Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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: $$ \{ \operatorname{image}f|_M \ : \ M \in \varphi\}$$

On the bottom of the post I included some notation, I hope this avoids notational issues.

At first we should show that $\varphi$ is already an ultrafilter.

Now we shall show that if $X=\mathbb{Z}$, then it follows that $\varphi$ is a one point filter.

For the first part my attempt was to use that a filter is a ultra filter when every set or its complement is in the filter. At first we look at the case where $x$ is independent of $f$ and then (somehow) conclude that $\varphi$ msut already be an ultrafilter. If this is done we could look at the equivalence relation $\sim$ given by \[ f \sim g \iff x_f=x_g \] Now we use the equivalence classes given by this equivalence relation. If I am right, this reduce the problem again to the case where $x$ is independent of $f$.

For the second part my only idea was looking the the properties of $\mathbb{Z}$ which could help us (that the ultrafilter is already a one point filter). I think one of the properties which help is that $\mathbb{Z}$ is countable, but I don't see how it helps.

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 $A\in F \wedge S\setminus A \in F$.

Maybe this question is not appropiate to MathOverflow cause it is actually not a research question. I faced this problem on the group exercises of my topology class I am in (so nothing which should be handed in). If it is not welcome I will delete it just leave a comment. I already posted that question on Math Stack Exchange

share|improve this question
1  
I wonder if there is a topological proof of this result or of similar results. Let $\mathcal{B}$ be the set $\mathcal{A}$ unioned with the set of all functions $f:P_{0}(X)\rightarrow X$ with finite range. Give $X$ the discrete uniformity and give $P_{0}(X)$ the coarsest uniformity such that every $f\in\mathcal{B}$ is uniformly continuous. Then the completion of $P_{0}(X)$ is precisely the set of all ultrafilters $U$ on $P_{0}(X)$ such that every mapping $f\in\mathcal{A}$ is constant on some set in $U$. Thus, it suffices to show that $P_{0}(X)$ is a complete uniform space. –  Joseph Van Name May 8 '13 at 4:20
    
In other words, one needs to show that $\\{(f(A))_{f\in\mathcal{B}}|A\in P_{0}(X)\\}$ is a closed subspace of the product space $X^{\mathcal{B}}$. –  Joseph Van Name May 8 '13 at 12:20
1  
I edited my answer to also include the topological proof that every ultrafilter such that each function $f:P_{0}(X)\rightarrow X$ is constant almost everywhere. I however should mention that the topological proof is very similar to the purely combinatorial proof. –  Joseph Van Name May 9 '13 at 23:22
add comment

1 Answer

up vote 7 down vote accepted

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.

$\textbf{Added 5/9/13}$

I mentioned in the comments how this problem can be phrased in terms of uniform spaces. I will now give a topological proof of the fact that every ultrafilter $\varphi$ on $P_{0}(\mathbb{N})$ where every $f\in\mathcal{A}$ is constant $\varphi$-a.e. is simply a principal ultrafilter. The topological proof of this fact that I am about to give is very similar to the purely combinatorial proof of this fact above. In order to translate this problem involving ultrafilters into a problem involving uniform spaces we will need the following result and definitions.

A uniform space $(X,\mathcal{U})$ is said to be non-Archimedean if $\mathcal{U}$ is generated by equivalence relations. For example, if $X$ is a compact space, then there is a unique uniformity $\mathcal{U}$ on $X$ compatible with the topology on $X$ and this uniformity is non-Archimedean if and only if $X$ is totally disconnected.

We define a partition space to be a pair $(X,F)$ where $X$ is a set and $F$ is a filter on the lattice of partitions of $X$. Clearly the partition spaces are in a one-to-one correspondence with the non-Archimedean uniform spaces since the partitions of a set are in a one-to-one correspondence with the filters on that set.

If $(X,F)$ is a partition space, then let $\mathfrak{B}^{*}(X,F)=\{\emptyset\}\cup\bigcup F$. Then $\mathfrak{B}^{*}(X,F)$ is a Boolean algebra. In particular, if $R\subseteq X$, then $R\in\mathfrak{B}^{*}(X,F)$ if and only if $\{R,R^{c}\}\setminus\{\emptyset\}\in F$ if and only if the characteristic function $\chi_{R}$ is uniformly continuous.

An ultrafilter $\mathcal{U}$ on $\mathfrak{B}^{*}(X,F)$ is said to be an $F$-ultrafilter if $\mathcal{U}\cap P\neq\emptyset$ for each $P\in F$.

$\mathbf{Theorem}$ Let $(X,F)$ be a separated partition space. Then the following are equivalent.

