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?


1 Answer 1


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 $V$.

  • $\begingroup$ 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 $E\cap\mathbb{N}^+=E+1$? $\endgroup$
    – user124910
    Jun 7 at 20:10
  • $\begingroup$ 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 $0$. $\endgroup$
    – user124910
    Jun 7 at 20:27
  • $\begingroup$ $G$ 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. $\endgroup$
    – John Doe
    Jun 8 at 7:01

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