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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 $C$?

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  • $\begingroup$ Does "$\sigma$-isomorphic" just mean "isomorphic as $\sigma$-algebras"? $\endgroup$ – Noah Schweber Nov 15 '18 at 16:34
  • $\begingroup$ Yes, a $\sigma$-isomorphism is an isomorphism which preserves countable supremas. $\endgroup$ – puzzled Nov 15 '18 at 16:40
  • $\begingroup$ Why isn't the obvious map from $C$ - that is, generated by sending the $\eta$th generator to the clopen set $\{f\in 2^{\omega_1}: f(\eta)=1\}$ - 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 $2$)? $\endgroup$ – Noah Schweber Nov 15 '18 at 16:40
  • $\begingroup$ 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... $\endgroup$ – puzzled Nov 15 '18 at 17:03
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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.

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  • $\begingroup$ 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. $\endgroup$ – puzzled Nov 15 '18 at 22:44

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