Filters and intersection of two binary relations - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T22:33:09Z http://mathoverflow.net/feeds/question/72401 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/72401/filters-and-intersection-of-two-binary-relations Filters and intersection of two binary relations porton 2011-08-08T21:18:28Z 2011-08-09T16:56:03Z <p>Let $\mathfrak{F}$ is the complete lattice of filters (including the improper filter) on some set, ordered inverse to set-theoretic inclusion.</p> <p>I will denote <code>$\left\langle f \right\rangle \mathcal{X} = \bigcap^{\mathfrak{F}} \left\{ f \left[ X \right] | X \in \mathcal{X} \right\}$</code> for every binary relation $f$ and filter $\mathcal{X}$.</p> <p>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.)</p> <p>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?</p> <p>You can read <a href="http://www.mathematics21.org/algebraic-general-topology.html" rel="nofollow">http://www.mathematics21.org/algebraic-general-topology.html</a> for my related research.</p> http://mathoverflow.net/questions/72401/filters-and-intersection-of-two-binary-relations/72409#72409 Answer by Andreas Blass for Filters and intersection of two binary relations Andreas Blass 2011-08-08T22:52:13Z 2011-08-08T22:52:13Z <p>No; non-Hausdorff ultrafilters give a counterexample. In detail, let $\mathcal B$ be a non-principal ultrafilter on an infinite set $N$. Let <code>$M=\{(x,y)\in N\times N:x\neq y\}$</code>. 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 <code>$f^{-1}(X)$</code> and <code>$g^{-1}(X)$</code> 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 $\mathcal B$.</p> http://mathoverflow.net/questions/72401/filters-and-intersection-of-two-binary-relations/72487#72487 Answer by porton for Filters and intersection of two binary relations porton 2011-08-09T16:56:03Z 2011-08-09T16:56:03Z <p>After too much thought I found a simple counter-example:</p> <p>Let $N$ is an infinite set.</p> <p>Let <code>$f = \{ (x;x) | x\in N \}$</code> and <code>$g = \{ (x;y) | x,y\in N, x\ne y \}$</code>.</p> <p>Let $\mathcal{A}=\mathcal{B}$ is a nontrivial ultrafilter on $N$.</p> <p>Using <a href="http://www.mathematics21.org/algebraic-general-topology.html" rel="nofollow">my theory</a> it is easy to show that this is a counter-example.</p> <p>Sorry that was trivial.</p>