On the intersection of finitely many ultrafilters

Question

I am interested in characterizing those filters that can be written as an intersection of finitely many ultrafilters. I would appreciate any reference on this topic.

Answer 1

Let me try to provide a helpful elementary answer.

Suppose that $F$ is a filter on a set $X$ arising as the intersection of finitely many ultrafilters $$F=\mu_1\cap\cdots\cap\mu_n.$$ We may assume that the $\mu_i$ are distinct. Since the filters $\mu_i$ are different, they must disagree on some subsets of $X$. For each pair $i\neq j$ we can find a set in $\mu_i$ and not in $\mu_j$, and by intersecting these sets, we can produce a partition $$X=X_1\sqcup X_2\sqcup\cdots\sqcup X_n$$ such that each $\mu_i$ concentrates on the $i$th piece $X_i$ (and therefore gives measure zero to the other pieces).

Note that a subset $Y\subseteq X$ is in $F$ just in case $Y\cap X_i\in\mu_i$ for every $i$.

Let me define that that a filter $F$ localizes to an ultrafilter on $Y\subseteq X$ if $\{Z\subseteq Y\mid Z\cup(X\setminus Y)\in F\}$ is an ultrafilter on $Y$.

What we've observed is that a filter $F$ is a finite intersection of ultrafilters if and only if there is a finite partition such that $F$ localizes to an ultrafilter on each piece.

A further conclusion is that if a filter $F$ is a finite intersection of ultrafilters $F=\mu_1\cap\cdots\cap\mu_n$, then the number $n$ is determined from $F$, and furthermore the ultrafilters $\mu_i$ are also determined from $F$.

To see this, if we partition $X$ into more than $n$ pieces, then some piece must not be in any $\mu_i$ and so will have measure $0$ with respect to $F$. So $n$ is determined.

For the measures $\mu_i$ being determined, I claim that $F\subseteq\nu$ for an ultrafilter $\nu$ if and only if $\nu=\mu_i$ for some $i$. If $F$ is contained in $\nu$ and $\nu\neq\mu_i$ for any $i$, then we can find a set $Y\in\nu$, $Y\notin\mu_i$ any $i$. But in this case the complement of $Y$ is in $F$, contrary to $F\subseteq\nu$.

So the ultrafilters that are used are exactly the ultrafilter completions of $F$. Putting this all together, what we have is:

Theorem. The following are equivalent for any filter $F$:

1. $F$ arises as a finite intersection of ultrafilters.
2. $F$ has only finitely many completions to an ultrafilter.
3. There is a finite partition of the underlying set such that $F$ localizes to an ultrafilter on each piece.
4. For some finite $n$, there is a partition of $F$ into $n$ pieces, none $F$-measure 0, but no such partition with $n+1$ pieces.

To see that 4 suffices for the others, if we have $X=X_1\sqcup\cdots X_n$ and each $X_i$ is not $F$-measure 0, then let $\mu_i$ be the localization of $F$ to $X_i$. If this is not an ultrafilter, then we can further partition $X_i$ into two pieces not of measure $0$, and make such a partition of size $n+1$, contrary to 4.

Answer 2

There is a category theoretic characterization of the filters that can be written as intersections of finitely many ultrafilters. I claim that a filter $Y$ on $G$ is the intersection of finitely many ultrafilters if and only if there are only finitely many monomorphisms $\iota:(X,F)\rightarrow(Y,G)$ up to composition with an isomorphism. I also claim that the filters that are intersections of finitely many ultrafilters are precisely the coproducts of finitely many ultrafilters in the category of all filters.

