# free ultrafilter question

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?