Filters and intersection of two binary relations - MathOverflow

Filters and intersection of two binary relations

2011-08-08T21:18:28Z

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.

Answer by Andreas Blass for Filters and intersection of two binary relations

2011-08-08T22:52:13Z

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 $\mathcal B$.

Answer by porton for Filters and intersection of two binary relations

2011-08-09T16:56:03Z

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.