Stone representation of the free $sigma$-algebra on $omega_1$ free generators
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Let $A$ be the free Boolean algebra on $omega$ free generators. Then $A$ is isomorphic to the field of clopen subsets of the Cantor space $2^omega$, which is the Stone space of $A$.
Let $B$ be the free (Boolean) $sigma$-algebra on $omega$ free generators this time. Then, I think, $B$ is $sigma$-isomorphic to the $sigma$-field generated by the clopen subsets (or Baire subsets) of the Cantor space $2^omega$, which is not the Stone space of $B$ since it is the Stone space of $A$.
Let $C$ be now the free (Boolean) $sigma$-algebra on $omega_1$ free generators. Is $C$ $sigma$-isomorphic the $sigma$-field generated by the clopen subsets (or Baire subsets) of the Cantor space $2^{omega_1}$? Is $2^{omega_1}$ the Stone space of $C$?
general-topology boolean-algebra
edited Nov 15 at 12:18
Asaf Karagila♦
300k32421751
asked Nov 15 at 12:16
puzzled
1676
add a comment |
up vote
4
down vote
favorite
Let $A$ be the free Boolean algebra on $omega$ free generators. Then $A$ is isomorphic to the field of clopen subsets of the Cantor space $2^omega$, which is the Stone space of $A$.
Let $B$ be the free (Boolean) $sigma$-algebra on $omega$ free generators this time. Then, I think, $B$ is $sigma$-isomorphic to the $sigma$-field generated by the clopen subsets (or Baire subsets) of the Cantor space $2^omega$, which is not the Stone space of $B$ since it is the Stone space of $A$.
Let $C$ be now the free (Boolean) $sigma$-algebra on $omega_1$ free generators. Is $C$ $sigma$-isomorphic the $sigma$-field generated by the clopen subsets (or Baire subsets) of the Cantor space $2^{omega_1}$? Is $2^{omega_1}$ the Stone space of $C$?
general-topology boolean-algebra
edited Nov 15 at 12:18
Asaf Karagila♦
300k32421751
asked Nov 15 at 12:16
puzzled
1676
Does "$sigma$-isomorphic" just mean "isomorphic as $sigma$-algebras"?
– Noah Schweber
Nov 15 at 16:34
Yes, a $sigma$-isomorphism is an isomorphism which preserves countable supremas.
– puzzled
Nov 15 at 16:40
Why isn't the obvious map from $C$ - that is, generated by sending the $eta$th generator to the clopen set ${fin 2^{omega_1}: f(eta)=1}$ - an isomorphism of $sigma$-algebras between $C$ and the $sigma$-algebra generated by the clopens in $2^{omega_1}$ with the usual Cantor topology (= product topology coming from the discrete topology on each factor $2$)?
– Noah Schweber
Nov 15 at 16:40
Sorry, @Noah, I am not sure I understand what you mean. I think that the map which sends every element of $C$ to the $sigma$-field generated by the clopens of $2^{omega_1}$ is indeed a $sigma$-isomorphism. I am just checking my facts, since I know that sometimes the devil is in the details. That would make $2^{omega_1}$ the Stone space of $C$, but, once again, I would like a confirmation...
– puzzled
Nov 15 at 17:03
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $A$ be the free Boolean algebra on $omega$ free generators. Then $A$ is isomorphic to the field of clopen subsets of the Cantor space $2^omega$, which is the Stone space of $A$.
Let $B$ be the free (Boolean) $sigma$-algebra on $omega$ free generators this time. Then, I think, $B$ is $sigma$-isomorphic to the $sigma$-field generated by the clopen subsets (or Baire subsets) of the Cantor space $2^omega$, which is not the Stone space of $B$ since it is the Stone space of $A$.
Let $C$ be now the free (Boolean) $sigma$-algebra on $omega_1$ free generators. Is $C$ $sigma$-isomorphic the $sigma$-field generated by the clopen subsets (or Baire subsets) of the Cantor space $2^{omega_1}$? Is $2^{omega_1}$ the Stone space of $C$?
general-topology boolean-algebra
edited Nov 15 at 12:18
Asaf Karagila♦
300k32421751
asked Nov 15 at 12:16
puzzled
1676
Let $A$ be the free Boolean algebra on $omega$ free generators. Then $A$ is isomorphic to the field of clopen subsets of the Cantor space $2^omega$, which is the Stone space of $A$.