  1. $(X,F)$ is complete (as a uniform space,i.e. Every Cauchy filter converges).

  2. Every $F$-ultrafilter on the Boolean algebra $\mathfrak{B}^{*}(X,F)$ is of the form $\{R\in\mathfrak{B}^{*}(X,F)|x\in R\}$ for some $x\in X$.

$\mathbf{Proof}$(Outline)

$1\rightarrow 2$. If $(X,F)$ is complete and $\mathcal{U}$ is an $F$-ultrafilter, then $\mathcal{U}$ is a Cauchy filter. Therefore $\mathcal{U}$ converges to some point $x\in X$. However, $\{R\in\mathfrak{B}^{*}(X,F)|x\in R\}$ is the only $F$-ultrafilter that converges to the point $x\in X$.

$2\rightarrow 1$. Assume that every $F$-ultrafilter on $\mathfrak{B}^{*}(X,F)$ is of the form $\{R\in\mathfrak{B}^{*}(X,F)|x\in R\}$. Let $Z$ be a Cauchy filter on $X$. Then $Z\cap\mathfrak{B}^{*}(X,F)$ is an $F$-ultrafilter, so $Z\cap\mathfrak{B}^{*}(X,F)=\{R\in\mathfrak{B}^{*}(X,F)|x\in R\}$ for some $x\in X$. It turns out that the filter $Z$ converges to the point $x$. $\mathbf{QED}$

We shall give the space $\mathbb{N}$ the discrete uniformity. Let $\mathcal{B}=\mathcal{A}\cup\{f:P_{0}(\mathbb{N})\rightarrow\mathbb{N}:|f[P_{0}(\mathbb{N})]|<\infty\}$. In other words, $\mathcal{B}$ is the union of the set $\mathcal{A}$ with the set of all functions $f:P_{0}(\mathbb{N})\rightarrow\mathbb{N}$ with bounded range. Now give $P_{0}(\mathbb{N})$ the coarsest uniformity such that every function $f\in\mathcal{B}$ is a uniformly continuous function. Let $(P_{0}(\mathbb{N}),F)$ be the corresponding partition space structure. Then $\mathfrak{B}^{*}(P_{0}(\mathbb{N}),F)=P(P_{0}(\mathbb{N}))$, and the $F$-ultrafilters are precisely the ultrafilters on the set $P_{0}(\mathbb{N})$ such that every function $f\in\mathcal{A}$ is constant almost everywhere. Therefore by the above theorem, every ultrafilter $\mathcal{U}$ on $P_{0}(\mathbb{N})$ such that every $f\in\mathcal{A}$ is constant $\mathcal{U}$-a.e. is principal if and only if the space $P_{0}(\mathbb{N})$ is a complete uniform space. We shall now prove that the uniform space $P_{0}(\mathbb{N})$ is complete.

Let $L:P_{0}(\mathbb{N})\rightarrow\mathbb{N}^{\mathcal{B}}$ be the function where $L(A)=(f(A))_{f\in\mathcal{B}}$ for each $A\in P_{0}(\mathbb{N})$. Give $\mathbb{N}^{\mathcal{B}}$ the product uniformity where $\mathbb{N}$ still has the discrete uniformity. Then $\mathbb{N}^{\mathcal{B}}$ is a complete uniform space since the product of complete uniform spaces is always a complete uniform space. Therefore the space $P_{0}(\mathbb{N})$ is complete if and only if the image $L[P_{0}(\mathbb{N})]$ is a complete subspace of $\mathbb{N}^{\mathcal{B}}$. However, the space $L[P_{0}(\mathbb{N})]$ is complete if and only if $L[P_{0}(\mathbb{N})]$ is a closed subspace of $\mathbb{N}^{\mathcal{B}}$. We therefore only need to show that $L[P_{0}(\mathbb{N})]$ is closed in $\mathbb{N}^{\mathcal{B}}$.

Let $(x_{f})_{f\in\mathcal{B}}\in \overline{L[P_{0}(\mathbb{N})]}$. Then since every neighborhood of $(x_{f})_{f\in\mathcal{B}}$ intersects $L[P_{0}(\mathbb{N})]$, whenever $h_{1},...,h_{n}\in\mathcal{B}$ there is some $A\in P_{0}(\mathbb{N})$ where $x_{h_{1}}=h_{1}(A),...,x_{h_{n}}=h_{n}(A)$.

