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Let $\mathfrak{A}$, $\mathfrak{B}$ be posets, $\mathfrak{A} \subseteq \mathfrak{B}$ (and $\mathfrak{A}$ is the induced order of $\mathfrak{B}$). Let $\mathcal{A}$ be a filter on $\mathfrak{A}$.

Help me to prove (or disprove): $$ \left\{ L \in \mathfrak{B} \mid \forall Y \in \mathfrak{A}: (Y \geq L \Rightarrow Y \in \mathcal{A}) \right\} = [\mathcal{A}]_{\mathfrak{B}} $$ where $[\mathcal{A}]_{\mathfrak{B}}$ is the filter on $\mathfrak{B}$ generated by the set $\mathcal{A}$.

If it does not hold for arbitrary posets $\mathfrak{A}$, $\mathfrak{B}$, does it holds for $\mathfrak{A}$, $\mathfrak{B}$ being booleans lattices of subsets of some sets?

By the way, is $$ \left\{ L \in \mathfrak{B} \mid \forall Y \in \mathfrak{A}: (Y \geq L \Rightarrow Y \in \mathcal{A}) \right\} $$ necessarily a filter on $\mathfrak{B}$?

I am also interested in the following special case:

  • $\mathfrak{B}$ is the boolean lattice $\mathscr{P}(U\times U)$ of all subsets of a set $U\times U$;
  • $\mathfrak{A}$ is the boolean lattice $\Gamma(U;U)$ of all finite unions of cartesian products of two subsets of the set $U$.

My main question is answered.

However, it is yet an open question whether $\left\{ L \in \mathfrak{B} \mid \forall Y \in \mathfrak{A}: (Y \geq L \Rightarrow Y \in \mathcal{A}) \right\}$ is always a filter.

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  • $\begingroup$ In one direction: Let $X \in [\mathcal{A}]_{\mathfrak{B}}$. Then $\exists Y' \in \mathcal{A} : X \geq Y'$. Thus if $Y \geq X$ for $Y \in \mathfrak{A}$ then $Y \geq Y'$ and thus $Y \in \mathcal{A}$. So $\forall Y \in \mathfrak{A}: (Y \geq X \Rightarrow Y \in \mathcal{A})$. $\endgroup$ – porton Dec 1 '14 at 19:11
  • $\begingroup$ The other direction is surprisingly difficult $\endgroup$ – porton Dec 1 '14 at 19:13
  • $\begingroup$ See also counter-examples in this thread: groups.google.com/forum/#!topic/sci.math/Plru0S8ePzs $\endgroup$ – porton Dec 2 '14 at 15:41
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Counterexample $\mathfrak{B}= \mathscr{P} (U \times U)$, $\mathfrak{A}= \Gamma (U ; U)$, $\mathcal{A} = \left\{ X \in \Gamma (U ; U) \mid X \supseteq {(\mathord{=})} |_U \right\}$ for $U=\mathbb{R}$. (It is easy to generalize for every infinite set $U$.)

Proof Let $L \in \mathscr{P} (U \times U)$.

Let's prove $$ \forall Y \in \mathfrak{A}: (Y \geq L \Rightarrow Y \in \mathcal{A}) \Leftrightarrow L \supseteq {(\mathord{=})} |_U $$ that is $$ \forall Y \in \Gamma (U ; U) : \left( Y \geq L \Rightarrow Y \supseteq {(\mathord{=})} |_U \right) \Leftrightarrow L \supseteq {(\mathord{=})} |_U . $$

$L \supseteq {(\mathord{=})} |_U \Rightarrow \forall Y \in \Gamma (U ; U) : \left( Y \geq L \Rightarrow Y \supseteq {(\mathord{=})} |_U \right)$ is obvious.

Let now $L \nsupseteq {(\mathord{=})} |_U$ that is $(x ; x) \not\ni L$. Then there exists $Y \in \Gamma (U ; U)$ such that $Y \geq L$ but $Y \nsupseteq {(\mathord{=})} |_U$ (take $Y = (\mathbb{R} \times \mathbb{R}) \setminus \{ (x ; x) \}$).

Thus we have proved the above equivalence.

Let $L_0 = \bigcup_{i \in \mathbb{Z}} ([i ; i + 1 [\times [i ; i + 1 [)$.

$L_0 \supseteq {(\mathord{=})} |_U$ but $L_0 \not\in [\mathcal{A}]_{\mathfrak{B}}$ because otherwise there would be $P \in \Gamma (U ; U)$ such that both $P \supseteq {(\mathord{=})} |_U$ and $P \subseteq L_0$. This is clearly impossible because $P$ must have a point in every $[i ; i + 1 [\times [i ; i + 1 [$ for $i \in \mathbb{Z}$ and thus isn't representable as a finite union of cartesian products.

Thus $\forall Y \in \mathfrak{A}: (Y \geq L_0 \Rightarrow Y \in \mathcal{A}) \nLeftrightarrow L_0 \in [\mathcal{A}]_{\mathfrak{B}}$.

It is yet an open question whether $\left\{ L \in \mathfrak{B} \mid \forall Y \in \mathfrak{A}: (Y \geq L \Rightarrow Y \in \mathcal{A}) \right\}$ is always a filter.

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