# Is this a free ultrafilter

Suppose $$U$$ is a free ultrafilter on $$\mathbb{N}$$. I'm trying to understand what is the ultrafilter generated by the collection $$G=\{E\subseteq\mathbb{N}: E=A+1, \ for \ some \ A\in U\}$$. I think this subset itself is not even a filter (for every set, the minimum element is at least 2, and so $$\mathbb{N}$$ can't be included). However, it can be extended to the filter $$V=\{E\subset\mathbb{N}: n\in\mathbb{N}, A_1,...,A_n\subset G, \cap_{j=1}^nA_j\subseteq E\}$$ since $$G$$ has the finite intersection property. I know $$V$$ can be extended to an ultrafilter, and the extension is a free ultrafilter if every finite subcollection of $$G$$ has infinite intersection. But these extensions are more unwieldy. Is $$V$$ itself a free ultrafilter?

Yes, $$V$$ is a free ultrafilter.

Denote $$\Bbb{N}^+ = \Bbb{N}\setminus\{0\}$$.

Let $$E\subseteq \Bbb{N}$$.

1. If $$0\in E$$, then either $$E\cap\Bbb{N}^+\in G$$ in which case $$E\in V$$, or $$\Bbb{N}\setminus E = \Bbb{N}^+\setminus E\in G$$ so $$\Bbb{N}\setminus E \in V$$.
2. If $$0\not\in E$$ then either $$E\in G$$ in which case $$E\in V$$, or $$\Bbb{N}^+\setminus E\in G$$ and then $$\Bbb{N}\setminus E = \{0\}\cup (\Bbb{N}^+\setminus E)\in V$$.

If $$V$$ were principal - either it is generated by $$\{x\}$$ for $$x > 0$$ in which case $$U$$ is generated by $$\{x-1\}$$ - a contradiction to $$U$$ being free, or $$V$$ is generated by $$\{0\}$$ - which is false as $$\{0\} \cap \Bbb{N}^+=\emptyset\not\in G$$ so $$\{0\}\not\in V$$ by definition. So $$V$$ is free.

$$V$$ is not equal to $$U$$, because either the odds or evens are in $$U$$ - so the other set would be in $$V$$.

• I'm not sure I understand why $E\cap\mathbb{N}^+$ or $\mathbb{N}\setminus E$ is in $G$. Is the argument that since $U$ is an ultrafilter, $E$ or $\mathbb{N}\setminus E$ is in $U$. Are you saying that if $E\in U$, then $E\cap\mathbb{N}^+=E+1$? Jun 7 at 20:10
• Also, could you explain why $V$ is generated by $x>0$ if it is principle? I know that if $V$ is principle, it is all subsets that contain some fixed $x\in\mathbb{N}$, but from the definition of $V$, I don't see why this element would have to be bigger than $0$. Jun 7 at 20:27
• $G$ is an ultrafilter on $\Bbb{N}^+$ and $E\in V$ if and only if $E\cap\Bbb{N}^+\in G$. For the second question - I handled the case of $\{0\}$ separately. Jun 8 at 7:01