Consequence of Los' theorem

Question

Let $D$ be a free ultrafilter on $\omega$ and let $\underline{M}$ be the structure $(\mathbb N,+,\times,<,0,1)$. Then for the element $u=[(1,2,3,\ldots,)]\in \underline{M}^\omega/ D$, we have $$\underline{M} \models 1+\ldots +1<u.$$

Can someone explain me what the above statement means? Apparently it is an illustration to differentiate between $\underline{M}$ and $\underline{M}^\omega/ D$ but I don't see what that means? I also don't see what is the role of the free ultrafiler.

Answer 1

Here is an explanation:

1.There is a canonical way to define the relation $<$ on the ultraproduct: $[((a_i))]<[((b_i))]$ iff $a_i<b_i$ a.s. It satisfies all properties of the linear order. It is also standard to define $+$ and $\cdot$ on the ultraproduct.

2. $\mathbb N$ embeds into the ultraproduct naturally: $n\mapsto [((n))]$, so we can identify $[((n))]$ with number $n$. Then in the ultraproduct $1+1+...+1$ ($n$ times) is equal to $n$.

3. According to the definition of $<$, we have that for every natural $n$, $n<[(1,2,3,...)]$.

4. The last property does not hold if the ultrafilter is not free in your terminology.

Answer 2

For each $n \in \mathbb{N}$, define $\overline{n}$ to be the term $1 + \ldots + 1$ where $1$ occurs $n$ times.

In other words, $\overline{0} = 0$ and $\overline{S(n)} = \overline{n} + 1$.

Then for all $n$, $\underline{M}^\omega / D \models \overline{n} < u$.

Proof: in particular, in the structure $\underline{M} ^ \omega / D$, the term $\overline{n}$ evaluates to $[(n, n, \ldots)]$. Take some element $J \in D$ such that $\forall j \in J, n < j$. This is possible because $D$ is free. We see that for all $n < j$, we have $n < j$ and in particular $(n, n, \ldots)_j = n < j + 1 = (1, 2, \ldots)_j$. Therefore, $\overline{n} < u$.