If $\mathcal{C}$ is a category, then recall that a morphism $f:X\rightarrow Y$ is a monomorphism precisely when for all objects $g_1,g_2:Q\rightarrow X$, we have the cancellation property $f\circ g_1=f\circ g_2\Rightarrow g_1=g_2$. If $\mathcal{C}$ is a category and $X$ is an object in $\mathcal{C}$, then let $M(Y)$ be the class of all monomorphisms $f:X\rightarrow Y$ for some object $X$ in $\mathcal{C}$ (we write $M_\mathcal{C}(Y)$ for $M(Y)$ to specify the category $\mathcal{C}$). Let $\simeq$ be the equivalence relation on $M(Y)$ where if $f_1:X_1\rightarrow Y,f_2:X_2\rightarrow Y$, then $f_1\simeq f_2$ if and only if there is an isomorphism $g:X_1\rightarrow X_2$ where $f_1=f_2\circ g$.

Let $\mathcal{F}$ be the category whose objects are the pairs $(X,F)$ where $X$ is a set and $F$ is a filter on $X$. If $(X,F),(Y,G)$ are objects in $\mathcal{F}$, then we say that a function $f:X\rightarrow Y$ is a morphism from $(X,F)$ to $(Y,G)$ if $R\in G\Rightarrow f^{-1}[R]\in F$. The composition in $\mathcal{F}$ is the ordinary composition of functions. Let $\mathcal{G}$ be the quotient category of $\mathcal{F}$ by the congruence $\simeq$ where if $f,g\in\text{Hom}((X,F),(Y,G))$, then $f\simeq g$ precisely when $\{x\in X:f(x)=g(x)\}\in F$. The paper Two Closed Categories of Filters by Andreas Blass gives more information on these categories $\mathcal{F},\mathcal{G}$.

Lemma: Suppose that $f\in\text{Hom}_\mathcal{F}((X,F),(Y,G))$. Then $[f]$ is a monomorphism in $\mathcal{G}$ if and only if there is some $R\in F$ where $f|_R$ is injective.

Proof:

$\leftarrow.$ Suppose that $f|_R$ is injective for some $R\in F$. Suppose that $[g_1],[g_2]:(Q,M)\rightarrow(X,F)$ are morphisms in $\mathcal{G}$ with $[f\circ g_1]=[f\circ g_2]$. Let $S=\{q\in Q:(f\circ g_1)(q)=(f\circ g_2)(q)\}.$ Then $S\in M$. Therefore, $g^{-1}_1[R]\cap S\cap g^{-1}_2[R]\in M$, but if $q\in S\cap g_1^{-1}[R]\cap g_2^{-1}[R]$, then $(f\circ g_1)(q)=(f\circ g_2)(q)$, but since $q\in g_1^{-1}[R]\cap g_2^{-1}[R]$, we know that $g_1(q),g_2(q)\in R$, so since $f|_R$ is injective, we conclude that $g_1(q)=g_2(q)$. We may therefore conclude that $[f\circ g_1]=[f\circ g_2].$

$\rightarrow.$ Suppose that $f|_R$ is non-injective whenever $R\in F$. Then let $Z=\ker(f)$, and let $\pi_1,\pi_2:Z\rightarrow X$ be the projection mappings. Then observe that $f\circ\pi_1=f\circ\pi_2$. Let $M$ be the filter on $Z$ generated by $\{\pi_1^{-1}[R]\cap\pi_2^{-1}[R]|R\in F\}$. Then the mappings $\pi_1,\pi_2$ are morphisms from $(Z,M)$ to $(X,F)$. On the other hand, if $R\in F$, then there are distinct $x,y\in R$ with $f(x)=f(y)$. Therefore, $(x,y)\in Z$ and also $(x,y)\in\pi_1^{-1}[R]\cap\pi_2^{-1}[R]$, but $\pi_1(x,y)\neq\pi_2(x,y)$. This means that $\pi_1|_T\neq\pi_2|_T$ for each $T\in M$. Therefore, $[\pi_1]\neq[\pi_2]$, so the morphism $[f]$ is not a monomorphism in $\mathcal{G}$.

Q.E.D.

Lemma: Suppose that $(X_i,F_i)$ is an object in $\mathcal{G}$ for $i\in I$. Then the coproduct of $(X_i,F_i)_{i\in I}$ is the object $(\bigcup_{i\in I}X_i,F)$ where $R\in F$ if and only if $R\cap X_i\in F_i$ for $i\in I$.

