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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.

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    $\begingroup$ I honestly do not know what your formulas mean... $\endgroup$
    – Mariano Suárez-Álvarez
    Commented 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"?) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented 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
    Commented Aug 8, 2011 at 21:46
  • $\begingroup$ And I guess «$\bigcap^{\mathfrak F}$» is exactly the same thing as $\bigcap$... $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Aug 8, 2011 at 21:50
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    $\begingroup$ Who receives the bounty if the asker has already accepted his own answer? $\endgroup$
    – The User
    Commented Jun 27, 2013 at 11:00

1 Answer 1

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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 $\mathcal B$.

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  • $\begingroup$ 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
    Commented Aug 9, 2011 at 10:42
  • $\begingroup$ @porton: Restrict the projections to have domain $M$. $\endgroup$
    – Andreas Blass
    Commented 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
    Commented 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) $\endgroup$
    – Andreas Blass
    Commented Aug 9, 2011 at 17:13
  • $\begingroup$ That shows $\mathcal B\subseteq f(\mathcal A)$; since both of these are ultrafilters, they are equal. $\endgroup$
    – Andreas Blass
    Commented Aug 9, 2011 at 17:14

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