Proof: $U$ is an ultrafilter of a Boolean algebra $B$ if and only if for all $x$ in $U$ exactly one of...
I have been stuck with this problem for a while now. I have a proof that letting $U$ be an ultrafilter, exactly one of $x,x^*$ belongs to $U$ for all $x$ in $B$, I did this by showing that both belong to $U$ implies that $U=B$ which cannot be the case, and if neither long to $U$ then a contradiction can be derived by de-Morgan's laws. However, I'm stuck with the reverse implication, I need to prove that:
"If $U$ is a subset of a Boolean algebra $B$ such that $forall x in B$, exactly one of $x in U$ and $x^* in U$ is true, U is an ultrafilter.".
Is this even the case? Any help would be greatly appreciated, thanks in advance.
elementary-set-theory set-theory boolean-algebra filters
asked 15 hours ago
Marmoset
62
New contributor
Marmoset is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I have been stuck with this problem for a while now. I have a proof that letting $U$ be an ultrafilter, exactly one of $x,x^*$ belongs to $U$ for all $x$ in $B$, I did this by showing that both belong to $U$ implies that $U=B$ which cannot be the case, and if neither long to $U$ then a contradiction can be derived by de-Morgan's laws. However, I'm stuck with the reverse implication, I need to prove that:
"If $U$ is a subset of a Boolean algebra $B$ such that $forall x in B$, exactly one of $x in U$ and $x^* in U$ is true, U is an ultrafilter.".
Is this even the case? Any help would be greatly appreciated, thanks in advance.
elementary-set-theory set-theory boolean-algebra filters
asked 15 hours ago
Marmoset
62
New contributor
Marmoset is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What do you denote by $x^*$? $neg x$?
– mathcounterexamples.net
15 hours ago
Indeed, my mistake.
– Marmoset
14 hours ago
For a given ultrafilter $usubset B$, define the set $u^ast = { 1-a: a in u}$ and consider the two questions: "Does this set meet the required hypothesis?" and "Is u^ast a filter?"
– Not Mike
14 hours ago
1
In general, such a subst will not even be a filter ...
– Hagen von Eitzen
12 hours ago
add a comment |
I have been stuck with this problem for a while now. I have a proof that letting $U$ be an ultrafilter, exactly one of $x,x^*$ belongs to $U$ for all $x$ in $B$, I did this by showing that both belong to $U$ implies that $U=B$ which cannot be the case, and if neither long to $U$ then a contradiction can be derived by de-Morgan's laws. However, I'm stuck with the reverse implication, I need to prove that:
"If $U$ is a subset of a Boolean algebra $B$ such that $forall x in B$, exactly one of $x in U$ and $x^* in U$ is true, U is an ultrafilter.".
Is this even the case? Any help would be greatly appreciated, thanks in advance.
elementary-set-theory set-theory boolean-algebra filters
asked 15 hours ago
Marmoset
62
New contributor
Marmoset is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have been stuck with this problem for a while now. I have a proof that letting $U$ be an ultrafilter, exactly one of $x,x^*$ belongs to $U$ for all $x$ in $B$, I did this by showing that both belong to $U$ implies that $U=B$ which cannot be the case, and if neither long to $U$ then a contradiction can be derived by de-Morgan's laws. However, I'm stuck with the reverse implication, I need to prove that:
"If $U$ is a subset of a Boolean algebra $B$ such that $forall x in B$, exactly one of $x in U$ and $x^* in U$ is true, U is an ultrafilter.".
Is this even the case? Any help would be greatly appreciated, thanks in advance.
elementary-set-theory set-theory boolean-algebra filters
elementary-set-theory set-theory boolean-algebra filters
asked 15 hours ago
Marmoset
62
New contributor
Marmoset is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 15 hours ago
Marmoset
62
New contributor
Marmoset is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 14 hours ago
asked 15 hours ago
Marmoset
62
New contributor
Marmoset is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 15 hours ago
Marmoset
62
asked 15 hours ago
Marmoset
62
62
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Marmoset is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Marmoset is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Check out our Code of Conduct.
What do you denote by $x^*$? $neg x$?
– mathcounterexamples.net
15 hours ago
Indeed, my mistake.
