# Showing a filter on the Power set of $\mathbb{Z}$ is a one point Filter

Let $\mathcal{P}_0(X)$ the Power set of $X$ without the empty set and let $\dot{x}:=\{A\subseteq X: x \in A\}$ the one point filter generated by $x$. Furtermore let $$\mathcal{A} := \{ f \in X^{\mathcal{P}_0(X)} : \ \forall A \in \mathcal{P}_0(X): f(A) \in A\}$$ be the set of the functions mapping subsets of the power set to elements of them. Let $\varphi$ be a filter on $\mathcal{P}_0(X)$ with $$\forall f \in \mathcal{A} \exists x_f \in X : f[\varphi]=\dot{x}_f.$$ Here $f[\varphi]$ stands for the filter generated from the filter basis: $$\{ \operatorname{image} f|_M \ : \ M \in \varphi\}$$

It seems like our definition of filter is not the standard definition, to avoid confusion $\varphi \subset \mathcal{P}(\mathcal{P}(X))$

Now we shall show that if $X=\mathbb{Z}$, then it follows that $\varphi$ is a one point filter. From the given property we already can conclude that $\varphi$ is an ultrafilter but I don't know what I shall do next, could someone give me a hint?

As it was asked for in the comments, our definition of a filter is:

Let $S$ be a set (an arbitrary one), and $A,B\subseteq S$. We call $\varphi\subset \mathcal{P}(S)$ a filter when the following holds

• $\varnothing\notin \varphi$
• $A,B\in \varphi \implies A\cap B \in \varphi$
• $A\subset B$ and $A\in \varphi \implies B\in \varphi$

An ultrafilter is a filter, such that there is no bigger filter, e.g. when $G$ is a filter and $F$ is an ultra filter and $G\supseteq F\implies G=F$. More convenient is the equivalence that for every subset $A$ of $S$ it holds that $A\in F \vee S\setminus A \in F$. The one point filter is already defined in the first sentence.

I posted the question on MathOverflow were someone wrote the answer.

• Isn't $f$\phi$$ as defined above just a subset of $X$? So how can it be a filter on $X$, i.e. a set of subsets of $X$? – Alexander Thumm May 3 '13 at 15:58
• @AlexanderThumm $f[\phi]$ is a family of subsets of $X$, I edited the post and hopes this clarifies it – Dominic Michaelis May 3 '13 at 19:04
• What is the origin of this question, what is its motivation? – yup May 4 '13 at 12:29
• @yup topology group exercise is its origin and the motivation is I didn't solved it while the group exercise – Dominic Michaelis May 4 '13 at 18:32
• – Martin May 9 '13 at 4:43

## 1 Answer

Please note that the following is not my answer nor my idea but the one of Joseph Van Name, who posted this answer on MathOverflow and I just copy his answer here for completeness.

For this problem, we shall use $\mathbb{N}$ instead of $\mathbb{Z}$ since $\mathbb{N}$ is easier to work with in this case. We shall say that $A$ has property $P$ almost everywhere (or for almost all $A$) abbreviated a.e. if $\{A\in P_{0}(X)|A\,\textrm{has property}\,P\}\in\varphi$.

For $n>0$, let $f_{n}\in\mathcal{A}$ be the function where if $A\in P_{0}(\mathbb{N})$ then $f_{n}(A)$ is the $n$-th element of $A$ whenever $|A|\geq n$ and $f_{n}(A)$ is the last element of $A$ whenever $|A|<n$. Let $y_{n}=x_{f_{n}}$. Then $f_{n}(A)=y_{n}$ for almost every $n$. If $i<j$, then clearly $y_{i}\leq y_{j}$. Furthermore, if $y_{n}=y_{n+1}$, then $f_{1}(A)=y_{1},...,f_{n+1}(A)=y_{n+1}$ for almost every $A\in P_{0}(\mathbb{N})$. However, if $f_{1}(A)=y_{1},...,f_{n+1}(A)=y_{n+1}$, then $A=\{y_{1},...,y_{n+1}\}$ making $\varphi$ a principal ultrafilter. We shall therefore assume that $y_{n}<y_{n+1}$ for all $n$.

We claim that $\varphi$ is an ultrafilter. Let $S=\{A\in P_{0}(X)|f_{1}(A)=y_{1},f_{2}(A)=y_{2}\}$. Then clearly $S\in\varphi$. Let $R\subseteq P_{0}(X)$. Let $h\in\mathcal{A}$ be a function where $h(A)=y_{1}$ for each $A\in R\cap S$ and $h(A)=y_{2}$ for each $A\in R^{c}\cap S$. Then the function $h$ is constant almost everywhere. It is clear that $h(A)=y_{1}$ a.e. or $h(A)=y_{2}$ a.e. If $h(A)=y_{1}$ almost everywhere, then $R\cap S\in\varphi$. If $h(A)=y_{2}$ a.e., then $R^{c}\cap S\in\varphi$. Therefore, we conclude that either $R\in\varphi$ or $R^{c}\in\varphi$. Therefore $\varphi$ is an ultrafilter.

We shall now prove that $\varphi$ is a principal ultrafilter. Assume for the sake of contradiction that $\varphi$ is a non-principal ultrafilter. Let $Y=\{y_{n}|n\in\mathbb{N}\}$. We claim that $Y\not\subseteq A$ for almost every $A$. To prove this assume that $Y\subseteq A$ for almost every $A$. Then the ultrafilter $\varphi$ is $\sigma$-complete. To see this, let $T=\{A\in P(X)|Y\subseteq A\}$, and let $P=\{R_{n}|n\in\mathbb{N}\}$ be a partition of $T$ into countably many pieces. Let $g\in\mathcal{A}$ be a function where if $A\in R_{n}$, then $g(A)=y_{n}$. Then the function $g$ is constant almost everywhere. In particular, $g(A)=y_{n}$ for almost every $A$. However, this implies that $R_{n}\in\varphi$ for some $n$ making the ultrafilter $\varphi$ $\sigma$-complete. On the other hand, it is well known that there are no non-principal $\sigma$-ultrafilters on $P_{0}(\mathbb{N})$ since $|P_{0}(\mathbb{N})|$ is far below the first measurable cardinal if one even exists. We conclude that $Y\not\subseteq A$ for almost every $A$.

Now let $t\in\mathcal{A}$ be a function where if $y_{1}\in A\subseteq\mathbb{N}$ and $Y\not\subseteq A$, then $t(A)=y_{n}$ where $y_{n+1}\not\in A$. Then $t(A)=y_{n}$ for almost all $A\in P_{0}(X)$. Then the function $t$ is constant almost everywhere. In particular, $t(A)=y_{n}$ for almost all $A\in P_{0}(X)$. However, this implies that $y_{n+1}\not\in A$ for almost all $n$. This contradicts the fact that $f_{n+1}(A)=y_{n+1}$ for almost every $A\in P_{0}(X)$. We therefore conclude that $\varphi$ can only be a principal ultrafilter.

• Maybe it would be better to post someone else's answer as a community wiki answer. – TMM May 13 '13 at 13:13
• @TMM made it Community wiki. I asked someone in chat but seems like there has been a missunterstanding. – Dominic Michaelis May 13 '13 at 13:24