Mixing
Do complete atomic boolean algebras form a full subcategory of boolean algebras?
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I would like to know if a boolean morphism (that is an application that respects $vee, wedge, neg, 0,1$) between two complete atomic boolean algebras is necessarily complete (it respects also infinite $vee$ and $wedge$)?
I am tempted to say yes because of the duality between complete atomic boolean algebras and sets, but I cannot find a proof.
Thank you!
boolean-algebra duality-theorems
asked Jan 28 at 14:18
L. De RudderL. De Rudder
335
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add a comment |
$begingroup$
I would like to know if a boolean morphism (that is an application that respects $vee, wedge, neg, 0,1$) between two complete atomic boolean algebras is necessarily complete (it respects also infinite $vee$ and $wedge$)?
I am tempted to say yes because of the duality between complete atomic boolean algebras and sets, but I cannot find a proof.
Thank you!
boolean-algebra duality-theorems
asked Jan 28 at 14:18
L. De RudderL. De Rudder
335
$endgroup$
add a comment |
$begingroup$
I would like to know if a boolean morphism (that is an application that respects $vee, wedge, neg, 0,1$) between two complete atomic boolean algebras is necessarily complete (it respects also infinite $vee$ and $wedge$)?
I am tempted to say yes because of the duality between complete atomic boolean algebras and sets, but I cannot find a proof.
Thank you!
boolean-algebra duality-theorems
asked Jan 28 at 14:18
L. De RudderL. De Rudder
335
$endgroup$
I would like to know if a boolean morphism (that is an application that respects $vee, wedge, neg, 0,1$) between two complete atomic boolean algebras is necessarily complete (it respects also infinite $vee$ and $wedge$)?
I am tempted to say yes because of the duality between complete atomic boolean algebras and sets, but I cannot find a proof.
Thank you!
boolean-algebra duality-theorems
boolean-algebra duality-theorems
asked Jan 28 at 14:18
L. De RudderL. De Rudder
335
asked Jan 28 at 14:18
L. De RudderL. De Rudder
335
asked Jan 28 at 14:18
L. De RudderL. De Rudder
335
asked Jan 28 at 14:18
L. De RudderL. De Rudder
335
asked Jan 28 at 14:18
L. De RudderL. De Rudder
335
335
add a comment |
add a comment |
2 Answers
2
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oldest
votes
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No. For instance, let $X$ be an infinite set and $U$ be a nonprincipal ultrafilter on $X$. Then there is a Boolean homomorphism $f:P(X)to{0,1}$ which maps the elements of $U$ to $1$ and the elements of $Xsetminus U$ to $0$. This homomorphism $f$ is not complete since it maps every singleton to $0$ but does not map every arbitrary join of singletons (i.e., every set) to $0$.
answered Jan 29 at 21:04
Eric WofseyEric Wofsey
191k14216349
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add a comment |
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Up to isomorphism, a complete and atomic Boolean algebra is a powerset algebra, and powerset algebras are always complete and atomic.
So, without loss of generality, we can think of complete and atomic Boolean algebras as being exactly the same as powerset algebras.
Now, let $mathbf A$ be one such algebra, that is, there exists a set $X$ such that $mathbf A = langle mathscr P(X), cap, cup, ', varnothing, X rangle$.
Let $Y$ be the set of ultrafilters of $mathbf A$;
for example, for each $x in X$, the set $U_x = {Z in mathscr P(X) : x in Z}$ is an ultrafilter (the principal ultrafilter generated by ${x}$, having ${x}$ as its infimum).
Let $mathbf B = langle mathscr P(Y), cap, cup, ', varnothing, Y rangle$ be the Boolean algebra of the powerset of $Y$.
Using properties of ultrafilters, one can easily prove that $Phi : mathbf A to mathbf B$ given by
$$Phi(a) = { U in Y : a in U }$$
is a Boolean algebra homomorphism.
Now consider the set $At_A$ of atoms of $mathbf A$.