Proof: Let $\iota_i:X_i\rightarrow\bigcup_{i\in I}X_i$ be the inclusion mapping. Then $\iota_i$ is a morphism in $\mathcal{F}$, and therefore $[\iota_i]$ is a morphism in $\mathcal{G}$.

Let $(Y,G)$ be an object in $\mathcal{G}$. Let $f_i:(X_i,F_i)\rightarrow(Y,G)$ be a morphism in $f_i$ for $i\in I$. Then let $f:\bigcup_{i\in I}X_i\rightarrow Y$ be the union $\bigcup_{i\in I}f_i$. To see that $f$ is a morphism in $\mathcal{F}$, suppose that $R\in G$. Then $f_i^{-1}[R]\in F_i$ for $i\in I$. Thus, $f^{-1}[R]=\bigcup_{i\in I}f_i^{-1}[R]\in F$ since $f_i^{-1}[R]\in F_i$ for $i\in I$. Observe that $f_i=f\circ\iota_i$.

I now claim that $[f]$ is the unique morphism in $\mathcal{G}$ where $[f_i]=[f]\circ[\iota_i]$ for $i\in I$. Suppose that $[f_i]=[g]\circ[\iota_i]$ for $i\in I$. Then $\{x\in X_i:f_i(x)=g\circ\iota_i(x)\}\in F_i$ for $i\in I$, so $\{x\in X:f(x)=g(x)\}=\bigcup_{i\in I}\{x\in X_i:f_i(x)=g(x)\}\in F$. Therefore, $[f]=[g]$. We conclude that $(\bigcup_{i\in I}X_i,F)$ is the coproduct of the filters $(X,F_i)_{i\in I}$

Q.E.D.

Proposition: Suppose that $f:(X,F)\rightarrow(Y,G)$ is a morphism in the category $\mathcal{F}$. Then let $H=\{S\subseteq Y:f^{-1}[S]\in F\}$. Then the filter $H$ is generated by $\{f[R]:R\in F\}.$

Proof: If $R\in F$, then $R\subseteq f^{-1}[f[R]]$, so $f^{-1}[f[R]]\in F.$ Therefore, $f[R]\in H$. On the other hand, if $S\in H$, then $f[f^{-1}[S]]\subseteq S$ and $f^{-1}[S]\in F$, so $f[f^{-1}[S]]\in H$. Therefore, $H$ is generated by the filterbase $\{f[R]:R\in F\}$.

Q.E.D.

Suppose that $f:(X,F)\rightarrow(Y,G)$ is a morphism in the category $\mathcal{F}$. Let $H=\{S\subseteq Y:f^{-1}[S]\in F\}$. Then define morphisms $f^e:(X,F)\rightarrow(Y,H),f^o:(Y,H)\rightarrow(Y,G)$ by setting $f^e(x)=f(x)$ for $x\in X$ and $f^o(y)=y$ for $y\in Y$. We observe that $f^o$ is a monomorphism.

Proposition: If $[f]$ is a monomorphism, then $[f^e]$ is an isomorphism.

Proof: If $[f]$ is a monomorphism, then there is some $R\in F$ where $f|_R$ is injective. Therefore, let $g:Y\rightarrow X$ be a function where $g(f(x))=x$ for $x\in R$. Then I claim that $g$ is a morphism in $\mathcal{F}$ from $(Y,H)$ to $(X,F)$. If $R\in F$, then $R=(g\circ f)^{-1}[R]=f^{-1}[g^{-1}[R]]$. Therefore, since $f^{-1}[g^{-1}[R]]\in F$, we know that $g^{-1}[R]\in H$. However, since $R\in F\Rightarrow g^{-1}[R]\in H$, we conclude that $g$ is a morphism from $(Y,H)$ to $(X,F)$. Clearly $[g]\circ[f^e]$ is the identity morphism. On the other hand, since $R\in F$, we know that $f[R]\in G$, but if $r\in R$, then $r=g(f(r))$, so $f(r)=f(g(f(r)))$. Therefore, $f\circ g|_{f[R]}$ is the identity mapping. We conclude that $[f^e]\circ[g]$ is the identity morphism. We conclude that $[f^e]$ is an isomorphism.