Let $B$ be the free (Boolean) $sigma$-algebra on $omega$ free generators this time. Then, I think, $B$ is $sigma$-isomorphic to the $sigma$-field generated by the clopen subsets (or Baire subsets) of the Cantor space $2^omega$, which is not the Stone space of $B$ since it is the Stone space of $A$.
Let $C$ be now the free (Boolean) $sigma$-algebra on $omega_1$ free generators. Is $C$ $sigma$-isomorphic the $sigma$-field generated by the clopen subsets (or Baire subsets) of the Cantor space $2^{omega_1}$? Is $2^{omega_1}$ the Stone space of $C$?
general-topology boolean-algebra
general-topology boolean-algebra
edited Nov 15 at 12:18
Asaf Karagila♦
300k32421751
asked Nov 15 at 12:16
puzzled
1676
edited Nov 15 at 12:18
Asaf Karagila♦
300k32421751
asked Nov 15 at 12:16
puzzled
1676
edited Nov 15 at 12:18
Asaf Karagila♦
300k32421751
edited Nov 15 at 12:18
Asaf Karagila♦
300k32421751
edited Nov 15 at 12:18
Asaf Karagila♦
300k32421751
300k32421751
asked Nov 15 at 12:16
puzzled
1676
asked Nov 15 at 12:16
puzzled
1676
asked Nov 15 at 12:16
puzzled
1676
1676
Does "$sigma$-isomorphic" just mean "isomorphic as $sigma$-algebras"?
– Noah Schweber
Nov 15 at 16:34
Yes, a $sigma$-isomorphism is an isomorphism which preserves countable supremas.
– puzzled
Nov 15 at 16:40
Why isn't the obvious map from $C$ - that is, generated by sending the $eta$th generator to the clopen set ${fin 2^{omega_1}: f(eta)=1}$ - an isomorphism of $sigma$-algebras between $C$ and the $sigma$-algebra generated by the clopens in $2^{omega_1}$ with the usual Cantor topology (= product topology coming from the discrete topology on each factor $2$)?
– Noah Schweber
Nov 15 at 16:40
Sorry, @Noah, I am not sure I understand what you mean. I think that the map which sends every element of $C$ to the $sigma$-field generated by the clopens of $2^{omega_1}$ is indeed a $sigma$-isomorphism. I am just checking my facts, since I know that sometimes the devil is in the details. That would make $2^{omega_1}$ the Stone space of $C$, but, once again, I would like a confirmation...
– puzzled
Nov 15 at 17:03
add a comment |
Does "$sigma$-isomorphic" just mean "isomorphic as $sigma$-algebras"?
– Noah Schweber
Nov 15 at 16:34
Yes, a $sigma$-isomorphism is an isomorphism which preserves countable supremas.
– puzzled
Nov 15 at 16:40
Why isn't the obvious map from $C$ - that is, generated by sending the $eta$th generator to the clopen set ${fin 2^{omega_1}: f(eta)=1}$ - an isomorphism of $sigma$-algebras between $C$ and the $sigma$-algebra generated by the clopens in $2^{omega_1}$ with the usual Cantor topology (= product topology coming from the discrete topology on each factor $2$)?
– Noah Schweber
Nov 15 at 16:40
Sorry, @Noah, I am not sure I understand what you mean. I think that the map which sends every element of $C$ to the $sigma$-field generated by the clopens of $2^{omega_1}$ is indeed a $sigma$-isomorphism. I am just checking my facts, since I know that sometimes the devil is in the details. That would make $2^{omega_1}$ the Stone space of $C$, but, once again, I would like a confirmation...
– puzzled
Nov 15 at 17:03
Does "$sigma$-isomorphic" just mean "isomorphic as $sigma$-algebras"?
– Noah Schweber
Nov 15 at 16:34
Does "$sigma$-isomorphic" just mean "isomorphic as $sigma$-algebras"?
