Proof: $U$ is an ultrafilter of a Boolean algebra $B$ if and only if for all $x$ in $U$ exactly one of...












1














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.










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  • What do you denote by $x^*$? $neg x$?
    – mathcounterexamples.net
    2 days ago












  • Indeed, my mistake.
    – Marmoset
    2 days 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
    2 days ago








  • 1




    In general, such a subst will not even be a filter ...
    – Hagen von Eitzen
    2 days ago
















1














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.










share|cite|improve this question









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
    2 days ago












  • Indeed, my mistake.
    – Marmoset
    2 days 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
    2 days ago








  • 1




    In general, such a subst will not even be a filter ...
    – Hagen von Eitzen
    2 days ago














1












1








1







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.










share|cite|improve this question









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 boolean-algebra filters






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edited 2 days ago









Andrés E. Caicedo

64.8k8158246




64.8k8158246






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asked 2 days ago









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Marmoset is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Check out our Code of Conduct.












  • What do you denote by $x^*$? $neg x$?
    – mathcounterexamples.net
    2 days ago












  • Indeed, my mistake.
    – Marmoset
    2 days 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
    2 days ago








  • 1




    In general, such a subst will not even be a filter ...
    – Hagen von Eitzen
    2 days ago


















  • What do you denote by $x^*$? $neg x$?
    – mathcounterexamples.net
    2 days ago












  • Indeed, my mistake.
    – Marmoset
    2 days 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
    2 days ago








  • 1




    In general, such a subst will not even be a filter ...
    – Hagen von Eitzen
    2 days ago
















What do you denote by $x^*$? $neg x$?
– mathcounterexamples.net
2 days ago






What do you denote by $x^*$? $neg x$?
– mathcounterexamples.net
2 days ago














Indeed, my mistake.
– Marmoset
2 days ago




Indeed, my mistake.
– Marmoset
2 days 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
2 days 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
2 days ago






1




1




In general, such a subst will not even be a filter ...
– Hagen von Eitzen
2 days ago




In general, such a subst will not even be a filter ...
– Hagen von Eitzen
2 days ago










1 Answer
1






active

oldest

votes


















1














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).






share|cite|improve this answer



















  • 2




    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
    2 days ago










  • @AndreasBlass Yes, I put that incorrectly, thank you for the correction!
    – spaceisdarkgreen
    2 days ago










  • Thank you very much! This was very helpful
    – Marmoset
    2 days ago











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1














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).






share|cite|improve this answer



















  • 2




    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
    2 days ago










  • @AndreasBlass Yes, I put that incorrectly, thank you for the correction!
    – spaceisdarkgreen
    2 days ago










  • Thank you very much! This was very helpful
    – Marmoset
    2 days ago
















1














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).






share|cite|improve this answer



















  • 2




    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
    2 days ago










  • @AndreasBlass Yes, I put that incorrectly, thank you for the correction!
    – spaceisdarkgreen
    2 days ago










  • Thank you very much! This was very helpful
    – Marmoset
    2 days ago














1












1








1






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).






share|cite|improve this answer














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).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























answered 2 days ago









spaceisdarkgreen

32.4k21753




32.4k21753








  • 2




    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
    2 days ago










  • @AndreasBlass Yes, I put that incorrectly, thank you for the correction!
    – spaceisdarkgreen
    2 days ago










  • Thank you very much! This was very helpful
    – Marmoset
    2 days ago














  • 2




    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
    2 days ago










  • @AndreasBlass Yes, I put that incorrectly, thank you for the correction!
    – spaceisdarkgreen
    2 days ago










  • Thank you very much! This was very helpful
    – Marmoset
    2 days ago








2




2




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
2 days 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
2 days ago












@AndreasBlass Yes, I put that incorrectly, thank you for the correction!
– spaceisdarkgreen
2 days ago




@AndreasBlass Yes, I put that incorrectly, thank you for the correction!
– spaceisdarkgreen
2 days ago












Thank you very much! This was very helpful
– Marmoset
2 days ago




Thank you very much! This was very helpful
– Marmoset
2 days ago










Marmoset is a new contributor. Be nice, and check out our Code of Conduct.










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