Q.E.D.

Proposition: Suppose that $f_1:(X_1,F)\rightarrow(Y,G)$ and $f_2:(X_2,F)\rightarrow(Y,G)$ are monomorphisms. Then $[f_1]\simeq[f_2]$ in $M_\mathcal{G}(Y,G)$ if and only if $[f_1^o]=[f_2^o]$.

Proof: We have already established that $[f_1^e],[f_2^e]$ are isomorphisms. Therefore, if $[f_1^o]=[f_2^o]$, then $[f_1]\simeq[f_1^o]=[f_2^o]\simeq[f_2]$.

Suppose now that $[f_1]\simeq[f_2]$. Then $[f_1^o]\simeq[f_1]\simeq[f_2]\simeq[f_2^o]$. Then there are filters $H_1,H_2\subseteq P(Y)$ with $G\subseteq H_1\cap H_2$ where $f_1^o:(Y,H_1)\rightarrow(Y,G),f_2^o:(Y,H_2)\rightarrow(Y,G)$ are isomorphisms.

There are therefore inverse isomorphisms $[p]:(H_1,Y)\rightarrow(H_2,Y),[q]:(H_2,Y)\rightarrow(H_1,Y)$ with $[f_1^o\circ p]=[f_2^o]$ and $[f_2^o\circ q]=[f_1^o].$ Therefore, there is some $R^\sharp\in H_1$ where $p(y)=f_2^o\circ p(y)=f_1^o(y)=y$ for $y\in R^\sharp\in H_1$. Similarly, there is some $S^\sharp\in H_2$ where $q(y)=y$ for $y\in S^\sharp$. If $R\in H_2$, then $p^{-1}[R]\in H_1$, but $R\cap R^\sharp=p^{-1}[R]\cap R^\sharp\in H_1$, so $R\in H_1$ as well. This means that $H_2\subseteq H_1$. For a similar reason, $H_1\subseteq H_2$. Q.E.D.

By the above proposition, the set $M_\mathcal{G}(Y,G)$ is in a one-to-one correspondence with the collection of all filters on the set $Y$ that extend $G$.

Proposition: Let $G$ be a filter on the set $Y$. Then the following are equivalent:

1. $G$ is the intersection of finitely many ultrafilters.

2. $G$ is a coproduct of finitely many ultrafilters in the category $\mathcal{G}$.

3. the set $M_\mathcal{G}(Y,G)/\simeq$ is finite.

Proof: $1\rightarrow 2.$ If $G$ is the intersection of finitely many ultrafilters, then there exists sets a partition $(X_1,\dots,X_n)$ of $X$ along with ultrafilters $U_1\subseteq P(X_1),\dots,U_n\subseteq P(X_n)$ where $R\in G$ if and only if $R\cap X_i\in U_i$ for $1\leq i\leq n$. But this means that $(X,G)$ is just the coproduct of $(X_1,U_1),\dots,(X_r,U_r)$.

$2\rightarrow 1.$ If $(X,G)$ is the coproduct of $(X_1,U_1),\dots,(X_r,U_r)$, then we can assume that $X=X_1\cup\dots\cup X_r$ and the sets $X_1,\dots,X_r$ are disjoint and that $R\in G$ if and only if $R\cap X_i\in U_i$ for $1\leq i\leq r$. Let $V_i$ be the ultrafilter on $X$ defined by setting $V_i=\{R\subseteq X:R\cap X_i\in U_i\}$. Then $G=V_1\cap\dots\cap V_r$.

$1\leftrightarrow 3.$ This follows from the above lemma since $M_\mathcal{G}(Y,G)$ corresponds with the collection of all filters on $Y$ that extend $G$.

Q.E.D.