– Noah Schweber
Nov 15 at 16:34
Yes, a $sigma$-isomorphism is an isomorphism which preserves countable supremas.
– puzzled
Nov 15 at 16:40
Yes, a $sigma$-isomorphism is an isomorphism which preserves countable supremas.
– puzzled
Nov 15 at 16:40
Why isn't the obvious map from $C$ - that is, generated by sending the $eta$th generator to the clopen set ${fin 2^{omega_1}: f(eta)=1}$ - an isomorphism of $sigma$-algebras between $C$ and the $sigma$-algebra generated by the clopens in $2^{omega_1}$ with the usual Cantor topology (= product topology coming from the discrete topology on each factor $2$)?
– Noah Schweber
Nov 15 at 16:40
Why isn't the obvious map from $C$ - that is, generated by sending the $eta$th generator to the clopen set ${fin 2^{omega_1}: f(eta)=1}$ - an isomorphism of $sigma$-algebras between $C$ and the $sigma$-algebra generated by the clopens in $2^{omega_1}$ with the usual Cantor topology (= product topology coming from the discrete topology on each factor $2$)?
– Noah Schweber
Nov 15 at 16:40
Sorry, @Noah, I am not sure I understand what you mean. I think that the map which sends every element of $C$ to the $sigma$-field generated by the clopens of $2^{omega_1}$ is indeed a $sigma$-isomorphism. I am just checking my facts, since I know that sometimes the devil is in the details. That would make $2^{omega_1}$ the Stone space of $C$, but, once again, I would like a confirmation...
– puzzled
Nov 15 at 17:03
Sorry, @Noah, I am not sure I understand what you mean. I think that the map which sends every element of $C$ to the $sigma$-field generated by the clopens of $2^{omega_1}$ is indeed a $sigma$-isomorphism. I am just checking my facts, since I know that sometimes the devil is in the details. That would make $2^{omega_1}$ the Stone space of $C$, but, once again, I would like a confirmation...
– puzzled
Nov 15 at 17:03
add a comment |
1 Answer
1
active
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votes
up vote
1
down vote
If $D$ is a free Boolean algebra on $kappa$ generators, its Stone space is indeed ${0,1}^kappa$: an ultrafilter is determined by which generators are in it, so by a function $f:kappa to {0,1}$, i.e. a member of this Cantor cube of weight $kappa$.
So ${0,1}^{omega_1}$ cannot be the Stone space of your $C$. You will want the Loomis-Sikorski theorem for the $sigma$-algebra case, I suppose.
answered Nov 15 at 21:00
Henno Brandsma
102k344108
Of course, I should have seen that. The Cantor cube ${0,1}^{omega_1}$ is the Stone space of the free Boolean algebra on $omega_1$ free generators, not the free $sigma$-algebra on $omega_1$ free generators. Thank you @Henno for your reply.
– puzzled
Nov 15 at 22:44
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
If $D$ is a free Boolean algebra on $kappa$ generators, its Stone space is indeed ${0,1}^kappa$: an ultrafilter is determined by which generators are in it, so by a function $f:kappa to {0,1}$, i.e. a member of this Cantor cube of weight $kappa$.
So ${0,1}^{omega_1}$ cannot be the Stone space of your $C$. You will want the Loomis-Sikorski theorem for the $sigma$-algebra case, I suppose.
answered Nov 15 at 21:00
Henno Brandsma
102k344108
Of course, I should have seen that. The Cantor cube ${0,1}^{omega_1}$ is the Stone space of the free Boolean algebra on $omega_1$ free generators, not the free $sigma$-algebra on $omega_1$ free generators. Thank you @Henno for your reply.
– puzzled
Nov 15 at 22:44
add a comment |
up vote
1
down vote
If $D$ is a free Boolean algebra on $kappa$ generators, its Stone space is indeed ${0,1}^kappa$: an ultrafilter is determined by which generators are in it, so by a function $f:kappa to {0,1}$, i.e. a member of this Cantor cube of weight $kappa$.
