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

Let $\mathfrak{F}$ is the complete lattice of filters (including the improper filter) on some set, ordered inverse to set-theoretic inclusion.

I will denote $\left\langle f \right\rangle \mathcal{X} = \bigcap^{\mathfrak{F}} \left\{ f \left[ X \right] | X \in \mathcal{X} \right\}$ for every binary relation $f$ and filter $\mathcal{X}$.

Let $\forall \mathcal{X}\in\mathfrak{F}:\left( \mathcal{X} \cap^{\mathfrak{F}} \mathcal{A} \neq 0^{\mathfrak{F}} \Rightarrow \left( \left\langle f \right\rangle \mathcal{X} \supseteq^{\mathfrak{F}} \mathcal{B} \wedge \left\langle g \right\rangle \mathcal{X} \supseteq^{\mathfrak{F}} \mathcal{B} \right) \right)$ for some binary relations $f$ and $g$ and filters $\mathcal{A}$ and $\mathcal{B}$. ($0^{\mathfrak{F}}$ is the filter which is the least in our order that is the biggest in set-theoretic order.)

Does the implication $\forall \mathcal{X}\in\mathfrak{F}:\left( \mathcal{X} \cap^{\mathfrak{F}} \mathcal{A} \neq 0^{\mathfrak{F}} \Rightarrow \left\langle f \cap g \right\rangle \mathcal{X} \supseteq^{\mathfrak{F}} \mathcal{B} \right)$ follow from the above assumption?

You can read http://www.mathematics21.org/algebraic-general-topology.html for my related research.

share|improve this question
    
I honestly do not know what your formulas mean... –  Mariano Suárez-Alvarez Aug 8 '11 at 21:42
    
(For example: is "the filter which is the least in our order that is the biggest in set-theoretic order" a circumlocution for "the filter of all sets"?) –  Mariano Suárez-Alvarez Aug 8 '11 at 21:44
    
@Mariano Suárez-Alvarez: Yes, this is the filter of all sets. Also $f[X] = \{ y | \exists x\in X: (x;y) \in f \}$. –  porton Aug 8 '11 at 21:46
    
And I guess «$\bigcap^{\mathfrak F}$» is exactly the same thing as $\bigcap$... –  Mariano Suárez-Alvarez Aug 8 '11 at 21:50
3  
Who receives the bounty if the asker has already accepted his own answer? –  The User Jun 27 '13 at 11:00
show 5 more comments

2 Answers 2

No; non-Hausdorff ultrafilters give a counterexample. In detail, let $\mathcal B$ be a non-principal ultrafilter on an infinite set $N$. Let $M=\{(x,y)\in N\times N:x\neq y\}$. Let $f$ and $g$ be the two projection functions from $M$ to $N$. Let $\mathcal A$ be any ultrafilter on $M$ containing all the sets $f^{-1}(X)$ and $g^{-1}(X)$ for $X\in\mathcal B$. I claim that $\mathcal A$ and $\mathcal B$ satisfy the hypothesis in your question. Indeed, if $\mathcal X$ is coherent with $\mathcal A$, then it is a subset of (i.e., higher in your ordering than) $\mathcal A$ because the latter is an ultrafilter. Therefore, the images of $\mathcal X$ under $f$ and under $g$ are subsets of the images of $\mathcal A$, both of which are $\mathcal B$. On the other hand, I also claim that your proposed conclusion fails. Indeed, $f\cap g$ is the empty relation (because the diagonal of $N\times N$ was removed in the definition of $M$), and therefore the image of any filter under $f\cap g$ is the improper filter, which is not a subset of $\mathcal B$.

share|improve this answer
    
What are projection functions from $M$ to $N$? I understand what is a projection function from $N\times N$ to $N$, but don't understand what is a projection function from $M$ to $N$. –  porton Aug 9 '11 at 10:42
    
@porton: Restrict the projections to have domain $M$. –  Andreas Blass Aug 9 '11 at 14:23
    
I don't understand why images of $\mathcal{A}$ under $f$ and $g$ are $\mathcal{B}$. Is $X$ a fixed arbitrary element of $\mathcal{B}$? –  porton Aug 9 '11 at 16:38
    
That the images of $\mathcal A$ under $f$ and $g$ are both $\mathcal B$ is immediate from the definition of images and of $\mathcal A$. In the latter definition, $X$ ranges over all $X\in\mathcal B$. In excessive but perhaps necessary detail: Since $f$ is a function and $\mathcal A$ is an ultrafilter, $f(\mathcal A)$ is an ultrafilter. If $X$ is any set in $\mathcal B$, then $f^{-1}(X)$ is in $\mathcal A$, so $f(f^{-1}(X))\in f(\mathcal A)$. But $f(f^{-1}(X))$ is a subset of $X$, so $X\in f(\mathcal A)$. (continued in next comment) –  Andreas Blass Aug 9 '11 at 17:13
    
That shows $\mathcal B\subseteq f(\mathcal A)$; since both of these are ultrafilters, they are equal. –  Andreas Blass Aug 9 '11 at 17:14
add comment
up vote -5 down vote accepted

After too much thought I found a simple counter-example:

Let $N$ is an infinite set.

Let $f = \{ (x;x) | x\in N \}$ and $g = \{ (x;y) | x,y\in N, x\ne y \}$.

Let $\mathcal{A}=\mathcal{B}$ is a nontrivial ultrafilter on $N$.

Using my theory it is easy to show that this is a counter-example.

Sorry that was trivial.

share|improve this answer
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.