How to prove de Vries algebras morphisms are dense and full if their duals are into ?
$begingroup$
Well, this is a quite short question but I think it will require some explainations.
Let's say that a de Vries (or compingent) algebra is a Boolean algebra $B=(B,0,1, wedge, vee, neg) $ with a binary relation $prec$ such that for each $a,b,c,d in B$ :
$0 prec 0$,
$a prec b Rightarrow a leq b$,
$ a leq c prec b Rightarrow a prec b$,
$a leq b, c leq d Rightarrow a wedge c prec b wedge d$,
$ a prec b Rightarrow neg b prec neg a$
$ a prec b neq 0 Rightarrow exists c in B setminus lbrace 0 rbrace : a prec c prec b$.
Now, let's say that a homomorphism $h$ between compingent algebras $B$ and $C$ is an application $h : B longrightarrow C$ such that
$ h(0) = 0$
$h( a wedge b) = h(a) wedge h(b)$
$ a prec b Rightarrow neg (h(neg a)) prec h(b)$.
Finally, a round (or compingent) filter $F$ in a de Vries algebra $B$ is a filter for Boolean algebra such that for $b in F$ there is $a in F$ with $ a prec b$. It also can be shown that a round filter $F$ is maximal if and only if for every $a, b in B$, $ a prec b$ implies $b in F$ or $neg a in F$.
And now, the question ! Suppose $h : B longrightarrow C$ is a homomorphism. Then $m(h)$ defined as
$$ m(h)(F) = lbrace a mid exists b in h^{-1}(F) : b prec a rbrace $$ is an application from the maximal round filters of $C$ to the maximal round filters of $B$.
What I want to prove is : if $m(h)$ is into then for every $c,d in C$ such that $c prec d $ there exist $a,b in B$ such that $$ (1) a prec neg b, c prec h(a) text{ and } neg d prec h(b).$$
According to de Vries thesis, this implication should derive from the fact that if $m(h)$ is into, then for every maximal round filtre $F$ of $C$ and every $d in F$, there is $a,b in B$ such that $$ (2) a prec neg b , a in m(h)(F) text{ and } neg d prec h(b).$$
While I have no problem to prove that $m(h)$ is into implies the existences of elements $a,b$ such that $(2)$, I can't prove that we can deduct $(1)$ from $(2)$ (this last step should be easy to show according to de Vries).
Thank you for any answer !
general-topology category-theory boolean-algebra duality-theorems
asked Dec 14 '18 at 11:37
L. De RudderL. De Rudder
335
$endgroup$
add a comment |
$begingroup$
Well, this is a quite short question but I think it will require some explainations.
Let's say that a de Vries (or compingent) algebra is a Boolean algebra $B=(B,0,1, wedge, vee, neg) $ with a binary relation $prec$ such that for each $a,b,c,d in B$ :
$0 prec 0$,
$a prec b Rightarrow a leq b$,
$ a leq c prec b Rightarrow a prec b$,
$a leq b, c leq d Rightarrow a wedge c prec b wedge d$,
$ a prec b Rightarrow neg b prec neg a$
$ a prec b neq 0 Rightarrow exists c in B setminus lbrace 0 rbrace : a prec c prec b$.
Now, let's say that a homomorphism $h$ between compingent algebras $B$ and $C$ is an application $h : B longrightarrow C$ such that
$ h(0) = 0$
$h( a wedge b) = h(a) wedge h(b)$
$ a prec b Rightarrow neg (h(neg a)) prec h(b)$.
Finally, a round (or compingent) filter $F$ in a de Vries algebra $B$ is a filter for Boolean algebra such that for $b in F$ there is $a in F$ with $ a prec b$. It also can be shown that a round filter $F$ is maximal if and only if for every $a, b in B$, $ a prec b$ implies $b in F$ or $neg a in F$.
And now, the question ! Suppose $h : B longrightarrow C$ is a homomorphism. Then $m(h)$ defined as
$$ m(h)(F) = lbrace a mid exists b in h^{-1}(F) : b prec a rbrace $$ is an application from the maximal round filters of $C$ to the maximal round filters of $B$.
What I want to prove is : if $m(h)$ is into then for every $c,d in C$ such that $c prec d $ there exist $a,b in B$ such that $$ (1) a prec neg b, c prec h(a) text{ and } neg d prec h(b).$$
According to de Vries thesis, this implication should derive from the fact that if $m(h)$ is into, then for every maximal round filtre $F$ of $C$ and every $d in F$, there is $a,b in B$ such that $$ (2) a prec neg b , a in m(h)(F) text{ and } neg d prec h(b).$$
While I have no problem to prove that $m(h)$ is into implies the existences of elements $a,b$ such that $(2)$, I can't prove that we can deduct $(1)$ from $(2)$ (this last step should be easy to show according to de Vries).
Thank you for any answer !
general-topology category-theory boolean-algebra duality-theorems
asked Dec 14 '18 at 11:37
L. De RudderL. De Rudder
335
$endgroup$
add a comment |
$begingroup$
Well, this is a quite short question but I think it will require some explainations.
