I am reading a W.A.J Luxemburg paper about nonstandard analysis (https://www.jstor.org/stable/3038221). He presents the following definition
I am stuck trying to prove the existence of delta-incomplete ultrafilters. I suppose I have to construct a chain of filters over an infinite set with a certain condition such that, at the moment of using Zorn's lemma, the maximal satisfies the definition of delta-incomplete.
I have this, and I am more confused as how can I proceed. Let $\mathcal{C}$ be a chain of filters over a set $I$ with a maximal $\mathscr{U}$ such that $\mathscr{U}$ is a $\delta$-incomplete ultrafilter. That is, there exists $\{I_n\}_{n\in\mathbb{N}}$, a countable partition of $I$ such that, for all $n$, $I_n \not\in \mathscr{U}$. So, we would have that same property for all $F \in\mathcal{C}$. If some $I_n\in F$ then $I_n\in\mathscr{U}$. that leads to $\mathcal{C} = \{\mathscr{U}\}$.
You don't need Zorn's lemma for this. The ultrafilter lemma is sufficient.
Let $X$ be an infinite set and $X=\cup_{n\in\Bbb{N}} U_n$ a partition of $X$ into non-empty mutually disjoint subsets. Let $$\mathcal{A} = \{U_n:n\in\Bbb{N}\}\cup\{S\subseteq X: S\text{ is finite}\}$$ and $$\mathcal{B} = \{X\setminus S: S\in \mathcal{A}\}$$
We show that $\mathcal{B}$ is a filter-subbase. Therefore $\mathcal{B}$ generates a filter $\mathcal{F}$, and $\mathcal{F}$ is contained in an ultrafilter $\mathcal{U}$ by the ultrafilter lemma. Then by definition for all $n\in\Bbb{N}$, $X\setminus U_n\in\mathcal{U}$ but $\cap_{n\in\Bbb{N}} (X\setminus U_n) = X\setminus(\cup_{n\in\Bbb{N}}U_n) = \emptyset$.
So let $I\subseteq \Bbb{N}$ be a finite set and $F\subseteq X$ a finite set. Then $$\cap_{i\in I}(X\setminus U_i) \cap (X\setminus F) = X\setminus (\cup_{i\in I}U_i \cup F)\tag{1}$$ However, note that $\Bbb{N}\setminus I$ is infinite, and since each $U_i$ is nonempty and they're mutually disjoint, $\cup_{i\in \Bbb{N}\setminus I}U_i$ is an infinite subset of $X$. Since $F$ is finite, it follows that $\cup_{i\in I}U_i \cup F$ is not all of $X$, so the right hand side of $(1)$ is non-empty. This shows the collection $\mathcal{B}$ is a filter-subbase as required.$\Box$