Number of ultrafilters in the free Boolean algebra on countably many generators

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Let $A$ be the free Boolean algebra on denumerably many generators. How many ultrafilter does $A$ contain? How to prove it?




boolean-algebra




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  • What are your thoughts about that? What have you tried so far?
    – Taroccoesbrocco
    Jul 28 at 10:58














up vote
1
down vote

favorite












Let $A$ be the free Boolean algebra on denumerably many generators. How many ultrafilter does $A$ contain? How to prove it?




boolean-algebra




share|cite|improve this question



















  • What are your thoughts about that? What have you tried so far?
    – Taroccoesbrocco
    Jul 28 at 10:58












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $A$ be the free Boolean algebra on denumerably many generators. How many ultrafilter does $A$ contain? How to prove it?




boolean-algebra




share|cite|improve this question











Let $A$ be the free Boolean algebra on denumerably many generators. How many ultrafilter does $A$ contain? How to prove it?





boolean-algebra





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 28 at 10:57









puzzled

1094




1094











  • What are your thoughts about that? What have you tried so far?
    – Taroccoesbrocco
    Jul 28 at 10:58
















  • What are your thoughts about that? What have you tried so far?
    – Taroccoesbrocco
    Jul 28 at 10:58















What are your thoughts about that? What have you tried so far?
– Taroccoesbrocco
Jul 28 at 10:58




What are your thoughts about that? What have you tried so far?
– Taroccoesbrocco
Jul 28 at 10:58










1 Answer
1






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oldest

votes

















up vote
2
down vote













Hint: remember that an ultrafilter on $B$ is essentially the same as a boolean algebra morphism $Bto 2$. How can you characterize morphisms from a free algebra to $2$ ?






share|cite|improve this answer





















  • Does that mean thus that $A$ has $2^omega$ ultrafilters?
    – puzzled
    Jul 28 at 11:13











  • Well what's your proof ?
    – Max
    Jul 28 at 11:38










  • Well, since $A$ has denumerably many elements and an ultrafilters is essentially the same as a Boolean epimorphism from $A$ to $2$, then $A$ has $2^omega$ ultrafilters.
    – puzzled
    Jul 28 at 11:45






  • 1




    That's not a correct proof (though the result is correct)
    – Max
    Jul 28 at 11:53










  • What would be your argument then?
    – puzzled
    Jul 28 at 12:06










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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote













Hint: remember that an ultrafilter on $B$ is essentially the same as a boolean algebra morphism $Bto 2$. How can you characterize morphisms from a free algebra to $2$ ?






share|cite|improve this answer





















  • Does that mean thus that $A$ has $2^omega$ ultrafilters?
    – puzzled
    Jul 28 at 11:13











  • Well what's your proof ?
    – Max
    Jul 28 at 11:38










  • Well, since $A$ has denumerably many elements and an ultrafilters is essentially the same as a Boolean epimorphism from $A$ to $2$, then $A$ has $2^omega$ ultrafilters.
    – puzzled
    Jul 28 at 11:45






  • 1




    That's not a correct proof (though the result is correct)
    – Max
    Jul 28 at 11:53










  • What would be your argument then?
    – puzzled
    Jul 28 at 12:06














up vote
2
down vote













Hint: remember that an ultrafilter on $B$ is essentially the same as a boolean algebra morphism $Bto 2$. How can you characterize morphisms from a free algebra to $2$ ?






share|cite|improve this answer





















  • Does that mean thus that $A$ has $2^omega$ ultrafilters?
    – puzzled
    Jul 28 at 11:13











  • Well what's your proof ?
    – Max
    Jul 28 at 11:38










  • Well, since $A$ has denumerably many elements and an ultrafilters is essentially the same as a Boolean epimorphism from $A$ to $2$, then $A$ has $2^omega$ ultrafilters.
    – puzzled
    Jul 28 at 11:45






  • 1




    That's not a correct proof (though the result is correct)
    – Max
    Jul 28 at 11:53










  • What would be your argument then?
    – puzzled
    Jul 28 at 12:06












up vote
2
down vote










up vote
2
down vote









Hint: remember that an ultrafilter on $B$ is essentially the same as a boolean algebra morphism $Bto 2$. How can you characterize morphisms from a free algebra to $2$ ?






share|cite|improve this answer













Hint: remember that an ultrafilter on $B$ is essentially the same as a boolean algebra morphism $Bto 2$. How can you characterize morphisms from a free algebra to $2$ ?







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 28 at 11:10









Max

9,8251635




9,8251635











  • Does that mean thus that $A$ has $2^omega$ ultrafilters?
    – puzzled
    Jul 28 at 11:13











  • Well what's your proof ?
    – Max
    Jul 28 at 11:38










  • Well, since $A$ has denumerably many elements and an ultrafilters is essentially the same as a Boolean epimorphism from $A$ to $2$, then $A$ has $2^omega$ ultrafilters.
    – puzzled
    Jul 28 at 11:45






  • 1




    That's not a correct proof (though the result is correct)
    – Max
    Jul 28 at 11:53










  • What would be your argument then?
    – puzzled
    Jul 28 at 12:06
















  • Does that mean thus that $A$ has $2^omega$ ultrafilters?
    – puzzled
    Jul 28 at 11:13











  • Well what's your proof ?
    – Max
    Jul 28 at 11:38










  • Well, since $A$ has denumerably many elements and an ultrafilters is essentially the same as a Boolean epimorphism from $A$ to $2$, then $A$ has $2^omega$ ultrafilters.
    – puzzled
    Jul 28 at 11:45






  • 1




    That's not a correct proof (though the result is correct)
    – Max
    Jul 28 at 11:53










  • What would be your argument then?
    – puzzled
    Jul 28 at 12:06















Does that mean thus that $A$ has $2^omega$ ultrafilters?
– puzzled
Jul 28 at 11:13





Does that mean thus that $A$ has $2^omega$ ultrafilters?
– puzzled
Jul 28 at 11:13













Well what's your proof ?
– Max
Jul 28 at 11:38




Well what's your proof ?
– Max
Jul 28 at 11:38












Well, since $A$ has denumerably many elements and an ultrafilters is essentially the same as a Boolean epimorphism from $A$ to $2$, then $A$ has $2^omega$ ultrafilters.
– puzzled
Jul 28 at 11:45




Well, since $A$ has denumerably many elements and an ultrafilters is essentially the same as a Boolean epimorphism from $A$ to $2$, then $A$ has $2^omega$ ultrafilters.
– puzzled
Jul 28 at 11:45




1




1




That's not a correct proof (though the result is correct)
– Max
Jul 28 at 11:53




That's not a correct proof (though the result is correct)
– Max
Jul 28 at 11:53












What would be your argument then?
– puzzled
Jul 28 at 12:06




What would be your argument then?
– puzzled
Jul 28 at 12:06












 

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