As in any atomic Boolean algebra, $bigvee At_A = top_A$ which is $X$, in this case.
For $a = {x}$, with $x in X$ (that is, $a in At_A$),
$$Phi(a) = {U in Y:{x} in U} = {U_x} in At_B,$$
whence $Phi(At_A) subseteq At_B$.
If $mathbf A$ is finite, then $Phi$ is an isomorphism and so it is complete.
But if $mathbf A$ is infinite, then it has at least one non-principal ultrafilter $U_0$, and so $Phi(At_A) neq At_B$, and therefore
$$bigvee Phi(At_A) < bigvee At_B = Y = Phi(X) = Phileft(bigvee At_Aright),$$
where the inequality holds because, in a Boolean algebra, the join of a proper subset of the atoms is less than the top element (it is the complement of the join of the remaining atoms).
We conclude that $Phi$ is not a complete homomorphism.
answered Jan 29 at 18:36
amrsaamrsa
3,8552618
$endgroup$
1
$begingroup$
$bigvee At_A = top_A$ is not true in an arbitrary Boolean algebra (though it is true in any atomic Boolean algebra).
$endgroup$
– Eric Wofsey
Jan 29 at 21:05
$begingroup$
@EricWofsey Yes, of course, here I went a bit too far. Thanks for pointing out; I'll edit it. Actually, when I wrote it, I actually mean that if a Boolean algebra has at least an atom, then it is atomic, which is not true!
$endgroup$
– amrsa
Jan 29 at 22:43
add a comment |
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2 Answers
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2 Answers
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$begingroup$
No. For instance, let $X$ be an infinite set and $U$ be a nonprincipal ultrafilter on $X$. Then there is a Boolean homomorphism $f:P(X)to{0,1}$ which maps the elements of $U$ to $1$ and the elements of $Xsetminus U$ to $0$. This homomorphism $f$ is not complete since it maps every singleton to $0$ but does not map every arbitrary join of singletons (i.e., every set) to $0$.
answered Jan 29 at 21:04
Eric WofseyEric Wofsey
191k14216349
$endgroup$
add a comment |
$begingroup$
No. For instance, let $X$ be an infinite set and $U$ be a nonprincipal ultrafilter on $X$. Then there is a Boolean homomorphism $f:P(X)to{0,1}$ which maps the elements of $U$ to $1$ and the elements of $Xsetminus U$ to $0$. This homomorphism $f$ is not complete since it maps every singleton to $0$ but does not map every arbitrary join of singletons (i.e., every set) to $0$.
answered Jan 29 at 21:04
Eric WofseyEric Wofsey
191k14216349
$endgroup$
add a comment |
$begingroup$
No. For instance, let $X$ be an infinite set and $U$ be a nonprincipal ultrafilter on $X$. Then there is a Boolean homomorphism $f:P(X)to{0,1}$ which maps the elements of $U$ to $1$ and the elements of $Xsetminus U$ to $0$. This homomorphism $f$ is not complete since it maps every singleton to $0$ but does not map every arbitrary join of singletons (i.e., every set) to $0$.
answered Jan 29 at 21:04
Eric WofseyEric Wofsey
191k14216349
$endgroup$
No. For instance, let $X$ be an infinite set and $U$ be a nonprincipal ultrafilter on $X$. Then there is a Boolean homomorphism $f:P(X)to{0,1}$ which maps the elements of $U$ to $1$ and the elements of $Xsetminus U$ to $0$. This homomorphism $f$ is not complete since it maps every singleton to $0$ but does not map every arbitrary join of singletons (i.e., every set) to $0$.
answered Jan 29 at 21:04
Eric WofseyEric Wofsey
191k14216349
answered Jan 29 at 21:04
Eric WofseyEric Wofsey
191k14216349
answered Jan 29 at 21:04
Eric WofseyEric Wofsey
191k14216349
answered Jan 29 at 21:04
Eric WofseyEric Wofsey
191k14216349
191k14216349
add a comment |
add a comment |
$begingroup$
Up to isomorphism, a complete and atomic Boolean algebra is a powerset algebra, and powerset algebras are always complete and atomic.