As before, let $f_{n}\in\mathcal{A}$ be the function where $f_{n}(A)$ 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 $y_{n}=x_{f_{n}}$ for all $n$. Then for all $n$, there is some $A\in P_{0}(\mathbb{N})$ with $y_{i}=x_{f_{i}}=f_{i}(A)$ for $1\leq i\leq n$. In particular, since the sequence $(f_{i}(A))_{i\in\mathbb{N}}$ is increasing for all $A$, we conclude that the sequence $(y_{n})_{n\in\mathbb{N}}$ is increasing.

Assume that $x_{f_{n}}=y_{n}=y_{n+1}=x_{f_{n+1}}$ for some $n$. Then there is an $A\in P_{0}(\mathbb{N})$ with $f_{i}(A)=x_{f_{i}}=y_{i}$ for $1\leq i\leq n+1$. Furthermore, one can clearly see that the $A\in P_{0}(\mathbb{N})$ with $f_{i}(A)=x_{f_{i}}=y_{i}$ for $1\leq i\leq n+1$ is unique. In particular $A=\{y_{1},...,y_{n}\}$. Let $U$ be the neighborhood of $(x_{f})_{f\in\mathcal{B}}$ consisting of all systems $(z_{f})_{f\in\mathbb{B}}$ such that $z_{f_{1}}=x_{f_{1}},...,z_{f_{n+1}}=x_{f_{n+1}}$. Then $U$ is an open set, but $U$ intersects $L[P_{0}(\mathbb{N})]$ at only one point, namely the point $L(\{y_{1},...,y_{n}\})=(f(\{y_{1},...,y_{n}\}))_{f\in\mathcal{B}}$. Therefore since $U$ is an open neighborhood of $(x_{f})_{f\in\mathcal{B}}$, $U$ intersects $L[P_{0}(\mathbb{N})]$ at only the point $L(\{y_{1},...,y_{n}\})$, and $(x_{f})_{f\in\mathcal{B}}\in\overline{L[P_{0}(\mathbb{N})]}$, we conclude that $(x_{f})_{f\in\mathcal{B}}=L(\{y_{1},...,y_{n}\})$.

Now assume that $y_{n}<y_{n+1}$ for all natural numbers $n$. Let $Y=\{y_{n}|n\in\mathbb{N}\}$. We shall now show that $(x_{f})_{f\in\mathcal{B}}=L(Y)$.

We first claim that there is a neighborhood $U$ of $(x_{f})_{f\in\mathcal{B}}$ such that if $L(A)\in U$, then $Y\subseteq A$.

Let $h\in\mathcal{B}$ be the function where $h(A)=1$ if $Y\subseteq A$ and $h(A)=2$ if $Y\not\subseteq A$. Let $t$ be a function where if $Y\not\subseteq A$ and $y_{1}\in A$, then $t(A)=y_{n-1}$ where $n$ is the least natural number with $y_{n}\not\in A$.

I claim that $x_{h}=1$. Suppose to the contrary that $x_{h}\neq 1$. Then there is some $A$ with $h(A)=x_{h},f_{1}(A)=x_{f_{1}}=y_{1},t(A)=x_{t}$.

Since $x_{h}\neq 1$, we have $h(A)=2$, so $Y\not\subseteq A$. Since $f_{1}(A)=y_{1}$, we have $y_{1}\in A$. Therefore by the definition of $t$, we have $x_{t}=t(A)=y_{n-1}$ where $n$ is the least natural number with $y_{n}\not\in A$.

Now, there is some $B\in P_{0}(\mathbb{N})$ with $h(B)=x_{h},t(B)=x_{t}=y_{n},f_{1}(B)=x_{f_{1}}=y_{1},f_{n+1}(B)=x_{f_{n+1}}=y_{n+1}$. Then since $f_{n+1}(B)=y_{n+1}$, we have $y_{n+1}\in B$. However, $f_{1}(B)=y_{1}$, so $y_{1}\in B$ and since $h(B)=x_{h}\neq 1$, we have $Y\not\subseteq B$. Therefore since $t(B)=y_{n}$, we have $y_{n+1}\not\in B$. This is a contradiction. Therefore, we conclude that $x_{h}=1$.

Let $g$ be the function where $g(A)$ is the least element of $A\setminus Y$ whenever $A\not\subseteq Y$ and $g(A)$ is the least element of $A$ whenever $A\subseteq Y$.

