

It’s easy to see that $\Bbb N^*$ must have an order type of the form $\omega+(\omega^*+\omega)\cdot\theta$, where $\theta$ is a dense linear order without endpoints, and hence that $\Bbb N^*/\sim$ must have an order type of the form $1+\theta$. Not all dense linear orders without endpoints can occur as $\theta$; e.g., it’s known that $\Bbb R$ cannot.

Answer 2

The hypernaturals have the same cardinality as the hyperrationals by the usual argument. But every real $x$ is infinitely close to a suitable hyperrational: for example, truncate $x$ at decimal rank $H$, where $H$ is a fixed hypernatural.

In fact, this is Martin Davis's favorite construction of the reals (starting with limited hyperrationals; see his 1977 book Davis, Martin Applied nonstandard analysis. Pure and Applied Mathematics. Wiley-Interscience [John Wiley & Sons], New York-London-Sydney, 1977).

At any rate, it follows that the map from limited hyperrationals to the reals is onto. This shows that the hypernaturals have the cardinality at least of the continuum. The ultraproduct construction you described starts with integer sequences hence no more than the continuum.

All the ordinals

ω, ω + 1, ω + 2, …, ω·2, ω·2 + 1, …, ω², …, ω³, …, ω^ω, …, ω^ωω, …,

are represented in a straightforward fashion by the corresponding hypernaturals using the transfer principle. By the time you get to

ε₀

you have a problem that needs to be overcome in a different way. Here you can use saturation to represent $\epsilon_0$ by a hypernatural larger than the countably many ordinals below it.

Moreover, saturation allows one to prove that all countable ordinals are representable in this way. Indeed, if there is a countable ordinal $\sigma$ such that ordinals up to $\sigma$ are not representable in the hypernaturals, then by well-ordering we can assume that $\sigma$ is the smallest such. Since there are countably many ordinals below $\sigma$, saturation on the hypernaturals allows one to find a hypernatural $n_\sigma$ greater than all the hypernaturals representing lower ordinals. Therefore $\sigma$ can also be represented in the hypernaturals in a way that respects the order, yielding a contradiction. Now use a suitable version of AC to find a maximal chain of such representatives in $\mathbb{N}^\ast$ to get the result.