So ${0,1}^{omega_1}$ cannot be the Stone space of your $C$. You will want the Loomis-Sikorski theorem for the $sigma$-algebra case, I suppose.
answered Nov 15 at 21:00
Henno Brandsma
102k344108
Of course, I should have seen that. The Cantor cube ${0,1}^{omega_1}$ is the Stone space of the free Boolean algebra on $omega_1$ free generators, not the free $sigma$-algebra on $omega_1$ free generators. Thank you @Henno for your reply.
– puzzled
Nov 15 at 22:44
add a comment |
up vote
1
down vote
up vote
1
down vote
If $D$ is a free Boolean algebra on $kappa$ generators, its Stone space is indeed ${0,1}^kappa$: an ultrafilter is determined by which generators are in it, so by a function $f:kappa to {0,1}$, i.e. a member of this Cantor cube of weight $kappa$.
So ${0,1}^{omega_1}$ cannot be the Stone space of your $C$. You will want the Loomis-Sikorski theorem for the $sigma$-algebra case, I suppose.
answered Nov 15 at 21:00
Henno Brandsma
102k344108
If $D$ is a free Boolean algebra on $kappa$ generators, its Stone space is indeed ${0,1}^kappa$: an ultrafilter is determined by which generators are in it, so by a function $f:kappa to {0,1}$, i.e. a member of this Cantor cube of weight $kappa$.
So ${0,1}^{omega_1}$ cannot be the Stone space of your $C$. You will want the Loomis-Sikorski theorem for the $sigma$-algebra case, I suppose.
answered Nov 15 at 21:00
Henno Brandsma
102k344108
answered Nov 15 at 21:00
Henno Brandsma
102k344108
answered Nov 15 at 21:00
Henno Brandsma
102k344108
answered Nov 15 at 21:00
Henno Brandsma
102k344108
102k344108
Of course, I should have seen that. The Cantor cube ${0,1}^{omega_1}$ is the Stone space of the free Boolean algebra on $omega_1$ free generators, not the free $sigma$-algebra on $omega_1$ free generators. Thank you @Henno for your reply.
– puzzled
Nov 15 at 22:44
add a comment |
Of course, I should have seen that. The Cantor cube ${0,1}^{omega_1}$ is the Stone space of the free Boolean algebra on $omega_1$ free generators, not the free $sigma$-algebra on $omega_1$ free generators. Thank you @Henno for your reply.
– puzzled
Nov 15 at 22:44
Of course, I should have seen that. The Cantor cube ${0,1}^{omega_1}$ is the Stone space of the free Boolean algebra on $omega_1$ free generators, not the free $sigma$-algebra on $omega_1$ free generators. Thank you @Henno for your reply.
– puzzled
Nov 15 at 22:44
Of course, I should have seen that. The Cantor cube ${0,1}^{omega_1}$ is the Stone space of the free Boolean algebra on $omega_1$ free generators, not the free $sigma$-algebra on $omega_1$ free generators. Thank you @Henno for your reply.
– puzzled
Nov 15 at 22:44
add a comment |
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Does "$sigma$-isomorphic" just mean "isomorphic as $sigma$-algebras"?
– Noah Schweber
Nov 15 at 16:34
Yes, a $sigma$-isomorphism is an isomorphism which preserves countable supremas.
– puzzled
Nov 15 at 16:40
Why isn't the obvious map from $C$ - that is, generated by sending the $eta$th generator to the clopen set ${fin 2^{omega_1}: f(eta)=1}$ - an isomorphism of $sigma$-algebras between $C$ and the $sigma$-algebra generated by the clopens in $2^{omega_1}$ with the usual Cantor topology (= product topology coming from the discrete topology on each factor $2$)?
– Noah Schweber
Nov 15 at 16:40
Sorry, @Noah, I am not sure I understand what you mean. I think that the map which sends every element of $C$ to the $sigma$-field generated by the clopens of $2^{omega_1}$ is indeed a $sigma$-isomorphism. I am just checking my facts, since I know that sometimes the devil is in the details. That would make $2^{omega_1}$ the Stone space of $C$, but, once again, I would like a confirmation...
– puzzled
Nov 15 at 17:03