We claim that $x_{g}\in Y$. To prove this claim, suppose to the contrary that $x_{g}\not\in Y$. Then $x_{g}<y_{n}$ for some $n$. Thus, there is some $A\in P_{0}(\mathbb{N})$ with $g(A)=x_{g},f_{1}(A)=x_{f_{1}}=y_{1},...,f_{n}(A)=x_{f_{n}}=y_{n}$. Therefore the first $n$ elements of $A$ are $y_{1},...,y_{n}$. This contradicts the fact that $x_{g}\in A\setminus Y$ and $x_{g}<y_{n}$. We therefore conclude that $x_{g}\in Y$.

Now let $U\subseteq\mathbb{N}^{\mathcal{B}}$ be the set where $(z_{f})_{f\in\mathcal{B}}$ if and only if $z_{g}=x_{g}$ and $z_{h}=x_{h}$. Then $U$ is an open neighborhood of $(x_{f})_{f\in\mathcal{B}}$.

If $A\in P_{0}(X)$ and $(f(A))_{f\in\mathcal{B}}=L(A)\in U$, then $g(A)=x_{g}=h(A)=x_{h}$. Therefore $h(A)=x_{h}=1$ and $g(A)=x_{g}\in Y$. Since $h(A)=x_{h}=1$, we conclude that $Y\subseteq A$. However, since $g(A)=x_{g}\in Y$, we conclude that $A\subseteq Y$ as well, so $A=Y$. We conclude that $L(Y)$ is the only possible point in $L[P_{0}(\mathbb{N})]\cap U$. Therefore since $U$ is a neighborhood of $(x_{f})_{f\in\mathcal{B}}$ and $(x_{f})_{f\in\mathcal{B}}\in\overline{L[P_{0}(\mathbb{N})]}$, we conclude that $(x_{f})_{f\in\mathcal{B}}=L(Y)$. Thus, the set $L[P_{0}(\mathbb{N})]$ is a closed set in $\mathbb{N}^{\mathcal{B}}$.

$\textbf{Added 5/10/13}$

I am now going to completely characterize the cardinalities $|X|$ such that if $\varphi$ is a filter where every function in $\mathcal{A}$ is constant almost everywhere then $\varphi$ is a principal ultrafilter. This characterization is a generalization of the proof that I gave at the top of this post. As one who is familiar with large cardinals might expect, the every such ultrafilter is a principal ultrafilter if and only if $|X|$ is non-measurable.

We say that a cardinal $\lambda$ is non-measurable if there is no non-principal $\sigma$-complete ultrafilter on $\lambda$. This is equivalent to saying that $\lambda$ is below the first measurable cardinal.

$\mathbf{Theorem}$ Let $X$ be a set.

I. Let $\varphi$ be a filter on $P_{0}(X)$ such that every function $f\in\mathcal{A}$ is constant $\varphi$-a.e. Then $\varphi$ is an $\sigma$-complete ultrafilter.

II. $|X|$ is non-measurable if and only if every filter $\varphi$ on $P_{0}(X)$ such that every function $f\in\mathcal{A}$ is constant almost everywhere is a principal ultrafilter.

$\mathbf{Proof}$. I. For each $f\in\mathcal{A}$, let $x_{f}$ be the constant with $x_{f}=f$ $\varphi$-a.e. Without loss of generality, assume that $X=\lambda$ for some cardinal $\lambda$. For all $n\in\omega$, let $f_{n}\in\mathcal{A}$ be the function where $f_{n}(A)$ 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 $y_{n}=x_{f_{n}}$ for all $n$. Let $Y=\{y_{n}|n>0\}$. Clearly the sequence $(y_{n})_{n}$ is increasing.

If $y_{1}=y_{2}$, then $\varphi$ is a principal ultrafilter since if $f_{1}(A)=y_{1},f_{2}(A)=y_{2}$, then $A=\{y_{1}\}$, so since $f_{1}(A)=y_{1},f_{2}(A)=y_{2}$ $\varphi$-a.e., we have $\varphi$ be the principal ultrafilter generated by $A$. Now assume that $y_{1}<y_{2}$.`

Let $S=\{A\in P_{0}(\lambda)|f_{1}(A)=y_{1},f_{2}(A)=y_{2}\}$. 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 $A\in R\cap S$ and $h(A)=y_{2}$ for $A\in R^{c}\cap S$. Then $h$ is constant almost everywhere and $h(A)\in\{y_{1},y_{2}\}$ for almost every $A\in P_{0}(\lambda)$ and $h(A)=y_{2}$. Therefore $h(A)=y_{1}$ for almost every $A\in P_{0}(\lambda)$ or $h(A)=y_{2}$ for almost every $A\in P_{0}(\lambda)$. If $h(A)=y_{1}$ for almost every $A\in P_{0}(\lambda)$, then $R\cap S\in\varphi$ and if $h(A)=y_{2}$ a.e., then $R^{c}\cap S\in\varphi$. Therefore either $R\in\varphi$ or $R^{c}\in\varphi$. Therefore $\varphi$ is an ultrafilter.