So, without loss of generality, we can think of complete and atomic Boolean algebras as being exactly the same as powerset algebras.
Now, let $mathbf A$ be one such algebra, that is, there exists a set $X$ such that $mathbf A = langle mathscr P(X), cap, cup, ', varnothing, X rangle$.
Let $Y$ be the set of ultrafilters of $mathbf A$;
for example, for each $x in X$, the set $U_x = {Z in mathscr P(X) : x in Z}$ is an ultrafilter (the principal ultrafilter generated by ${x}$, having ${x}$ as its infimum).
Let $mathbf B = langle mathscr P(Y), cap, cup, ', varnothing, Y rangle$ be the Boolean algebra of the powerset of $Y$.
Using properties of ultrafilters, one can easily prove that $Phi : mathbf A to mathbf B$ given by
$$Phi(a) = { U in Y : a in U }$$
is a Boolean algebra homomorphism.
Now consider the set $At_A$ of atoms of $mathbf A$.
As in any atomic Boolean algebra, $bigvee At_A = top_A$ which is $X$, in this case.
For $a = {x}$, with $x in X$ (that is, $a in At_A$),
$$Phi(a) = {U in Y:{x} in U} = {U_x} in At_B,$$
whence $Phi(At_A) subseteq At_B$.
If $mathbf A$ is finite, then $Phi$ is an isomorphism and so it is complete.
But if $mathbf A$ is infinite, then it has at least one non-principal ultrafilter $U_0$, and so $Phi(At_A) neq At_B$, and therefore
$$bigvee Phi(At_A) < bigvee At_B = Y = Phi(X) = Phileft(bigvee At_Aright),$$
where the inequality holds because, in a Boolean algebra, the join of a proper subset of the atoms is less than the top element (it is the complement of the join of the remaining atoms).
We conclude that $Phi$ is not a complete homomorphism.
answered Jan 29 at 18:36
amrsaamrsa
3,8552618
$endgroup$
1
$begingroup$
$bigvee At_A = top_A$ is not true in an arbitrary Boolean algebra (though it is true in any atomic Boolean algebra).
$endgroup$
– Eric Wofsey
Jan 29 at 21:05
$begingroup$
@EricWofsey Yes, of course, here I went a bit too far. Thanks for pointing out; I'll edit it. Actually, when I wrote it, I actually mean that if a Boolean algebra has at least an atom, then it is atomic, which is not true!
$endgroup$
– amrsa
Jan 29 at 22:43
add a comment |
$begingroup$
Up to isomorphism, a complete and atomic Boolean algebra is a powerset algebra, and powerset algebras are always complete and atomic.
So, without loss of generality, we can think of complete and atomic Boolean algebras as being exactly the same as powerset algebras.
Now, let $mathbf A$ be one such algebra, that is, there exists a set $X$ such that $mathbf A = langle mathscr P(X), cap, cup, ', varnothing, X rangle$.
Let $Y$ be the set of ultrafilters of $mathbf A$;
for example, for each $x in X$, the set $U_x = {Z in mathscr P(X) : x in Z}$ is an ultrafilter (the principal ultrafilter generated by ${x}$, having ${x}$ as its infimum).
Let $mathbf B = langle mathscr P(Y), cap, cup, ', varnothing, Y rangle$ be the Boolean algebra of the powerset of $Y$.
Using properties of ultrafilters, one can easily prove that $Phi : mathbf A to mathbf B$ given by
$$Phi(a) = { U in Y : a in U }$$
is a Boolean algebra homomorphism.
Now consider the set $At_A$ of atoms of $mathbf A$.
As in any atomic Boolean algebra, $bigvee At_A = top_A$ which is $X$, in this case.
