-8
$\begingroup$

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.

$\endgroup$
10
  • 1
    $\begingroup$ I honestly do not know what your formulas mean... $\endgroup$ Aug 8, 2011 at 21:42
  • $\begingroup$ (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"?) $\endgroup$ Aug 8, 2011 at 21:44
  • $\begingroup$ @Mariano Suárez-Alvarez: Yes, this is the filter of all sets. Also $f[X] = \{ y | \exists x\in X: (x;y) \in f \}$. $\endgroup$
    – porton
    Aug 8, 2011 at 21:46
  • $\begingroup$ And I guess «$\bigcap^{\mathfrak F}$» is exactly the same thing as $\bigcap$... $\endgroup$ Aug 8, 2011 at 21:50
  • 5
    $\begingroup$ Who receives the bounty if the asker has already accepted his own answer? $\endgroup$
    – The User
    Jun 27, 2013 at 11:00

1 Answer 1

6
$\begingroup$

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 $\mathcal B$.

$\endgroup$
5
  • $\begingroup$ 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 $N$. $\endgroup$
    – porton
    Aug 9, 2011 at 10:42
  • $\begingroup$ @porton: Restrict the projections to have domain $M$. $\endgroup$ Aug 9, 2011 at 14:23
  • $\begingroup$ I don't understand why images of $\mathcal{A}$ under $f$ and $g$ are $\mathcal{B}$. Is $X$ a fixed arbitrary element of $\mathcal{B}$? $\endgroup$
    – porton
    Aug 9, 2011 at 16:38
  • $\begingroup$ 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) $\endgroup$ Aug 9, 2011 at 17:13
  • $\begingroup$ That shows $\mathcal B\subseteq f(\mathcal A)$; since both of these are ultrafilters, they are equal. $\endgroup$ Aug 9, 2011 at 17:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.