If $y_{n}=y_{n+1}$, and $f_{1}(A)=y_{1},...,f_{n+1}(A)=y_{n+1}$, then clearly $A=\{y_{1},...,y_{n}\}$, but since $f_{1}(A)=y_{1},...,f_{n+1}(A)=y_{n+1}$ for almost every $A$, we conclude that $\varphi$ is the principal ultrafilter generated by $A$.

Now assume that $y_{n}<y_{n+1}$ for all $n$. Let $Y=\{y_{n}|n\in\mathbb{N},n>0\}$. We claim that $Y\subseteq A$ for almost every $A\in P_{0}(\mathbb{N})$. Suppose to the contrary that $Y\not\subseteq A$ for almost every $A\in P_{0}(\mathbb{N})$. Let $t\in\mathcal{A}$ be a function such that if $Y\not\subseteq A$ and $y_{1}\in A$, then $t(A)=y_{n-1}$ where $n$ is the least natural number where $y_{n}\not\in A$. Since $Y\not\subseteq A$ and $y_{1}\in A$ for almost every $A\in P_{0}(\lambda)$, we have $t(A)\in Y$ for almost every $A\in Y$. Since the function $t$ is constant almost everywhere, we conclude that $x_{t}=y_{n}$ for some $n$. Therefore, $t(A)=y_{n},f_{1}(A)=y_{1},f_{n+1}(A)=y_{n+1}$ and $Y\not\subseteq A$ for almost every $A\in P_{0}(\lambda)$. However, this is a contradiction since for such $A$ we have that $t(A)=y_{n},f_{1}(A)=y_{1},Y\not\subseteq A$ implies that $y_{n+1}\not\in A$, but clearly $y_{n+1}=f_{n+1}(A)\in A$. Therefore, we conclude that $Y\subseteq A$ for almost every $A\in P_{0}(\mathbb{N})$.

Since $Y$ is a countable set and $Y\subseteq A$ for almost every $A$, it is now fairly easy to show that the ultrafilter $\varphi$ is $\sigma$-complete. If $P=\{R_{n}|n\in\mathbb{N},n>0\}$ is a partition of $P_{0}(\lambda)$ into countably many pieces, then let $g$ be the function where if $Y\subseteq A$ and $A\in R_{n}$, then $g(A)=y_{n}$. Then $g$ is constant almost everywhere. Furthermore, since $Y\subseteq A$ for almost every $A$, we have $g(A)\in Y$ for almost every $A$. Therefore there is some $n$ where $g(A)=y_{n}$ for almost every $A$. Therefore since $g(A)=y_{n}$ and $Y\subseteq A$ for almost every $A$, and if $g(A)=y_{n}$ and $Y\subseteq A$, then $A\in R_{n}$. Therefore $A\in R_{n}$ for almost every $A$, so $R_{n}\in\varphi$. Since the ultrafilter $\varphi$ selects an element from every countable partition of $P_{0}(\lambda)$, then ultrafilter $\varphi$ is $\sigma$-complete.

II. $\rightarrow.$ If $X$ is of non-measurable cardinality, then $P_{0}(X)$ is of non-measurable cardinality as well. Furthermore, since every filter $\varphi$ such that every function $f\in\mathcal{A}$ is constant almost everywhere is a $\sigma$-complete ultrafilter, we have $\varphi$ be a principal ultrafilter.

$\leftarrow.$ If $X=\lambda$ is of measurable cardinality, then there is some measurable cardinal $\mu$ with $\mu\leq\lambda$. Since every measurable cardinal $\mu$ is $\mu$-supercompact, there is a normal ultrafilter $U$ over $[\mu]^{<\mu}=\{A\subseteq \mu:0<|A|<\mu\}$. Therefore if $\varphi$ is the ultrafilter over $P_{0}(\lambda)$ where $R\in\varphi$ if and only if $R\cap[\mu]^{<\mu}\in U$, then every $f\in\mathcal{A}$ is constant $\varphi$-a.e., but $\varphi$ is a non-principal ultrafilter. $\mathbf{QED}$

share|improve this answer
    
@Joseph Van Name are you on Math Stack Exchange too? Because if so, I can give you the bounty there. –  Dominic Michaelis May 9 '13 at 16:23
1  
I am not on Math Stack Exchange. –  Joseph Van Name May 9 '13 at 18:04
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.