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Question on the free BA on countably many generators
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Let $A$ be the free Boolean algebra of countably many generators. As a Boolean algebra, it is isomorphic to the field $F$ of clopen subsets of its Stone space, namely, the Cantor space $X=2^omega$.
It is known that $A$ is atomless, countably infinite, and incomplete. Since $A$ and $F$ are isomorphic, $F$ is also atomless, countably infinite, and incomplete.
Consider now the smallest subfield $F_0=emptyset,X$ of clopen subsets of $Z$. Does $F_0$ inherit the properties of $F$, that is, is $F_0$ also atomless (trivially), countably infinite, and incomplete?
This question may be naive, but I just cannot decide whether or not $F_0$ inherits the properties of $F$, especially, if I consider that there exists a Boolean epimorphism of $A$ into $F_0$.
boolean-algebra
asked Jul 31 at 14:53
puzzled
1094
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up vote
0
down vote
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Let $A$ be the free Boolean algebra of countably many generators. As a Boolean algebra, it is isomorphic to the field $F$ of clopen subsets of its Stone space, namely, the Cantor space $X=2^omega$.
It is known that $A$ is atomless, countably infinite, and incomplete. Since $A$ and $F$ are isomorphic, $F$ is also atomless, countably infinite, and incomplete.
Consider now the smallest subfield $F_0=emptyset,X$ of clopen subsets of $Z$. Does $F_0$ inherit the properties of $F$, that is, is $F_0$ also atomless (trivially), countably infinite, and incomplete?
This question may be naive, but I just cannot decide whether or not $F_0$ inherits the properties of $F$, especially, if I consider that there exists a Boolean epimorphism of $A$ into $F_0$.
boolean-algebra
asked Jul 31 at 14:53
puzzled
1094
$F$ is completely irrelevant to the question. You just have a (very simple!) Boolean algebra $F_0=emptyset, X$. Have you tried simply answering any of your questions about this Boolean algebra directly?
– Eric Wofsey
Jul 31 at 18:57
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $A$ be the free Boolean algebra of countably many generators. As a Boolean algebra, it is isomorphic to the field $F$ of clopen subsets of its Stone space, namely, the Cantor space $X=2^omega$.
It is known that $A$ is atomless, countably infinite, and incomplete. Since $A$ and $F$ are isomorphic, $F$ is also atomless, countably infinite, and incomplete.
Consider now the smallest subfield $F_0=emptyset,X$ of clopen subsets of $Z$. Does $F_0$ inherit the properties of $F$, that is, is $F_0$ also atomless (trivially), countably infinite, and incomplete?
This question may be naive, but I just cannot decide whether or not $F_0$ inherits the properties of $F$, especially, if I consider that there exists a Boolean epimorphism of $A$ into $F_0$.
boolean-algebra
asked Jul 31 at 14:53
puzzled
1094
Let $A$ be the free Boolean algebra of countably many generators. As a Boolean algebra, it is isomorphic to the field $F$ of clopen subsets of its Stone space, namely, the Cantor space $X=2^omega$.
It is known that $A$ is atomless, countably infinite, and incomplete. Since $A$ and $F$ are isomorphic, $F$ is also atomless, countably infinite, and incomplete.
Consider now the smallest subfield $F_0=emptyset,X$ of clopen subsets of $Z$. Does $F_0$ inherit the properties of $F$, that is, is $F_0$ also atomless (trivially), countably infinite, and incomplete?
This question may be naive, but I just cannot decide whether or not $F_0$ inherits the properties of $F$, especially, if I consider that there exists a Boolean epimorphism of $A$ into $F_0$.
boolean-algebra
asked Jul 31 at 14:53
puzzled
1094
asked Jul 31 at 14:53
puzzled
1094
asked Jul 31 at 14:53
puzzled
1094
asked Jul 31 at 14:53
puzzled
1094
1094
$F$ is completely irrelevant to the question. You just have a (very simple!) Boolean algebra $F_0=emptyset, X$. Have you tried simply answering any of your questions about this Boolean algebra directly?
– Eric Wofsey
Jul 31 at 18:57
add a comment |Â
$F$ is completely irrelevant to the question. You just have a (very simple!) Boolean algebra $F_0=emptyset, X$. Have you tried simply answering any of your questions about this Boolean algebra directly?
– Eric Wofsey
Jul 31 at 18:57
$F$ is completely irrelevant to the question. You just have a (very simple!) Boolean algebra $F_0=emptyset, X$. Have you tried simply answering any of your questions about this Boolean algebra directly?
– Eric Wofsey
Jul 31 at 18:57
$F$ is completely irrelevant to the question. You just have a (very simple!) Boolean algebra $F_0=emptyset, X$. Have you tried simply answering any of your questions about this Boolean algebra directly?
– Eric Wofsey
Jul 31 at 18:57
add a comment |Â
1 Answer
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No, for all three parts. $emptyset,X$ has an atom, namely $X$. It is not countably infinite because it is finite; its cardinality is $2$. And it is complete: Any subset containing $X$ has $X$ as its least upper bound, and any subset not containing $X$ has $emptyset$ as its least upper bound.
answered Jul 31 at 15:47
Andreas Blass
47.4k348104
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
No, for all three parts. $emptyset,X$ has an atom, namely $X$. It is not countably infinite because it is finite; its cardinality is $2$. And it is complete: Any subset containing $X$ has $X$ as its least upper bound, and any subset not containing $X$ has $emptyset$ as its least upper bound.
answered Jul 31 at 15:47
Andreas Blass
47.4k348104
add a comment |Â
up vote
1
down vote
No, for all three parts. $emptyset,X$ has an atom, namely $X$. It is not countably infinite because it is finite; its cardinality is $2$. And it is complete: Any subset containing $X$ has $X$ as its least upper bound, and any subset not containing $X$ has $emptyset$ as its least upper bound.
answered Jul 31 at 15:47
Andreas Blass
47.4k348104
add a comment |Â
up vote
1
down vote
up vote
1
down vote
No, for all three parts. $emptyset,X$ has an atom, namely $X$. It is not countably infinite because it is finite; its cardinality is $2$. And it is complete: Any subset containing $X$ has $X$ as its least upper bound, and any subset not containing $X$ has $emptyset$ as its least upper bound.
answered Jul 31 at 15:47
Andreas Blass
47.4k348104
No, for all three parts. $emptyset,X$ has an atom, namely $X$. It is not countably infinite because it is finite; its cardinality is $2$. And it is complete: Any subset containing $X$ has $X$ as its least upper bound, and any subset not containing $X$ has $emptyset$ as its least upper bound.
answered Jul 31 at 15:47
Andreas Blass
47.4k348104
answered Jul 31 at 15:47
Andreas Blass
47.4k348104
answered Jul 31 at 15:47
Andreas Blass
47.4k348104
answered Jul 31 at 15:47
Andreas Blass
47.4k348104
47.4k348104
add a comment |Â
add a comment |Â
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$F$ is completely irrelevant to the question. You just have a (very simple!) Boolean algebra $F_0=emptyset, X$. Have you tried simply answering any of your questions about this Boolean algebra directly?
– Eric Wofsey
Jul 31 at 18:57