For $a = {x}$, with $x in X$ (that is, $a in At_A$),
$$Phi(a) = {U in Y:{x} in U} = {U_x} in At_B,$$
whence $Phi(At_A) subseteq At_B$.
If $mathbf A$ is finite, then $Phi$ is an isomorphism and so it is complete.
But if $mathbf A$ is infinite, then it has at least one non-principal ultrafilter $U_0$, and so $Phi(At_A) neq At_B$, and therefore
$$bigvee Phi(At_A) < bigvee At_B = Y = Phi(X) = Phileft(bigvee At_Aright),$$
where the inequality holds because, in a Boolean algebra, the join of a proper subset of the atoms is less than the top element (it is the complement of the join of the remaining atoms).
We conclude that $Phi$ is not a complete homomorphism.
answered Jan 29 at 18:36
amrsaamrsa
3,8552618
$endgroup$
1
$begingroup$
$bigvee At_A = top_A$ is not true in an arbitrary Boolean algebra (though it is true in any atomic Boolean algebra).
$endgroup$
– Eric Wofsey
Jan 29 at 21:05
$begingroup$
@EricWofsey Yes, of course, here I went a bit too far. Thanks for pointing out; I'll edit it. Actually, when I wrote it, I actually mean that if a Boolean algebra has at least an atom, then it is atomic, which is not true!
$endgroup$
– amrsa
Jan 29 at 22:43
add a comment |
$begingroup$
Up to isomorphism, a complete and atomic Boolean algebra is a powerset algebra, and powerset algebras are always complete and atomic.
So, without loss of generality, we can think of complete and atomic Boolean algebras as being exactly the same as powerset algebras.
Now, let $mathbf A$ be one such algebra, that is, there exists a set $X$ such that $mathbf A = langle mathscr P(X), cap, cup, ', varnothing, X rangle$.
Let $Y$ be the set of ultrafilters of $mathbf A$;
for example, for each $x in X$, the set $U_x = {Z in mathscr P(X) : x in Z}$ is an ultrafilter (the principal ultrafilter generated by ${x}$, having ${x}$ as its infimum).
Let $mathbf B = langle mathscr P(Y), cap, cup, ', varnothing, Y rangle$ be the Boolean algebra of the powerset of $Y$.
Using properties of ultrafilters, one can easily prove that $Phi : mathbf A to mathbf B$ given by
$$Phi(a) = { U in Y : a in U }$$
is a Boolean algebra homomorphism.
Now consider the set $At_A$ of atoms of $mathbf A$.
As in any atomic Boolean algebra, $bigvee At_A = top_A$ which is $X$, in this case.
For $a = {x}$, with $x in X$ (that is, $a in At_A$),
$$Phi(a) = {U in Y:{x} in U} = {U_x} in At_B,$$
whence $Phi(At_A) subseteq At_B$.
If $mathbf A$ is finite, then $Phi$ is an isomorphism and so it is complete.
But if $mathbf A$ is infinite, then it has at least one non-principal ultrafilter $U_0$, and so $Phi(At_A) neq At_B$, and therefore
$$bigvee Phi(At_A) < bigvee At_B = Y = Phi(X) = Phileft(bigvee At_Aright),$$
where the inequality holds because, in a Boolean algebra, the join of a proper subset of the atoms is less than the top element (it is the complement of the join of the remaining atoms).
We conclude that $Phi$ is not a complete homomorphism.
answered Jan 29 at 18:36
amrsaamrsa
3,8552618
$endgroup$
Up to isomorphism, a complete and atomic Boolean algebra is a powerset algebra, and powerset algebras are always complete and atomic.
So, without loss of generality, we can think of complete and atomic Boolean algebras as being exactly the same as powerset algebras.
Now, let $mathbf A$ be one such algebra, that is, there exists a set $X$ such that $mathbf A = langle mathscr P(X), cap, cup, ', varnothing, X rangle$.