– Marmoset
14 hours ago
For a given ultrafilter $usubset B$, define the set $u^ast = { 1-a: a in u}$ and consider the two questions: "Does this set meet the required hypothesis?" and "Is u^ast a filter?"
– Not Mike
14 hours ago
1
In general, such a subst will not even be a filter ...
– Hagen von Eitzen
12 hours ago
add a comment |
What do you denote by $x^*$? $neg x$?
– mathcounterexamples.net
15 hours ago
Indeed, my mistake.
– Marmoset
14 hours ago
For a given ultrafilter $usubset B$, define the set $u^ast = { 1-a: a in u}$ and consider the two questions: "Does this set meet the required hypothesis?" and "Is u^ast a filter?"
– Not Mike
14 hours ago
1
In general, such a subst will not even be a filter ...
– Hagen von Eitzen
12 hours ago
What do you denote by $x^*$? $neg x$?
– mathcounterexamples.net
15 hours ago
What do you denote by $x^*$? $neg x$?
– mathcounterexamples.net
15 hours ago
Indeed, my mistake.
– Marmoset
14 hours ago
Indeed, my mistake.
– Marmoset
14 hours ago
For a given ultrafilter $usubset B$, define the set $u^ast = { 1-a: a in u}$ and consider the two questions: "Does this set meet the required hypothesis?" and "Is u^ast a filter?"
– Not Mike
14 hours ago
For a given ultrafilter $usubset B$, define the set $u^ast = { 1-a: a in u}$ and consider the two questions: "Does this set meet the required hypothesis?" and "Is u^ast a filter?"
– Not Mike
14 hours ago
1
1
In general, such a subst will not even be a filter ...
– Hagen von Eitzen
12 hours ago
In general, such a subst will not even be a filter ...
– Hagen von Eitzen
12 hours ago
add a comment |
1 Answer
1
active
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votes
A correct statement would be
If $Usubseteq B$ is a filter and for all $xin B$ exactly one of $xin U$ or $x^*in U$ is true, then $U$ is an ultrafilter.
The converse holds as well, and this is sometimes taken as the definition of an ultrafilter. I presume the one you've been given is that an ultrafilter is a maximal (proper) filter.
To see that the statement holds, note first that the fact that both $x$ and $x^*$ aren't in $U$ implies $U$ is a proper filter. Assume that $F$ is a filter extending $U$ and let $xin Fsetminus U.$ Then $x^*in U,$ so $x^*in F$ and since both $x$ and $x^*$ are in $F,$ $F$ is improper.
Without the provision that $U$ is a filter, the statement becomes very badly false. Just randomly choose elements of a Boolean algebra, randomly label with zero or one (indicators as to whether the element is inside or outside $U$), and assign the complement the opposite label. Odds are you hit a counterexample. (To guarantee you do so, just make sure you start out labeling the $0$ element with a one and the $1$ element with a zero).
answered 11 hours ago
spaceisdarkgreen
32.4k21753
1
Although it's easy to get counterexamples, it's not quite as easy as you made it sound. You need to arrange for each complementary pair $x$ and $neg x$ to get opposite values. So randomly pick one element from each complementary pair, label the chosen elements $1$ and the unchosen ones $0$. As you said, you have a good chance of getting a counterexample, and you can guarantee a counterexample by choosing $0$ from the pair ${0,1}$. (Also, if you have bad luck and get an ultrafilter, just take its complement to get a counterexample.)
– Andreas Blass
8 hours ago
@AndreasBlass Yes, I put that incorrectly, thank you for the correction!
– spaceisdarkgreen
8 hours ago
add a comment |
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1 Answer
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1 Answer
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A correct statement would be
If $Usubseteq B$ is a filter and for all $xin B$ exactly one of $xin U$ or $x^*in U$ is true, then $U$ is an ultrafilter.
The converse holds as well, and this is sometimes taken as the definition of an ultrafilter. I presume the one you've been given is that an ultrafilter is a maximal (proper) filter.
To see that the statement holds, note first that the fact that both $x$ and $x^*$ aren't in $U$ implies $U$ is a proper filter. Assume that $F$ is a filter extending $U$ and let $xin Fsetminus U.$ Then $x^*in U,$ so $x^*in F$ and since both $x$ and $x^*$ are in $F,$ $F$ is improper.
