# Stone representation of the free $\sigma$-algebra on $\omega_1$ free generators

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 $$C$$?

• Does "$\sigma$-isomorphic" just mean "isomorphic as $\sigma$-algebras"? – Noah Schweber Nov 15 '18 at 16:34
• Yes, a $\sigma$-isomorphism is an isomorphism which preserves countable supremas. – Beginner Nov 15 '18 at 16:40
• 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 $2$)? – Noah Schweber Nov 15 '18 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... – Beginner Nov 15 '18 at 17:03

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.
• 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. – Beginner Nov 15 '18 at 22:44