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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asked Jul 28 at 10:57

puzzled

1094

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

asked Jul 28 at 10:57

puzzled

1094

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

asked Jul 28 at 10:57

puzzled

1094

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

asked Jul 28 at 10:57

puzzled

1094

share|cite|improve this question

share|cite|improve this question

asked Jul 28 at 10:57

puzzled

1094

asked Jul 28 at 10:57

puzzled

1094

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
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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

add a comment | 

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

answered Jul 28 at 11:10

Max

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
  

  
  
  
  

 | 
show 2 more comments

Your Answer

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1 Answer
1

active

oldest

votes

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

answered Jul 28 at 11:10

Max

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
  

  
  
  
  

 | 
show 2 more comments

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

answered Jul 28 at 11:10

Max

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
  

  
  
  
  

 | 
show 2 more comments

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

answered Jul 28 at 11:10

Max

9,8251635

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

answered Jul 28 at 11:10

Max

9,8251635

share|cite|improve this answer

share|cite|improve this answer

answered Jul 28 at 11:10

Max

9,8251635

answered Jul 28 at 11:10

Max

9,8251635

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
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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

 | 
show 2 more comments

 

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