Without the provision that $U$ is a filter, the statement becomes very badly false. Just randomly choose elements of a Boolean algebra, randomly label with zero or one (indicators as to whether the element is inside or outside $U$), and assign the complement the opposite label. Odds are you hit a counterexample. (To guarantee you do so, just make sure you start out labeling the $0$ element with a one and the $1$ element with a zero).
answered 11 hours ago
spaceisdarkgreen
32.4k21753
1
Although it's easy to get counterexamples, it's not quite as easy as you made it sound. You need to arrange for each complementary pair $x$ and $neg x$ to get opposite values. So randomly pick one element from each complementary pair, label the chosen elements $1$ and the unchosen ones $0$. As you said, you have a good chance of getting a counterexample, and you can guarantee a counterexample by choosing $0$ from the pair ${0,1}$. (Also, if you have bad luck and get an ultrafilter, just take its complement to get a counterexample.)
– Andreas Blass
8 hours ago
@AndreasBlass Yes, I put that incorrectly, thank you for the correction!
– spaceisdarkgreen
8 hours ago
add a comment |
A correct statement would be
If $Usubseteq B$ is a filter and for all $xin B$ exactly one of $xin U$ or $x^*in U$ is true, then $U$ is an ultrafilter.
The converse holds as well, and this is sometimes taken as the definition of an ultrafilter. I presume the one you've been given is that an ultrafilter is a maximal (proper) filter.
To see that the statement holds, note first that the fact that both $x$ and $x^*$ aren't in $U$ implies $U$ is a proper filter. Assume that $F$ is a filter extending $U$ and let $xin Fsetminus U.$ Then $x^*in U,$ so $x^*in F$ and since both $x$ and $x^*$ are in $F,$ $F$ is improper.
Without the provision that $U$ is a filter, the statement becomes very badly false. Just randomly choose elements of a Boolean algebra, randomly label with zero or one (indicators as to whether the element is inside or outside $U$), and assign the complement the opposite label. Odds are you hit a counterexample. (To guarantee you do so, just make sure you start out labeling the $0$ element with a one and the $1$ element with a zero).
answered 11 hours ago
spaceisdarkgreen
32.4k21753
1
Although it's easy to get counterexamples, it's not quite as easy as you made it sound. You need to arrange for each complementary pair $x$ and $neg x$ to get opposite values. So randomly pick one element from each complementary pair, label the chosen elements $1$ and the unchosen ones $0$. As you said, you have a good chance of getting a counterexample, and you can guarantee a counterexample by choosing $0$ from the pair ${0,1}$. (Also, if you have bad luck and get an ultrafilter, just take its complement to get a counterexample.)
– Andreas Blass
8 hours ago
@AndreasBlass Yes, I put that incorrectly, thank you for the correction!
– spaceisdarkgreen
8 hours ago
add a comment |
A correct statement would be
If $Usubseteq B$ is a filter and for all $xin B$ exactly one of $xin U$ or $x^*in U$ is true, then $U$ is an ultrafilter.
The converse holds as well, and this is sometimes taken as the definition of an ultrafilter. I presume the one you've been given is that an ultrafilter is a maximal (proper) filter.
To see that the statement holds, note first that the fact that both $x$ and $x^*$ aren't in $U$ implies $U$ is a proper filter. Assume that $F$ is a filter extending $U$ and let $xin Fsetminus U.$ Then $x^*in U,$ so $x^*in F$ and since both $x$ and $x^*$ are in $F,$ $F$ is improper.
Without the provision that $U$ is a filter, the statement becomes very badly false. Just randomly choose elements of a Boolean algebra, randomly label with zero or one (indicators as to whether the element is inside or outside $U$), and assign the complement the opposite label. Odds are you hit a counterexample. (To guarantee you do so, just make sure you start out labeling the $0$ element with a one and the $1$ element with a zero).
answered 11 hours ago
spaceisdarkgreen
32.4k21753
A correct statement would be
If $Usubseteq B$ is a filter and for all $xin B$ exactly one of $xin U$ or $x^*in U$ is true, then $U$ is an ultrafilter.
The converse holds as well, and this is sometimes taken as the definition of an ultrafilter. I presume the one you've been given is that an ultrafilter is a maximal (proper) filter.