Let's say that a de Vries (or compingent) algebra is a Boolean algebra $B=(B,0,1, wedge, vee, neg) $ with a binary relation $prec$ such that for each $a,b,c,d in B$ :
$0 prec 0$,
$a prec b Rightarrow a leq b$,
$ a leq c prec b Rightarrow a prec b$,
$a leq b, c leq d Rightarrow a wedge c prec b wedge d$,
$ a prec b Rightarrow neg b prec neg a$
$ a prec b neq 0 Rightarrow exists c in B setminus lbrace 0 rbrace : a prec c prec b$.
Now, let's say that a homomorphism $h$ between compingent algebras $B$ and $C$ is an application $h : B longrightarrow C$ such that
$ h(0) = 0$
$h( a wedge b) = h(a) wedge h(b)$
$ a prec b Rightarrow neg (h(neg a)) prec h(b)$.
Finally, a round (or compingent) filter $F$ in a de Vries algebra $B$ is a filter for Boolean algebra such that for $b in F$ there is $a in F$ with $ a prec b$. It also can be shown that a round filter $F$ is maximal if and only if for every $a, b in B$, $ a prec b$ implies $b in F$ or $neg a in F$.
And now, the question ! Suppose $h : B longrightarrow C$ is a homomorphism. Then $m(h)$ defined as
$$ m(h)(F) = lbrace a mid exists b in h^{-1}(F) : b prec a rbrace $$ is an application from the maximal round filters of $C$ to the maximal round filters of $B$.
What I want to prove is : if $m(h)$ is into then for every $c,d in C$ such that $c prec d $ there exist $a,b in B$ such that $$ (1) a prec neg b, c prec h(a) text{ and } neg d prec h(b).$$
According to de Vries thesis, this implication should derive from the fact that if $m(h)$ is into, then for every maximal round filtre $F$ of $C$ and every $d in F$, there is $a,b in B$ such that $$ (2) a prec neg b , a in m(h)(F) text{ and } neg d prec h(b).$$
While I have no problem to prove that $m(h)$ is into implies the existences of elements $a,b$ such that $(2)$, I can't prove that we can deduct $(1)$ from $(2)$ (this last step should be easy to show according to de Vries).
Thank you for any answer !
general-topology category-theory boolean-algebra duality-theorems
asked Dec 14 '18 at 11:37
L. De RudderL. De Rudder
335
$endgroup$
Well, this is a quite short question but I think it will require some explainations.
Let's say that a de Vries (or compingent) algebra is a Boolean algebra $B=(B,0,1, wedge, vee, neg) $ with a binary relation $prec$ such that for each $a,b,c,d in B$ :
$0 prec 0$,
$a prec b Rightarrow a leq b$,
$ a leq c prec b Rightarrow a prec b$,
$a leq b, c leq d Rightarrow a wedge c prec b wedge d$,
$ a prec b Rightarrow neg b prec neg a$
$ a prec b neq 0 Rightarrow exists c in B setminus lbrace 0 rbrace : a prec c prec b$.
Now, let's say that a homomorphism $h$ between compingent algebras $B$ and $C$ is an application $h : B longrightarrow C$ such that
$ h(0) = 0$
$h( a wedge b) = h(a) wedge h(b)$
$ a prec b Rightarrow neg (h(neg a)) prec h(b)$.
Finally, a round (or compingent) filter $F$ in a de Vries algebra $B$ is a filter for Boolean algebra such that for $b in F$ there is $a in F$ with $ a prec b$. It also can be shown that a round filter $F$ is maximal if and only if for every $a, b in B$, $ a prec b$ implies $b in F$ or $neg a in F$.
And now, the question ! Suppose $h : B longrightarrow C$ is a homomorphism. Then $m(h)$ defined as
$$ m(h)(F) = lbrace a mid exists b in h^{-1}(F) : b prec a rbrace $$ is an application from the maximal round filters of $C$ to the maximal round filters of $B$.
What I want to prove is : if $m(h)$ is into then for every $c,d in C$ such that $c prec d $ there exist $a,b in B$ such that $$ (1) a prec neg b, c prec h(a) text{ and } neg d prec h(b).$$
According to de Vries thesis, this implication should derive from the fact that if $m(h)$ is into, then for every maximal round filtre $F$ of $C$ and every $d in F$, there is $a,b in B$ such that $$ (2) a prec neg b , a in m(h)(F) text{ and } neg d prec h(b).$$
While I have no problem to prove that $m(h)$ is into implies the existences of elements $a,b$ such that $(2)$, I can't prove that we can deduct $(1)$ from $(2)$ (this last step should be easy to show according to de Vries).
Thank you for any answer !
general-topology category-theory boolean-algebra duality-theorems
general-topology category-theory boolean-algebra duality-theorems
asked Dec 14 '18 at 11:37
L. De RudderL. De Rudder
335
asked Dec 14 '18 at 11:37
L. De RudderL. De Rudder
335
asked Dec 14 '18 at 11:37
L. De RudderL. De Rudder
335
asked Dec 14 '18 at 11:37
L. De RudderL. De Rudder
335
asked Dec 14 '18 at 11:37
L. De RudderL. De Rudder
335
335
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