Is a set closed under finite intersections? (about filters)

In my research I was faced with the problem (as a special example and a pattern for more general problems) whether the family $\operatorname{GR} ( \Delta \times^{\mathsf{FCD}} \Delta)$ of sets is closed regarding finite intersections.

To make the problem accessible for these who have not read my book, I will define it in elementary terms:

$\operatorname{GR} ( \Delta \times^{\mathsf{FCD}} \Delta)$ is a set of binary relations on $\mathbb{R}$ defined by the formula:

$$f \in \operatorname{GR} ( \Delta \times^{\mathsf{FCD}} \Delta) \Leftrightarrow \forall X \in \mathscr{P} \mathbb{R} : ( (\forall D \in \Delta : X \cap D \neq \varnothing) \Rightarrow f [X] \in \Delta)$$

where $\Delta$ is the filter of neighborhoods of zero on real line, that is the filter generated by the set $\{(-\epsilon;\epsilon) \,|\, \epsilon\in\mathbb{R}, \epsilon>0 \}$ and $f[X] = \{ y \,|\, \exists x\in X: (x,y) \in f \}$.

Is $\operatorname{GR} ( \Delta \times^{\mathsf{FCD}} \Delta)$ closed under finite intersections?

• It is enough to prove that for every $f \in \operatorname{GR} ( \Delta \times^{\mathsf{FCD}} \Delta)$ there is a positive $\varepsilon$ such that $\forall x \in ( - \varepsilon ; \varepsilon) : f[\{x\}] \in \Delta$. Nov 17, 2013 at 20:17
• The above comment is proved in portonmath.tiddlyspace.com/… Dec 12, 2013 at 0:20
• @KarlKronenfeld: I added more parentheses to the formula. It seems you misunderstood me Dec 12, 2013 at 0:54
• @AlexRavsky: It is not proved that there exists positive $\epsilon_1$ such that $f[\{x_1\}]\in\Delta$ for each $x_1\in(-\epsilon_1,\epsilon_1)$. Dec 17, 2013 at 19:34
• @porton Thanks for your answer. Then I do a second try. It seems the following. Put $f_1=\{(x,y)\in\mathbb R^2:|x|\le |y|$ or $y=0\}$ and $f_2=\{(x,y)\in\mathbb R^2:|x|\ge |y|$ or $x=0\}$. Then $f_1, f_2\in\operatorname{GR} ( \Delta \times^{\mathsf{FCD}} \Delta)$ but $f_1\cap f_2=\{(x,y)\in\mathbb R^2:|x|=|y|$ or $x=0$ or $y=0\}\not\in\operatorname{GR} ( \Delta \times^{\mathsf{FCD}} \Delta)$. Is this example OK? Dec 17, 2013 at 20:58

It seems the following. Put $f_1=\{(x,y)\in\mathbb R^2:|x|\le |y|$ or $y=0\}$ and $f_2=\{(x,y)\in\mathbb R^2:|x|\ge |y|$ or $x=0\}$. Then $f_1, f_2\in\operatorname{GR} ( \Delta \times^{\mathsf{FCD}} \Delta)$ but $f_1\cap f_2=\{(x,y)\in\mathbb R^2:|x|=|y|$ or $x=0$ or $y=0\}\not\in\operatorname{GR} ( \Delta \times^{\mathsf{FCD}} \Delta)$.
• I am not convinced that $f_1$ and $f_2$ are in $\operatorname{GR} ( \Delta \times^{\mathsf{FCD}} \Delta)$. Note that $f\in \operatorname{GR} ( \Delta \times^{\mathsf{FCD}} \Delta)$ iff for every sequence $s_n\to 0$, $f[\{s_n\}]\in\Delta$. But, try $f_2$ on, say, $\{1/n:n\in\mathbb N\}$. Dec 18, 2013 at 3:11
• @KarlKronenfeld: Due funcoid magic $f_1\in\operatorname{GR} ( \Delta \times^{\mathsf{FCD}} \Delta) \Leftrightarrow f_2\in\operatorname{GR} ( \Delta \times^{\mathsf{FCD}} \Delta)$ and moreover $\operatorname{GR} ( \Delta \times^{\mathsf{FCD}} \Delta)$ is symmetric, see mathematics21.org/algebraic-general-topology.html Dec 18, 2013 at 3:25
• @porton Forget my deleted comment, $0\notin f_1[\{1/n:n\in\mathbb N\}]$. So, indeed, neither $f_1$ nor $f_2$ is in $GR(\dots)$. Dec 18, 2013 at 3:28
• @KarlKronenfeld: $0\in f_1(1/n)$ for every $n\in\mathbb{N}$ because $(1/n,0)\in f_1$ Dec 18, 2013 at 3:30
• @KarlKronenfeld: You missed $y=0$ in the definition of $f_1$ Dec 18, 2013 at 3:31