Let $Y$ be the set of ultrafilters of $mathbf A$;
for example, for each $x in X$, the set $U_x = {Z in mathscr P(X) : x in Z}$ is an ultrafilter (the principal ultrafilter generated by ${x}$, having ${x}$ as its infimum).
Let $mathbf B = langle mathscr P(Y), cap, cup, ', varnothing, Y rangle$ be the Boolean algebra of the powerset of $Y$.
Using properties of ultrafilters, one can easily prove that $Phi : mathbf A to mathbf B$ given by
$$Phi(a) = { U in Y : a in U }$$
is a Boolean algebra homomorphism.
Now consider the set $At_A$ of atoms of $mathbf A$.
As in any atomic Boolean algebra, $bigvee At_A = top_A$ which is $X$, in this case.
For $a = {x}$, with $x in X$ (that is, $a in At_A$),
$$Phi(a) = {U in Y:{x} in U} = {U_x} in At_B,$$
whence $Phi(At_A) subseteq At_B$.
If $mathbf A$ is finite, then $Phi$ is an isomorphism and so it is complete.
But if $mathbf A$ is infinite, then it has at least one non-principal ultrafilter $U_0$, and so $Phi(At_A) neq At_B$, and therefore
$$bigvee Phi(At_A) < bigvee At_B = Y = Phi(X) = Phileft(bigvee At_Aright),$$
where the inequality holds because, in a Boolean algebra, the join of a proper subset of the atoms is less than the top element (it is the complement of the join of the remaining atoms).
We conclude that $Phi$ is not a complete homomorphism.
answered Jan 29 at 18:36
amrsaamrsa
3,8552618
edited Jan 29 at 22:46
answered Jan 29 at 18:36
amrsaamrsa
3,8552618
answered Jan 29 at 18:36
amrsaamrsa
3,8552618
answered Jan 29 at 18:36
amrsaamrsa
3,8552618
3,8552618
1
$begingroup$
$bigvee At_A = top_A$ is not true in an arbitrary Boolean algebra (though it is true in any atomic Boolean algebra).
$endgroup$
– Eric Wofsey
Jan 29 at 21:05
$begingroup$
@EricWofsey Yes, of course, here I went a bit too far. Thanks for pointing out; I'll edit it. Actually, when I wrote it, I actually mean that if a Boolean algebra has at least an atom, then it is atomic, which is not true!
$endgroup$
– amrsa
Jan 29 at 22:43
add a comment |
1
$begingroup$
$bigvee At_A = top_A$ is not true in an arbitrary Boolean algebra (though it is true in any atomic Boolean algebra).
$endgroup$
– Eric Wofsey
Jan 29 at 21:05
$begingroup$
@EricWofsey Yes, of course, here I went a bit too far. Thanks for pointing out; I'll edit it. Actually, when I wrote it, I actually mean that if a Boolean algebra has at least an atom, then it is atomic, which is not true!
$endgroup$
– amrsa
Jan 29 at 22:43
1
1
$begingroup$
$bigvee At_A = top_A$ is not true in an arbitrary Boolean algebra (though it is true in any atomic Boolean algebra).
$endgroup$
– Eric Wofsey
Jan 29 at 21:05
$begingroup$
$bigvee At_A = top_A$ is not true in an arbitrary Boolean algebra (though it is true in any atomic Boolean algebra).
$endgroup$
– Eric Wofsey
Jan 29 at 21:05
$begingroup$
@EricWofsey Yes, of course, here I went a bit too far. Thanks for pointing out; I'll edit it. Actually, when I wrote it, I actually mean that if a Boolean algebra has at least an atom, then it is atomic, which is not true!
$endgroup$
– amrsa
Jan 29 at 22:43
$begingroup$
@EricWofsey Yes, of course, here I went a bit too far. Thanks for pointing out; I'll edit it. Actually, when I wrote it, I actually mean that if a Boolean algebra has at least an atom, then it is atomic, which is not true!
$endgroup$
– amrsa
Jan 29 at 22:43
add a comment |
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