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

Let $U$ is a set. I will speak about filters on this set.

If $f$ is a function and $a$ is a filter then I define $f \left[ a \right]$ as the filter whose base is $\lbrace f[A] | A \in a \rbrace$.

I will call super-embedding-1 of filter $a$ into filter $b$ a function $f$ such that $f \left[ a \right] \subseteq b$ and super-embedding-2 of filter $a$ into filter $b$ a function $f$ such that $f \left[ a \right] = b$.

Let define preorders $\leqslant_1$ and $\leqslant_2$ on the set of filters:

$b \leqslant_1 a$ if there are super-embedding-1 from $a$ to $b$ and $b \leqslant_2 a$ if there are super-embedding-2 from $a$ to $b$.

Question 1: $\leqslant_1$ is the same as $\leqslant_2$?

Filters $a$ and $b$ are isomorphic if exists a bijective super-embedding-2 $f$ from $a$ to $b$ such that $f^{- 1}$ is super-embedding-2 from $b$ to $a$. For two other equivalent characterizations of isomorphic filters see this blog post and this blog post (the second blog post requires this article).

Being isomorphic is an equivalence relation. I will call classes of filters equivalence classes under the being isomorphic relation. I will call classes of ultrafilters these classes of filters which contain ultrafilters.

Further I will denote $i = 1, 2$. So every open problem below is in fact two problems.

Question 2: Is $\leqslant_i$ for ultrafilters the same as Rudin-Keisler order (paragraph 9 of Comfort and Negrepontis ``The Theory of Ultrafilters'') of ultrafilters? If not, how they are related?

Question 3: Is the preorder of classes of filters induced by $\leqslant_i$ a partial order?

Question 4: Is the preorder of classes of ultrafilters induced by $\leqslant_i$ a partial order?

Question 5: If it is a partial order, is it a linear order?

Question 6: If it is a linear order, is it a well-order (or maybe anti-well-order)?

Question 7: If in the above definition of isomorphic filters super-embedding-2 is replaced with super-embedding-1, does it remain equivalent to the above definition?

share|improve this question
4  
I don't understand the downvotes (at least two, since I gave +1), for what appears to be a reasonable question about modified versions of the Rudin-Keisler order. –  Joel David Hamkins Jan 24 '11 at 15:42

1 Answer 1

up vote 8 down vote accepted

I understand your question better now.

First, in your general context of filters the relations $\leq_1$ and $\leq_2$ are not the same. To see this, let $G=\{I\}$ be the trivial filter on a set $I$ with at least two points, and let $\mu$ be any nonprincipal ultrafilter on $I$. Since $G\subset \mu$, we see that $\mu\leq_1 G$ as witnessed by the identity function $i$ on $I$. (Details: since $i[I]=I$, it follows that $i[G]$ is the filter with base $\{I\}$, which is the same as $G$. So $i[G]=G$, which is a subset of $\mu$, and so $\mu\leq_1 G$.) Meanwhile, I claim that $\mu\not\leq_2 G$. To see this, observe that for any function $f:I\to I$, we have $f[G]$ is the filter with base $\{f[I]\}$, and so $f[G]\neq\mu$ since $\mu$ is nonprincipal.
So the relations are different.

Note also that if $\mu$ is an ultrafilter on $I$ and $F\leq_1 \mu$ via the function $f$ for a filter $F$, then $F$ is an ultrafilter. The reason is that if $Y\notin F$, then $f^{-1}Y\notin\mu$ and so $f^{-1}(I-Y)\in\mu$, which implies $f[f^{-1}(I-Y)]\in F$, which implies $I-Y\in F$, so $F$ is an ultrafilter.

Next, I claim that for ultrafilters, the relations are the same.

Theorem. If $\nu$ is an ultrafilter, then $F\leq_1\nu\iff F\leq_2\nu$.

Proof. It suffices to prove the forward direction. Suppose $\nu$ is an ultrafilter on a set $J$ and $F$ is a filter on $I$ and $F\leq_1\nu$ as witnessed by $f:J\to I$. So $f[\nu]\subset F$. Consider any $X\in F$. If $f^{-1}X\in\nu$, then we get $X\supset f[f^{-1}X]\in f[\nu]$ and so $X\in f[\nu]$. Otherwise, since $\nu$ is an ultrafilter, we have $f^{-1}(I-X)\in\nu$ and so $I-X\supset f[f^{-1}(I-X)]\in f[\nu]\subset F$, which would put disjoint sets in $F$, a contradiction. QED

Finally, I claim that for ultrafilters, the relation $\leq_2$ is the same as the Rudin-Keisler order. The usual definition of this order is that if $F$ is a filter on $J$ and $f:J\to I$ is any function, then one we may define a filter $G=f*F$ on $I$ by $X\in G\leftrightarrow f^{-1}X\in F$. The Rudin-Keisler order is defined so that $G\leq_{RK} F$ if and only if there is $f$ for which $G=f*F$.

Suppose $F$ is a filter on $J$ and $f:J\to I$. I claim generally that $f*F=f[F]$. This is because $Y\subset f^{-1}f[Y]$ for $Y\subset J$ shows that $f[F]\subset f*F$; and conversely $f[f^{-1}X]\subset X$ for $X\subset I$ shows $f*F\subset f[F]$.

It follows that $\leq_2$ is the same as the Rudin-Keisler order.

share|improve this answer
    
I don't understand you. What you say about measures? In my original problem measures were not mentioned. What is "the filter generated by $a$ on $J$"? My notion of isomorphism is not too strong, it is carefully balanced. You say "there is no surjection from $I$ to $J$" what is irrelevant because I require existence of a function not a surjection. It seems for me that your entire answer is pointless (or do I misunderstand?) –  porton Jan 24 '11 at 11:41
    
If $b=f[a]$, then $f$ must be surjective, in order that $J\in f[a]$, since $J\in b$ for sure. So if there is no surjection from $I$ to $J$, then $b=f[a]$ is impossible. (My remarks about measures are just about countably complete ultrafilters, which are the same as 2-valued measures measuring all sets.) –  Joel David Hamkins Jan 24 '11 at 12:29
    
You misunderstood me. By $b=f[a]$ where $a$ is a filter I mean that $b$ is the filter generated by the base $\lbrace f[A] | A \in a \rbrace$. Sorry, my notation is ambiguous. $f[a]$ for $a$ a filter on $U$ is not the same as $f[a]$ where $a$ is a subset of $U$. You should infer the meaning from context. –  porton Jan 24 '11 at 12:35
    
Yes, I misunderstood you, and how have now edited my answer. –  Joel David Hamkins Jan 24 '11 at 14:43
    
First correct "G\leq_{RK]" F into "G\leq_{RK}". Second, I don't understand what $f*F$ is because it is not mentioned in the formula $X\in G\leftrightarrow f^{-1}X\in F$ which should define it. Third, where you've got that definition of Rudin-Kiesler order? I am now reading Comfort and Negrepontis ``The Theory of Ultrafilters''. There that order is defined in a different way and sadly it is not shown that it is equivalent to your definition. What you'd recommend me to read about this order? –  porton Jan 24 '11 at 15:09

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.