To see that the statement holds, note first that the fact that both $x$ and $x^*$ aren't in $U$ implies $U$ is a proper filter. Assume that $F$ is a filter extending $U$ and let $xin Fsetminus U.$ Then $x^*in U,$ so $x^*in F$ and since both $x$ and $x^*$ are in $F,$ $F$ is improper.
Without the provision that $U$ is a filter, the statement becomes very badly false. Just randomly choose elements of a Boolean algebra, randomly label with zero or one (indicators as to whether the element is inside or outside $U$), and assign the complement the opposite label. Odds are you hit a counterexample. (To guarantee you do so, just make sure you start out labeling the $0$ element with a one and the $1$ element with a zero).
answered 11 hours ago
spaceisdarkgreen
32.4k21753
edited 8 hours ago
answered 11 hours ago
spaceisdarkgreen
32.4k21753
answered 11 hours ago
spaceisdarkgreen
32.4k21753
answered 11 hours ago
spaceisdarkgreen
32.4k21753
32.4k21753
1
Although it's easy to get counterexamples, it's not quite as easy as you made it sound. You need to arrange for each complementary pair $x$ and $neg x$ to get opposite values. So randomly pick one element from each complementary pair, label the chosen elements $1$ and the unchosen ones $0$. As you said, you have a good chance of getting a counterexample, and you can guarantee a counterexample by choosing $0$ from the pair ${0,1}$. (Also, if you have bad luck and get an ultrafilter, just take its complement to get a counterexample.)
– Andreas Blass
8 hours ago
@AndreasBlass Yes, I put that incorrectly, thank you for the correction!
– spaceisdarkgreen
8 hours ago
add a comment |
1
Although it's easy to get counterexamples, it's not quite as easy as you made it sound. You need to arrange for each complementary pair $x$ and $neg x$ to get opposite values. So randomly pick one element from each complementary pair, label the chosen elements $1$ and the unchosen ones $0$. As you said, you have a good chance of getting a counterexample, and you can guarantee a counterexample by choosing $0$ from the pair ${0,1}$. (Also, if you have bad luck and get an ultrafilter, just take its complement to get a counterexample.)
– Andreas Blass
8 hours ago
@AndreasBlass Yes, I put that incorrectly, thank you for the correction!
– spaceisdarkgreen
8 hours ago
1
1
Although it's easy to get counterexamples, it's not quite as easy as you made it sound. You need to arrange for each complementary pair $x$ and $neg x$ to get opposite values. So randomly pick one element from each complementary pair, label the chosen elements $1$ and the unchosen ones $0$. As you said, you have a good chance of getting a counterexample, and you can guarantee a counterexample by choosing $0$ from the pair ${0,1}$. (Also, if you have bad luck and get an ultrafilter, just take its complement to get a counterexample.)
– Andreas Blass
8 hours ago
Although it's easy to get counterexamples, it's not quite as easy as you made it sound. You need to arrange for each complementary pair $x$ and $neg x$ to get opposite values. So randomly pick one element from each complementary pair, label the chosen elements $1$ and the unchosen ones $0$. As you said, you have a good chance of getting a counterexample, and you can guarantee a counterexample by choosing $0$ from the pair ${0,1}$. (Also, if you have bad luck and get an ultrafilter, just take its complement to get a counterexample.)
– Andreas Blass
8 hours ago
@AndreasBlass Yes, I put that incorrectly, thank you for the correction!
– spaceisdarkgreen
8 hours ago
@AndreasBlass Yes, I put that incorrectly, thank you for the correction!
– spaceisdarkgreen
8 hours ago
add a comment |
Marmoset is a new contributor. Be nice, and check out our Code of Conduct.
Marmoset is a new contributor. Be nice, and check out our Code of Conduct.
Marmoset is a new contributor. Be nice, and check out our Code of Conduct.
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What do you denote by $x^*$? $neg x$?
– mathcounterexamples.net
15 hours ago
Indeed, my mistake.
– Marmoset
14 hours ago
For a given ultrafilter $usubset B$, define the set $u^ast = { 1-a: a in u}$ and consider the two questions: "Does this set meet the required hypothesis?" and "Is u^ast a filter?"
– Not Mike
14 hours ago
1
In general, such a subst will not even be a filter ...
– Hagen von Eitzen
12 hours ago