I asked a question here Is this a free ultrafilter, and I'm attempting to write up the solution.

$\mathbb{N}=\{1,2,...\}$ and for any $A\subseteq\mathbb{N}$, $A+1=\{a+1:a\in A\}$.

Suppose $U$ is a free ultrafilter. Consider the collection of sets $G=\{A+1:A\in U\}$. I'd like to show $V=\{E:A+1\subseteq E, \ for \ some \ A\in U\}$ is a free ultrafilter.

It has the finite intersection property, since $U$ does, so it is a filter. It is an ultra filter. If $G\subset\mathbb{N}$

case 1: $1\in G$. Since $U$ is an ultrafilter, either $G-1\subset\mathbb{N}$ is in $U$ or $(G-1)^c=G^c-1\in U$. If $G-1\in U$, then $G=G-1+1\in V$. If $G^c-1\in U$, then $G^c=G^c-1+1\in V$.

case 2. $1\not\in G$. Then $1\in G^c$, and apply the same argument.

So $V$ is an ultrafilter. It is a free ultrafilter. Suppose It is a fixed ultrafilter. Then $\cap V=\{x\}$. Since $\mathbb{N}\in U$, $\mathbb{N}+1\in V$, so $x\not=1$. Hence, $x>1$. So for each $E\in V$, $x\in E\implies$. In partiuclar, for $E=A+1$, $A\in U$, $x\in E$. So $1\leq x-1\in A$ for each $A\in U$, and so $U$ is fixed. Contradiction.

Is this correct?



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