Representation of free sigma-algebras

In his Lectures on Boolean Algebras, Halmos states the following theorem (p. 102):

Theorem 14 For every set $$I$$, there exists a free $$\sigma$$-algebra generated by $$I$$, and, in fact, that algebra is isomorphic to the $$\sigma$$-field of all Baire sets in the Cantor space $$2^I$$.

I don't understand this theorem.

By a theorem of Loomis, every (Boolean) $$\sigma$$-algebra is isomorphic to the quotient of a $$\sigma$$-field by a $$\sigma$$-ideal (a theorem that Halmos states himself on the same page). So, why is the free $$\sigma$$-algebra generated by $$I$$, as a $$\sigma$$-algebra, not isomorphic to the quotient of a $$\sigma$$-field by a $$\sigma$$-ideal?

PS: Halmos calls a Baire set an element of the $$\sigma$$-field generated by the clopen sets.

• Can you say precisely what it is that you don't understand about the theorem? None of the things you have written give a reason to doubt the theorem... Commented Oct 22, 2018 at 16:20
• It seems you have some deeper misunderstanding about things, but it is impossible to tell what this misunderstanding is from what you have written. Commented Oct 22, 2018 at 16:28
• Thanks @Eric for your comments. I rephrased my question. Commented Oct 22, 2018 at 16:52

1 Answer

So, why is the free $$\sigma$$-algebra generated by $$I$$, as a $$\sigma$$-algebra, not isomorphic to the quotient of a $$\sigma$$-field by a $$\sigma$$-ideal?

Theorem 14 says no such thing at all. It says the free $$\sigma$$-algebra generated by $$I$$ is isomorphic to a certain $$\sigma$$-field. This in no way means that it can't also be isomorphic to the quotient of a $$\sigma$$-field by a $$\sigma$$-ideal.

(In fact, Theorem 14 immediately implies that the free $$\sigma$$-algebra generated by $$I$$ is isomorphic to the quotient of a $$\sigma$$-field by a $$\sigma$$-ideal, since it is isomorphic to the quotient of itself by the trivial $$\sigma$$-ideal $$\{0\}$$.)